{% extends 'marketing/base.html' %} {% load staticfiles %} {% load paginate %} {% block content %} {{ block.super }}
| Campaign | Receipents | Sent | Unique Opens | Unique Clicks | Actions |
|---|---|---|---|---|---|
| Primary {{ campaign.title }} | 5478 | {{ campaign.sent_on_format }} | 20.6% 125 Opened | 15.5% 78 Clicked | |
| No records found! |


Peregrine CMS - a head optional content management system based on Apache Sling and Vue.JS.
Peregrine CMS is an open source project sponsored by headwire.com, Inc. Commercial support is available upon request. We encourage contributions to the project. Please let us know what you think about this product.


PAF Museum puts Abhinandan’s likeness, remains of his jet on display
by Tufail AhmedKARACHI : The Pakistan Air Force Museum in Karachi has long housed some of the most prized war trophies the force has gathered in the course of its 72-year history.
There is, most famously, the Folland Gnat fighter jet of the Indian Air Force that was forced to land in Pakistan by PAF interceptors during the air battles of the 1965 war.
And then there are two Afghan Air Force aircraft, including a MiG-21 fighter jet, which were captured after the Soviet Union invaded Afghanistan.
There is even a set of flight overalls worn by an Israeli fighter pilot whose Mirage aircraft was shot down by a PAF pilot on deputation in a friendly Arab nation.
None of them, however, are as exciting as the museum’s latest centerpiece, an entire section dedicated to Operation Swift Retort, the PAF response to India’s botched air strikes near Balakot earlier this year.
The February 27 air action, which resulted in dogfights over Jammu and Kashmir for the first time in nearly 50 years, ended with the PAF shooting down two Indian warplanes for no loss and Pakistani forces capturing one of the downed pilots, Wing Commander Abhinandan Varthaman.
Varthaman’s likeness and the remains of the aircraft he piloted form the main attractions of the exhibit, which The Express Tribune was allowed to see before it is formally opened to public.
The tail section and other debris of his MiG-21 fighter jet which was downed over Azad Jammu and Kashmir are displayed front and centre at the very entrance of the newly inaugurated section.
Inside, a mannequin painted to resemble the wing commander and dressed in his flight overalls has put on display alongside another bearing the likeness of his Pakistan Army ‘saviour’.
In a cheeky bit of detail, a tea mug emblazoned with the viral receipt for the ‘fantastic tea’ the India pilot enjoyed in captivity has been placed next to his effigy.
Other displays in the section include a diorama of the entire aerial operation as well as a wall dedicated to the pilots who played a key role in it.
There are also displays chronicling the events that made Operation Swift Retort a necessity as well as a wall highlighting media coverage of the entire episode.
The new gallery was inaugurated by Air Chief Marshal Mujahid Anwar Khan on November 6 and will be opened to the public very soon.


{% extends "layout.html" %} {% block brand %} {% include "includes/v2/brandChange.html" %} {% endblock %} {% block content %} {% include "includes/v2/navV5.html" %} Agent Sandra Nash Edit Skillset Skill sets Benefit ESA Change name Handover type Maintenance Change name Handover Reasons All Change name Back {% endblock %}


- getAbsRow() - Method in class com.psec.excel.WriteExcel.Area
-
Return next absolute row.
- getBaseCol() - Method in class com.psec.excel.WriteExcel.Area
-
Return absolute column of start of Area on sheet.
- getBaseRow() - Method in class com.psec.excel.WriteExcel.Area
-
Return absolute row of start of Area on sheet.
- getColAsStr(Row, ReadExcelFile.ColMap) - Method in class com.psec.excel.ReadExcelFile
-
Return a stringized version of the Cell at the specified location.
- getColCell(Row, ReadExcelFile.ColMap) - Method in class com.psec.excel.ReadExcelFile
-
Return Cell at the specified location.
- getColMap() - Method in class com.psec.excel.ReadExcelRecord
-
The ColMap string definition is returned.
- getColMap() - Method in class com.psec.run.DemoExcel.OpenBalRec
-
- getColMap() - Method in class com.psec.run.DemoExcel.ReadCounts
-
- getDataRow() - Method in class com.psec.excel.WriteExcel.Area
-
Return absolute row of first data row.
- getHdrCount() - Method in class com.psec.excel.WriteExcel.Area
-
Return size of Header array.
- getHdrExtras() - Method in class com.psec.excel.WriteExcel
-
Gets the standard header builtins styles.
- getPrintArea(String) - Method in class com.psec.excel.ReadExcelFile
-
Returns the String version of a Sheet print area.
- getProp(String) - Method in class com.psec.util.Parm.Opt
-
Return property value for sKey
or null;
- getProp(String, String) - Method in class com.psec.util.Parm.Opt
-
Return property value for sKey
or sDef;
- getProps() - Method in class com.psec.util.Parm.Opt
-
Return an array of property value names;
- getRowCount() - Method in class com.psec.excel.WriteExcel.Area
-
Return size of Rows array.
- getRowCount(String) - Method in class com.psec.excel.WriteExcel
-
Gets the Row count on the specified Sheet.
- getRows() - Method in class com.psec.excel.WriteExcel.Area
-
Gets the current row array contents.
- getSheet(String) - Method in class com.psec.excel.ReadExcelFile
-
Returns the named Sheet.
- getSheet() - Method in class com.psec.excel.WriteExcel.Area
-
Return Sheet name area.
- getStdExtras() - Method in class com.psec.excel.WriteExcel
-
Gets the standard builtins styles.
- getStrValue(String, int, int) - Method in class com.psec.excel.WriteExcel
-
Gets the String value of the specified Cell.
- getWorkbook() - Method in class com.psec.excel.ReadExcelFile
-
Gets the Workbook
being accessed.
- getWorkbook() - Method in class com.psec.excel.WriteExcel
-
Gets the Workbook
being constructed.
- getWriter() - Method in class com.psec.excel.WriteExcel.Area
-
Return WriteExcel for this area.


#1105 james park, johannesburg, southafrica. Compliments Of The Day, Permit me to introduce myself to you. My name is Dr Jerry Jerome, A Top management Official of one of the finance corporation in johannesburg south Africa.I came to know about you in my private search for a reliable person/company to handle an important assingment on behalf of my Colleagues and myself.As a matter of fact, I got your information from Miss Joyce Brown of The International Affairs Department Of The South African Chamber Of Business. Proposition A German Mr. peter khan, 66 years of age and a prosperous oil/Diamond merchant had in our Bank the sum of $35.5million in a domiciliary account. Unfortunately, he was killed in 2002 plane crash involving concord AF4878 in Gonesse, France. Efforts has been made by the management of my bank through the German Embassy in johannesburg to contact any of the deceased children but to no avail,as we are made to understand he had no children.Given the skeletal information available to the bank, it has so far been impossible to reach any of the relatives. The situation left for the Management is to declare the deceased account dormant and revert the Amount to trading on behalf of the interest of the Bank. In order to avoid this development, my Colleagues and I now seek your approval to have you stand as a distant relative to Mr. peter khan,so that the money would be released to you. All documents and proofs to enable you get the money will be carefully packaged once we receive your consent on this proposal.I shall Apply for a leave as soon as the funds gets into your reliable account to come to your country for possible sharing and investment opportunity which you will assist me in doing. May I assure you that this is safe and risk-free, as we have taken care of all necessary modalities to ensure a hitch-free movement. I have the authority of my partners involved in this assignment to propose that should you be willing to assist us in this,your share would be 40%. I have reposed my confidence in you and hope that you will not disappoint me. Please Best Regards


THE MATRIX (1999)
Directed by the Wachowski Brothers Starring : Keanu Reeves (Anderson / Neo) Laurence Fishburne (Morpheus) Carrie Anne-Moss (Trinity) Running Time : 2hrs
Reviewed by Jack Choo
Rating : ****1/2 out of *****
THE MATRIX is possibly one of the most imaginative Hollywood film of this decade. This kung-fu laced sci-fi fantasy has the enough action to drive any action fan crazy (even put to shame many kung-fu based movies from Hong Kong) and one of the most imaginative yet surprisingly believable plot ever conceived for commercial film. This review contain spoilers that may ruin your enjoyment of the film by knowing too much but this is one film which is so hard to review without giving away too much. Do discontinue your reading now and watch the film immediately! I guess by now, word-of-mouth would generated on this movie would have seep through the crevices of every movie-goer in this world. So does it matter really? Oh well.
The story brings us the revelations concerning `The One' Thomas Anderson (Reeves), a computer-geek who when not working in his white-collar computing job in a big firm, he will be spending time at home, in front of his computer, grabbing `contraband' information from the net for sale. Call it a bit of `side-income' but Thomas is indeed onto something when he is mysteriously led to the man he has been seeking for so long; the elusive yet almost godly Morpheus. Thomas's obsession is fueled by his curiosity on THE MATRIX and it this inner obsession that inadvertently led him, almost by fate to Morpheus, the guru on the subject.
The review above was posted to the
rec.arts.movies.reviews newsgroup (de.rec.film.kritiken for German reviews).
The Internet Movie Database accepts no responsibility for the contents of the
review and has no editorial control. Unless stated otherwise, the copyright
belongs to the author.
Please direct comments/criticisms of the review to relevant newsgroups.
Broken URLs inthe reviews are the responsibility of the author.
The formatting of the review is likely to differ from the original due
to ASCII to HTML conversion.
Related links: index of all rec.arts.movies.reviews reviews


If SimpleService instance is a singleton object, then simpleServiceId from feature modules should be the same.
However, this demo demostrates there could be multiple instances being created.


If you want to be in business, consider registering for an ABN.
You’ll need an Australian Business Number (ABN) for government purposes if you:
If you already have an ABN, you don’t need to register for a new one. If you’re changing the type of activity you do, you can use your existing ABN. Your ABN stays with you, although you need to notify the Australian Taxation Office if you stop using it.
If you change your business structure e.g. changing from a sole trader to a company, you’ll need to register a new ABN.
If you sell goods or services to businesses, they may ask you for an ABN when they pay you. It’s not compulsory to have an ABN. You can use a Statement by a supplier form. to avoid the business withholding an amount from the payment to you for not having an ABN. The statement lets the business know that you are selling the goods or services as your hobby.


{extend name='./common/father' /} {block name='mid'}
has conflit: two people changes the file at same time
The recent update version
your version


Before Change
sources=["direct_dependency.py"],
dependencies=[":transitive_dependency"],
)
source = FileContent(
path=f"{self.source_root}/target.py",
content=dedent(
\
"""Code is not executed, but Pylint will check that variables exist and are used"""
from direct_dependency import THIS_VARIABLE_EXISTS
print(THIS_VARIABLE_EXISTS)
).encode(),
)
self.write_file(source)
self.create_python_library(sources=["target.py"], dependencies=[":direct_dependency"])
// We must re-write the files because `create_python_library` will have over-written the
After Change
def test_includes_direct_dependencies(self) -> None:
self.make_target_with_origin(source_files=[], name="transitive_dependency")
direct_dependency_content = dedent(
\
// No docstring because Pylint doesn"t lint dependencies
from transitive_dep import doesnt_matter_if_variable_exists
THIS_VARIABLE_EXISTS = ""
)
self.make_target_with_origin(
source_files=[
FileContent(
f"{self.source_root}/direct_dependency.py", direct_dependency_content.encode()


Personal petcare for Gedling, carlton and surrounding areas.
Full website coming soon. Meanwhile, please call Claire on 07968 074138 or visit our facebook page.
Full website coming soon. Meanwhile, please call Claire on 07968 074138 or visit our facebook page.


Mesos + Marathon
Edit (pull request)Mesos is like a kernel for the datacenter. It provides fine-grained abstraction of the datacenter resources, isolation and native support for Docker containers. Marathon is an open-source meta-framework for Mesos dedicated to container orchestration. It is developed and maintained by Mesosphere. Marathon is a production-ready container orchestration platform for Mesos with fist-class Docker support. It features automated health-checks and failure recovery, allowing seamless execution of services or long-running jobs. Being a meta-framework, Marathon is also proficient in running other Mesos frameworks, such as Chronos.
Combined, Mesos and Marathon can turn any datacenter into a highly available, scalable and fault-tolerant PaaS for cloud applications. We developed a Marathon orchestrator plugin for Alien 4 Cloud, as part of our 1.4.0 roadmap to achieve Docker support.
This project is at an alpha stage and still undergoing development. We might add, change, or delete any functionality described in this document. Any feedback would be highly appreciated !
Alien 4 Cloud Marathon support
The plugin features deployment and management of complex topologies with containers in Marathon by leveraging docker-tosca-types. Currently, we only support running Docker containers.
Topologies deployed with the plugin benefit from Marathon’s fault-recovery features. This means that Marathon will gracefully re-schedule and restart (possibly, on a different agent) a failing container.
We made a demonstration video showcasing the deployment of Nodecellar using the plugin.
Service discovery
TL;DR: Service discovery is pretty much automatic using the Marathon plugin with MesosDNS & MarathonLB running on your cluster.
Service discovery between containers launched by the plugin through Marathon is achieved using MesosDNS and MarathonLB, respectively a DNS service and a HAProxy load balancer that provides port-based service discovery. Both are running in the cluster as Marathon tasks.
Containers launched with Marathon are all resolvable through DNS resolution using the app_name.marathon.mesos pattern (but you will still need to know the containers’ allocated port, which should be randomly assigned by Marathon), or using MarathonLB as a reverse proxy, using a well-know service port. When assigned a service port, containers running in Marathon can be accessed by reaching MarathonLB on the said service port. Because MarathonLB itself is running on Marathon, it’s IP address is also resolvable through MesosDNS. This means that containers with service ports can be accessed using the pattern *
This whole mechanism being relatively complex, the plugin will automatically assign a service port to containers that are targeted by a ConnectsTo relationship from at least another container in the topology. Respectively, the plugin will also replace any reference to the target’s endpoint port and ip_address attributes with the service port and MarathonLB DNS name, respectively.
External Storage support
TL;DR: We added experimental support of the external storage feature from Marathon. We currently use REX-Ray as a Docker Volume Driver. The REX-Ray service needs IAAS credentials to operate.
Please note that while REX-Ray is able to dynamically provision volumes on your provider, those will NOT be cleaned up upon undeployment. REX-Ray currently only supports AWS, although it is moving forward quite rapidly, and new features or providers are likely to be added in the near future. As of today, REX-Ray does not support high availability.
REX-Ray is built on a client-server architecture, with clients operating as Docker Volume Drivers that use the libStorage service. The libStorage service (which we also call RexRayServer for simplicity) runs alongside the Master(s) node(s) and is able to provision and mount storage resources on the fly. It needs to be configured and given proper credentials to manage storage providers. Storage providers configuration is best described in libStorage’s documentation.
HTTP Health checks
In addition to Mesos tasks states, Marathon features automatic HTTP health checks against running containers. A health check is considered passing if its HTTP response code is between 200 and 399 inclusive, and its response is received within the a timeout period.
The plugin adds a default health check to all the containers in the topology.
Known limitations and caveats
- It is not possible to scale Docker containers. This is due to Marathon only allowing singletons when using external volumes in conjunction with containers.
- We did not exactly follow the TOSCA model for Docker containers as it is still incubating.
- It not possible to stop the deployment of an application. Wait for it to be deployed then hit un-deploy.
- The connexion to Marathon is NOT secured.
- Health checks events are not parsed. However, the health of each instance is polled when refreshing the runtime view.


Insurance Group 17, SAVE 5661 off new list price, Sat. Nav.,Heated Seats,Cruise Control,Climate Control,1/2 Leather Trim,Reversing Sensors Rear,17 inch Wheels,Electric Folding Mirrors,Bluetooth,Alloy Wheels,Reverse Parking Camera,Solid paint, 53.3 MPG,All vehicles come with the balance of the manufacturer's warranty, are inspected by trained technicians and are history checked. Test drives and part-exchange are available, and you can choose, buy and drive away the same day. For more details on this or one of over 4000 other cars at Motorpoint, the UKs leading Car Supermarket, search Motorpoint.co.uk. (If you wish to enquire about this vehicle please quote 667249)
ViewSome quick example text to build on the card title and make up the bulk of the card's content.
ButtonInsurance Group 17, SAVE 5661 off new list price, Sat. Nav.,Heated Seats,Cruise Control,Climate Control,1/2 Leather Trim,Reversing Sensors Rear,17 inch Wheels,Electric Folding Mirrors,Bluetooth,Alloy Wheels,Reverse Parking Camera,Solid paint, 53.3 MPG,All vehicles come with the balance of the manufacturer's warranty, are inspected by trained technicians and are history checked. Test drives and part-exchange are available, and you can choose, buy and drive away the same day. For more details on this or one of over 4000 other cars at Motorpoint, the UKs leading Car Supermarket, search Motorpoint.co.uk. (If you wish to enquire about this vehicle please quote 667249)
View


This is a public page, but you can navigate to protected pages to trigger authentication.
Different content is shown depending on the role of the authenticated user.
Click here to see a greeting. (administrators only)
Click here to see current time. (users only)
Click here to see tolldata. (administrators with read role only)
Click here to see users. (administrators only)


Team Name |
Team Id |
Division |
Head Coach |
Points in total |
Rebounds in total |
Assists in total |
Steals in total |
|---|---|---|---|---|---|---|---|
| {{team[0]}} | {{team[1]}} | {{team[2]}} | {{team[3]}} | {{team[4]}} | {{team[5]}} | {{team[6]}} | {{team[7]}} |


f(x)
is graphed below.
What is f(x)
?
f(x)=
VSCALE
FN
(\
1 / HSCALE
\ x)
The function starts at its maximum value (ie, f(0)=VSCALE
), so what kind of function is it?
The cosine function, \cos(x)
, starts at 1 (ie, \cos(0)=1
), so f(x)
must be a scaled version of the cosine function.
The function starts at zero (ie, f(0)=0
), so what kind of function is it?
The sine function, \sin(x)
, starts at 0 (ie, \sin(0)=0
), so f(x)
must be a scaled version of the sine function.
The distance from peak to peak is piFraction( PERIOD )
, so the period of f(x)
is piFraction( PERIOD )
.
The distance between every other zero is piFraction( PERIOD )
, so the period of f(x)
is piFraction( PERIOD )
.
The period of a normal FNS function is 2\pi
, and the period we want is piFraction( PERIOD )
, so we don't need to worry about scaling the function horizontally.
The period of a normal FNS function is 2\pi
, and the period we want is piFraction( PERIOD )
, so we need to scale the FNS function horizontally by decFrac( PERIOD / 2 / PI )
.
To horizontally scale \FN(x)
by decFrac( PERIOD / 2 / PI )
, we need to substitute decFrac( 2 * PI / PERIOD )x
in for x
to get \FN(decFrac( 2 * PI / PERIOD )x)
.
The height of f(x)
is decFrac( VSCALE )
, so the amplitude of f(x)
is decFrac( VSCALE )
.
The amplitude of f(x)
is 1, so we don't need to worry about scaling the function vertically.
The amplitude of f(x)
is decFrac( VSCALE )
, so we need to scale FNS function vertically by decFrac( VSCALE )
.
To scale the function vertically, multiply the whole thing by decFrac( VSCALE )
.
So the resulting function (after we perform all these manipulations) is plus( toFractionTex( VSCALE ) + "\\" + FN + "(" + plus( toFractionTex( 1 / HSCALE ) + "x" ) + ")" )
.


Batch normalization.
tf.compat.v1.nn.fused_batch_norm( x, scale, offset, mean=None, variance=None, epsilon=0.001, data_format='NHWC', is_training=True, name=None )
See Source: [Batch Normalization: Accelerating Deep Network Training by Reducing Internal Covariate Shift; S. Ioffe, C. Szegedy] (http://arxiv.org/abs/1502.03167).
x
: Input Tensor
of 4 dimensions.scale
: A Tensor
of 1 dimension for scaling.offset
: A Tensor
of 1 dimension for bias.mean
: A Tensor
of 1 dimension for population mean used for inference.variance
: A Tensor
of 1 dimension for population variance
used for inference.epsilon
: A small float number added to the variance of x.data_format
: The data format for x. Either "NHWC" (default) or "NCHW".is_training
: A bool value to specify if the operation is used for
training or inference.name
: A name for this operation (optional).y
: A 4D Tensor for the normalized, scaled, offsetted x.batch_mean
: A 1D Tensor for the mean of x.batch_var
: A 1D Tensor for the variance of x.ValueError
: If mean or variance is not None when is_training is True.


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Sugar Baby Porkchops
Yields: 6
Preparation time: 15
Cooking time: 30
Ingredients
-
1/2" thick pork chops 4-6Bread crumbs 2 cupsBrown sugar 1/2 cupSalt 1 tspDry Italian Seasoning 2 Tbsp.Butter or Olive Oil to cover the bottom of a cast iron pan or saute panApple wedges
Directions
-
Mix all dry ingredients in a bowl.Cover each pork chop in the dry mixture.Put butter or oil in the frying pan, place pork chops into warm butter or oil allowing it to come up to temperature with the meat in it (this will lock the moisture in).Fry each side of the pork chops until golden brown


|
qml-ar
|
This template class wraps an object adding additional information: id, timestamp. More...
#include <QObject>
#include <QDateTime>
#include "pipelinecontainerinfo.h"
#include "pipelinecontainer.cpp"
Go to the source code of this file.
Classes | |
| class | PipelineContainer< T > |
| This template class wraps an object adding additional information: id, timestamp. More... | |
This template class wraps an object adding additional information: id, timestamp.


Hello {% include "user.snippet" %} — you're almost there.
Now what?
{{ APP_NAME }} is a two-player deal. You need to find another Foursquare user
to intersect your checkins with. You can:
-
Give out the link {{ short_url }} to
others to have them intersect their checkins with yours.
-
If you use Twitter, you can
out that link and get your followers to join in.
-
If you're on the go, you can get a QR code
so that someone else can intersect their checkins without having to type
in that long URL.
{% include "mihai-intersect.snippet" %}
{% if user.email_address %}
{% endif %}


The expression for the standard Mean Squared Error (MSE) which we used to define our cost function and the equations for the ordinary least squares (OLS) method, that is our optimization problem is $$ {\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}. $$ or we can state it as $$ {\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2, $$ where we have used the definition of a norm-2 vector, that is $$ \vert\vert \boldsymbol{x}\vert\vert_2 = \sqrt{\sum_i x_i^2}. $$
By minimizing the above equation with respect to the parameters \( \boldsymbol{\beta} \) we could then obtain an analytical expression for the parameters \( \boldsymbol{\beta} \). We can add a regularization parameter \( \lambda \) by defining a new cost function to be optimized, that is $$ {\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_2^2 $$
which leads to the Ridge regression minimization problem where we require that \( \vert\vert \boldsymbol{\beta}\vert\vert_2^2\le t \), where \( t \) is a finite number larger than zero. By defining $$ C(\boldsymbol{X},\boldsymbol{\beta})=\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1, $$
we have a new optimization equation $$ {\displaystyle \min_{\boldsymbol{\beta}\in {\mathbb{R}}^{p}}}\frac{1}{n}\vert\vert \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\vert\vert_2^2+\lambda\vert\vert \boldsymbol{\beta}\vert\vert_1 $$ which leads to Lasso regression. Lasso stands for least absolute shrinkage and selection operator.
Here we have defined the norm-1 as $$ \vert\vert \boldsymbol{x}\vert\vert_1 = \sum_i \vert x_i\vert. $$
When the repetitive splitting of the data set is done randomly, samples may accidently end up in a fast majority of the splits in either training or test set. Such samples may have an unbalanced influence on either model building or prediction evaluation. To avoid this \( k \)-fold cross-validation structures the data splitting. The samples are divided into \( k \) more or less equally sized exhaustive and mutually exclusive subsets. In turn (at each split) one of these subsets plays the role of the test set while the union of the remaining subsets constitutes the training set. Such a splitting warrants a balanced representation of each sample in both training and test set over the splits. Still the division into the \( k \) subsets involves a degree of randomness. This may be fully excluded when choosing \( k=n \). This particular case is referred to as leave-one-out cross-validation (LOOCV).
For the various values of \( k \)
The code here uses Ridge regression with cross-validation (CV) resampling and \( k \)-fold CV in order to fit a specific polynomial.
import numpy as np
import matplotlib.pyplot as plt
from sklearn.model_selection import KFold
from sklearn.linear_model import Ridge
from sklearn.model_selection import cross_val_score
from sklearn.preprocessing import PolynomialFeatures
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(3155)
# Generate the data.
nsamples = 100
x = np.random.randn(nsamples)
y = 3*x**2 + np.random.randn(nsamples)
## Cross-validation on Ridge regression using KFold only
# Decide degree on polynomial to fit
poly = PolynomialFeatures(degree = 6)
# Decide which values of lambda to use
nlambdas = 500
lambdas = np.logspace(-3, 5, nlambdas)
# Initialize a KFold instance
k = 5
kfold = KFold(n_splits = k)
# Perform the cross-validation to estimate MSE
scores_KFold = np.zeros((nlambdas, k))
i = 0
for lmb in lambdas:
ridge = Ridge(alpha = lmb)
j = 0
for train_inds, test_inds in kfold.split(x):
xtrain = x[train_inds]
ytrain = y[train_inds]
xtest = x[test_inds]
ytest = y[test_inds]
Xtrain = poly.fit_transform(xtrain[:, np.newaxis])
ridge.fit(Xtrain, ytrain[:, np.newaxis])
Xtest = poly.fit_transform(xtest[:, np.newaxis])
ypred = ridge.predict(Xtest)
scores_KFold[i,j] = np.sum((ypred - ytest[:, np.newaxis])**2)/np.size(ypred)
j += 1
i += 1
estimated_mse_KFold = np.mean(scores_KFold, axis = 1)
## Cross-validation using cross_val_score from sklearn along with KFold
# kfold is an instance initialized above as:
# kfold = KFold(n_splits = k)
estimated_mse_sklearn = np.zeros(nlambdas)
i = 0
for lmb in lambdas:
ridge = Ridge(alpha = lmb)
X = poly.fit_transform(x[:, np.newaxis])
estimated_mse_folds = cross_val_score(ridge, X, y[:, np.newaxis], scoring='neg_mean_squared_error', cv=kfold)
# cross_val_score return an array containing the estimated negative mse for every fold.
# we have to the the mean of every array in order to get an estimate of the mse of the model
estimated_mse_sklearn[i] = np.mean(-estimated_mse_folds)
i += 1
## Plot and compare the slightly different ways to perform cross-validation
plt.figure()
plt.plot(np.log10(lambdas), estimated_mse_sklearn, label = 'cross_val_score')
plt.plot(np.log10(lambdas), estimated_mse_KFold, 'r--', label = 'KFold')
plt.xlabel('log10(lambda)')
plt.ylabel('mse')
plt.legend()
plt.show()
When you are comparing your own code with for example Scikit-Learn's library, there are some technicalities to keep in mind. The examples here demonstrate some of these aspects with potential pitfalls.
The discussion here focuses on the role of the intercept, how we can set up the design matrix, what scaling we should use and other topics which tend confuse us.
The intercept can be interpreted as the expected value of our target/output variables when all other predictors are set to zero. Thus, if we cannot assume that the expected outputs/targets are zero when all predictors are zero (the columns in the design matrix), it may be a bad idea to implement a model which penalizes the intercept. Furthermore, in for example Ridge and Lasso regression, the default solutions from the library Scikit-Learn (when not shrinking \( \beta_0 \)) for the unknown parameters \( \boldsymbol{\beta} \), are derived under the assumption that both \( \boldsymbol{y} \) and \( \boldsymbol{X} \) are zero centered, that is we subtract the mean values.
If our predictors represent different scales, then it is important to standardize the design matrix \( \boldsymbol{X} \) by subtracting the mean of each column from the corresponding column and dividing the column with its standard deviation. Most machine learning libraries do this as a default. This means that if you compare your code with the results from a given library, the results may differ.
The Standadscaler function in Scikit-Learn does this for us. For the data sets we have been studying in our various examples, the data are in many cases already scaled and there is no need to scale them. You as a user of different machine learning algorithms, should always perform a survey of your data, with a critical assessment of them in case you need to scale the data.
If you need to scale the data, not doing so will give an unfair penalization of the parameters since their magnitude depends on the scale of their corresponding predictor.
Suppose as an example that you you have an input variable given by the heights of different persons. Human height might be measured in inches or meters or kilometers. If measured in kilometers, a standard linear regression model with this predictor would probably give a much bigger coefficient term, than if measured in millimeters. This can clearly lead to problems in evaluating the cost/loss functions.
Keep in mind that when you transform your data set before training a model, the same transformation needs to be done on your eventual new data set before making a prediction. If we translate this into a Python code, it would could be implemented as follows
#Model training, we compute the mean value of y and X
y_train_mean = np.mean(y_train)
X_train_mean = np.mean(X_train,axis=0)
X_train = X_train - X_train_mean
y_train = y_train - y_train_mean
# The we fit our model with the training data
trained_model = some_model.fit(X_train,y_train)
#Model prediction, we need also to transform our data set used for the prediction.
X_test = X_test - X_train_mean #Use mean from training data
y_pred = trained_model(X_test)
y_pred = y_pred + y_train_mean
Let us try to understand what this may imply mathematically when we subtract the mean values, also known as zero centering. For simplicity, we will focus on ordinary regression, as done in the above example.
The cost/loss function for regression is $$ C(\beta_0, \beta_1, ... , \beta_{p-1}) = \frac{1}{n}\sum_{i=0}^{n} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij}\beta_j\right)^2,. $$ Recall also that we use the squared value since this leads to an increase of the penalty for higher differences between predicted and output/target values.
What we have done is to single out the \( \beta_0 \) term in the definition of the mean squared error (MSE). The design matrix \( X \) does in this case not contain any intercept column. When we take the derivative with respect to \( \beta_0 \), we want the derivative to obey $$ \frac{\partial C}{\partial \beta_j} = 0, $$
for all \( j \). For \( \beta_0 \) we have $$ \frac{\partial C}{\partial \beta_0} = -\frac{2}{n}\sum_{i=0}^{n-1} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij} \beta_j\right). $$ Multiplying away the constant \( 2/n \), we obtain $$ \sum_{i=0}^{n-1} \beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} \sum_{j=1}^{p-1} X_{ij} \beta_j. $$
Let us special first to the case where we have only two parameters \( \beta_0 \) and \( \beta_1 \). Our result for \( \beta_0 \) simplifies then to $$ n\beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} X_{i1} \beta_1. $$ We obtain then $$ \beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \beta_1\frac{1}{n}\sum_{i=0}^{n-1} X_{i1}. $$ If we define $$ \mu_1=\frac{1}{n}\sum_{i=0}^{n-1} (X_{i1}, $$ and if we define the mean value of the outputs as $$ \mu_y=\frac{1}{n}\sum_{i=0}^{n-1}y_i, $$ we have $$ \beta_0 = \mu_y - \beta_1\mu_{1}. $$ In the general case, that is we have more parameters than \( \beta_0 \) and \( \beta_1 \), we have $$ \beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \frac{1}{n}\sum_{i=0}^{n-1}\sum_{j=1}^{p-1} X_{ij}\beta_j. $$
Replacing \( y_i \) with \( y_i - y_i - \overline{\boldsymbol{y}} \) and centering also our design matrix results in a cost function (in vector-matrix disguise) $$ C(\boldsymbol{\beta}) = (\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta})^T(\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta}). $$
If we minimize with respect to \( \boldsymbol{\beta} \) we have then $$ \hat{\boldsymbol{\beta}} = (\tilde{X}^T\tilde{X})^{-1}\tilde{X}^T\boldsymbol{\tilde{y}}, $$
where \( \boldsymbol{\tilde{y}} = \boldsymbol{y} - \overline{\boldsymbol{y}} \) and \( \tilde{X}_{ij} = X_{ij} - \frac{1}{n}\sum_{k=0}^{n-1}X_{kj} \).
For Ridge regression we need to add \( \lambda \boldsymbol{\beta}^T\boldsymbol{\beta} \) to the cost function and get then $$ \hat{\boldsymbol{\beta}} = (\tilde{X}^T\tilde{X} + \lambda I)^{-1}\tilde{X}^T\boldsymbol{\tilde{y}}. $$
What does this mean? And why do we insist on all this? Let us look at some examples.
This code shows a simple first-order fit to a data set using the above transformed data, where we consider the role of the intercept first, by either excluding it or including it (code example thanks to Øyvind Sigmundson Schøyen). Here our scaling of the data is done by subtracting the mean values only. Note also that we do not split the data into training and test.
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
np.random.seed(2021)
def MSE(y_data,y_model):
n = np.size(y_model)
return np.sum((y_data-y_model)**2)/n
def fit_beta(X, y):
return np.linalg.pinv(X.T @ X) @ X.T @ y
true_beta = [2, 0.5, 3.7]
x = np.linspace(0, 1, 11)
y = np.sum(
np.asarray([x ** p * b for p, b in enumerate(true_beta)]), axis=0
) + 0.1 * np.random.normal(size=len(x))
degree = 3
X = np.zeros((len(x), degree))
# Include the intercept in the design matrix
for p in range(degree):
X[:, p] = x ** p
beta = fit_beta(X, y)
# Intercept is included in the design matrix
skl = LinearRegression(fit_intercept=False).fit(X, y)
print(f"True beta: {true_beta}")
print(f"Fitted beta: {beta}")
print(f"Sklearn fitted beta: {skl.coef_}")
ypredictOwn = X @ beta
ypredictSKL = skl.predict(X)
print(f"MSE with intercept column")
print(MSE(y,ypredictOwn))
print(f"MSE with intercept column from SKL")
print(MSE(y,ypredictSKL))
plt.figure()
plt.scatter(x, y, label="Data")
plt.plot(x, X @ beta, label="Fit")
plt.plot(x, skl.predict(X), label="Sklearn (fit_intercept=False)")
# Do not include the intercept in the design matrix
X = np.zeros((len(x), degree - 1))
for p in range(degree - 1):
X[:, p] = x ** (p + 1)
# Intercept is not included in the design matrix
skl = LinearRegression(fit_intercept=True).fit(X, y)
# Use centered values for X and y when computing coefficients
y_offset = np.average(y, axis=0)
X_offset = np.average(X, axis=0)
beta = fit_beta(X - X_offset, y - y_offset)
intercept = np.mean(y_offset - X_offset @ beta)
print(f"Manual intercept: {intercept}")
print(f"Fitted beta (wiothout intercept): {beta}")
print(f"Sklearn intercept: {skl.intercept_}")
print(f"Sklearn fitted beta (without intercept): {skl.coef_}")
ypredictOwn = X @ beta
ypredictSKL = skl.predict(X)
print(f"MSE with Manual intercept")
print(MSE(y,ypredictOwn+intercept))
print(f"MSE with Sklearn intercept")
print(MSE(y,ypredictSKL))
plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
plt.plot(x, skl.predict(X), "--", label="Sklearn (fit_intercept=True)")
plt.grid()
plt.legend()
plt.show()
The intercept is the value of our output/target variable when all our features are zero and our function crosses the \( y \)-axis (for a one-dimensional case).
Printing the MSE, we see first that both methods give the same MSE, as they should. However, when we move to for example Ridge regression, the way we treat the intercept may give a larger or smaller MSE, meaning that the MSE can be penalized by the value of the intercept. Not including the intercept in the fit, means that the regularization term does not include \( \beta_0 \). For different values of \( \lambda \), this may lead to differeing MSE values.
To remind the reader, the regularization term, with the intercept in Ridge regression is given by $$ \lambda \vert\vert \boldsymbol{\beta} \vert\vert_2^2 = \lambda \sum_{j=0}^{p-1}\beta_j^2, $$ but when we take out the intercept, this equation becomes $$ \lambda \vert\vert \boldsymbol{\beta} \vert\vert_2^2 = \lambda \sum_{j=1}^{p-1}\beta_j^2. $$
For Lasso regression we have $$ \lambda \vert\vert \boldsymbol{\beta} \vert\vert_1 = \lambda \sum_{j=1}^{p-1}\vert\beta_j\vert. $$
It means that, when scaling the design matrix and the outputs/targets, by subtracting the mean values, we have an optimization problem which is not penalized by the intercept. The MSE value can then be smaller since it focuses only on the remaining quantities. If we however bring back the intercept, we will get a MSE which then contains the intercept.
Armed with this wisdom, we attempt first to simply set the intercept equal to False in our implementation of Ridge regression for our well-known vanilla data set.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn import linear_model
def MSE(y_data,y_model):
n = np.size(y_model)
return np.sum((y_data-y_model)**2)/n
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(3155)
n = 100
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
Maxpolydegree = 20
X = np.zeros((n,Maxpolydegree))
#We include explicitely the intercept column
for degree in range(Maxpolydegree):
X[:,degree] = x**degree
# We split the data in test and training data
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
p = Maxpolydegree
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 6
MSEOwnRidgePredict = np.zeros(nlambdas)
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 2, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
OwnRidgeBeta = np.linalg.pinv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
# Note: we include the intercept column and no scaling
RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
RegRidge.fit(X_train,y_train)
# and then make the prediction
ytildeOwnRidge = X_train @ OwnRidgeBeta
ypredictOwnRidge = X_test @ OwnRidgeBeta
ytildeRidge = RegRidge.predict(X_train)
ypredictRidge = RegRidge.predict(X_test)
MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
print("Beta values for own Ridge implementation")
print(OwnRidgeBeta)
print("Beta values for Scikit-Learn Ridge implementation")
print(RegRidge.coef_)
print("MSE values for own Ridge implementation")
print(MSEOwnRidgePredict[i])
print("MSE values for Scikit-Learn Ridge implementation")
print(MSERidgePredict[i])
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'r', label = 'MSE own Ridge Test')
plt.plot(np.log10(lambdas), MSERidgePredict, 'g', label = 'MSE Ridge Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
The results here agree when we force Scikit-Learn's Ridge function to include the first column in our design matrix. We see that the results agree very well. Here we have thus explicitely included the intercept column in the design matrix. What happens if we do not include the intercept in our fit? Let us see how we can change this code by zero centering (thanks to Stian Bilek for inpouts here).
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn import linear_model
from sklearn.preprocessing import StandardScaler
def MSE(y_data,y_model):
n = np.size(y_model)
return np.sum((y_data-y_model)**2)/n
# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(315)
n = 100
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
Maxpolydegree = 20
X = np.zeros((n,Maxpolydegree-1))
for degree in range(1,Maxpolydegree): #No intercept column
X[:,degree-1] = x**(degree)
# We split the data in test and training data
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
#For our own implementation, we will need to deal with the intercept by centering the design matrix and the target variable
X_train_mean = np.mean(X_train,axis=0)
#Center by removing mean from each feature
X_train_scaled = X_train - X_train_mean
X_test_scaled = X_test - X_train_mean
#The model intercept (called y_scaler) is given by the mean of the target variable (IF X is centered)
#Remove the intercept from the training data.
y_scaler = np.mean(y_train)
y_train_scaled = y_train - y_scaler
p = Maxpolydegree-1
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 6
MSEOwnRidgePredict = np.zeros(nlambdas)
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 2, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
OwnRidgeBeta = np.linalg.pinv(X_train_scaled.T @ X_train_scaled+lmb*I) @ X_train_scaled.T @ (y_train_scaled)
intercept_ = y_scaler - X_train_mean@OwnRidgeBeta #The intercept can be shifted so the model can predict on uncentered data
#Add intercept to prediction
ypredictOwnRidge = X_test_scaled @ OwnRidgeBeta + y_scaler
RegRidge = linear_model.Ridge(lmb)
RegRidge.fit(X_train,y_train)
ypredictRidge = RegRidge.predict(X_test)
MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
print("Beta values for own Ridge implementation")
print(OwnRidgeBeta) #Intercept is given by mean of target variable
print("Beta values for Scikit-Learn Ridge implementation")
print(RegRidge.coef_)
print('Intercept from own implementation:')
print(intercept_)
print('Intercept from Scikit-Learn Ridge implementation')
print(RegRidge.intercept_)
print("MSE values for own Ridge implementation")
print(MSEOwnRidgePredict[i])
print("MSE values for Scikit-Learn Ridge implementation")
print(MSERidgePredict[i])
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'b--', label = 'MSE own Ridge Test')
plt.plot(np.log10(lambdas), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
We see here, when compared to the code which includes explicitely the intercept column, that our MSE value is actually smaller. This is because the regularization term does not include the intercept value \( \beta_0 \) in the fitting. This applies to Lasso regularization as well. It means that our optimization is now done only with the centered matrix and/or vector that enter the fitting procedure. Note also that the problem with the intercept occurs mainly in these type of polynomial fitting problem.
The next example is indeed an example where all these discussions about the role of intercept are not present.
The one-dimensional Ising model with nearest neighbor interaction, no external field and a constant coupling constant \( J \) is given by $$ \begin{align} H = -J \sum_{k}^L s_k s_{k + 1}, \label{_auto1} \end{align} $$
where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins in the system is determined by \( L \). For the one-dimensional system there is no phase transition.
We will look at a system of \( L = 40 \) spins with a coupling constant of \( J = 1 \). To get enough training data we will generate 10000 states with their respective energies.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable
import seaborn as sns
import scipy.linalg as scl
from sklearn.model_selection import train_test_split
import tqdm
sns.set(color_codes=True)
cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
L = 40
n = int(1e4)
spins = np.random.choice([-1, 1], size=(n, L))
J = 1.0
energies = np.zeros(n)
for i in range(n):
energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
Here we use ordinary least squares regression to predict the energy for the nearest neighbor one-dimensional Ising model on a ring, i.e., the endpoints wrap around. We will use linear regression to fit a value for the coupling constant to achieve this.
A more general form for the one-dimensional Ising model is $$ \begin{align} H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. \label{_auto2} \end{align} $$
Here we allow for interactions beyond the nearest neighbors and a state dependent coupling constant. This latter expression can be formulated as a matrix-product $$ \begin{align} \boldsymbol{H} = \boldsymbol{X} J, \label{_auto3} \end{align} $$
where \( X_{jk} = s_j s_k \) and \( J \) is a matrix which consists of the elements \( -J_{jk} \). This form of writing the energy fits perfectly with the form utilized in linear regression, that is $$ \begin{align} \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}, \label{_auto4} \end{align} $$
We split the data in training and test data as discussed in the previous example
X = np.zeros((n, L ** 2))
for i in range(n):
X[i] = np.outer(spins[i], spins[i]).ravel()
y = energies
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
In the ordinary least squares method we choose the cost function $$ \begin{align} C(\boldsymbol{X}, \boldsymbol{\beta})= \frac{1}{n}\left\{(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})\right\}. \label{_auto5} \end{align} $$
We then find the extremal point of \( C \) by taking the derivative with respect to \( \boldsymbol{\beta} \) as discussed above. This yields the expression for \( \boldsymbol{\beta} \) to be $$ \boldsymbol{\beta} = \frac{\boldsymbol{X}^T \boldsymbol{y}}{\boldsymbol{X}^T \boldsymbol{X}}, $$
which immediately imposes some requirements on \( \boldsymbol{X} \) as there must exist an inverse of \( \boldsymbol{X}^T \boldsymbol{X} \). If the expression we are modeling contains an intercept, i.e., a constant term, we must make sure that the first column of \( \boldsymbol{X} \) consists of \( 1 \). We do this here
X_train_own = np.concatenate(
(np.ones(len(X_train))[:, np.newaxis], X_train),
axis=1
)
X_test_own = np.concatenate(
(np.ones(len(X_test))[:, np.newaxis], X_test),
axis=1
)
def ols_inv(x: np.ndarray, y: np.ndarray) -> np.ndarray:
return scl.inv(x.T @ x) @ (x.T @ y)
beta = ols_inv(X_train_own, y_train)
Doing the inversion directly turns out to be a bad idea since the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is singular. An alternative approach is to use the singular value decomposition. Using the definition of the Moore-Penrose pseudoinverse we can write the equation for \( \boldsymbol{\beta} \) as $$ \boldsymbol{\beta} = \boldsymbol{X}^{+}\boldsymbol{y}, $$
where the pseudoinverse of \( \boldsymbol{X} \) is given by $$ \boldsymbol{X}^{+} = \frac{\boldsymbol{X}^T}{\boldsymbol{X}^T\boldsymbol{X}}. $$
Using singular value decomposition we can decompose the matrix \( \boldsymbol{X} = \boldsymbol{U}\boldsymbol{\Sigma} \boldsymbol{V}^T \), where \( \boldsymbol{U} \) and \( \boldsymbol{V} \) are orthogonal(unitary) matrices and \( \boldsymbol{\Sigma} \) contains the singular values (more details below). where \( X^{+} = V\Sigma^{+} U^T \). This reduces the equation for \( \omega \) to $$ \begin{align} \boldsymbol{\beta} = \boldsymbol{V}\boldsymbol{\Sigma}^{+} \boldsymbol{U}^T \boldsymbol{y}. \label{_auto6} \end{align} $$
Note that solving this equation by actually doing the pseudoinverse (which is what we will do) is not a good idea as this operation scales as \( \mathcal{O}(n^3) \), where \( n \) is the number of elements in a general matrix. Instead, doing \( QR \)-factorization and solving the linear system as an equation would reduce this down to \( \mathcal{O}(n^2) \) operations.
def ols_svd(x: np.ndarray, y: np.ndarray) -> np.ndarray:
u, s, v = scl.svd(x)
return v.T @ scl.pinv(scl.diagsvd(s, u.shape[0], v.shape[0])) @ u.T @ y
beta = ols_svd(X_train_own,y_train)
When extracting the \( J \)-matrix we need to make sure that we remove the intercept, as is done here
J = beta[1:].reshape(L, L)
A way of looking at the coefficients in \( J \) is to plot the matrices as images.
fig = plt.figure(figsize=(20, 14))
im = plt.imshow(J, **cmap_args)
plt.title("OLS", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
plt.show()
It is interesting to note that OLS considers both \( J_{j, j + 1} = -0.5 \) and \( J_{j, j - 1} = -0.5 \) as valid matrix elements for \( J \). In our discussion below on hyperparameters and Ridge and Lasso regression we will see that this problem can be removed, partly and only with Lasso regression.
In this case our matrix inversion was actually possible. The obvious question now is what is the mathematics behind the SVD?
Let us bring back the Ising model again, but now with an additional focus on Ridge and Lasso regression as well. We repeat some of the basic parts of the Ising model and the setup of the training and test data. The one-dimensional Ising model with nearest neighbor interaction, no external field and a constant coupling constant \( J \) is given by $$ \begin{align} H = -J \sum_{k}^L s_k s_{k + 1}, \label{_auto7} \end{align} $$ where \( s_i \in \{-1, 1\} \) and \( s_{N + 1} = s_1 \). The number of spins in the system is determined by \( L \). For the one-dimensional system there is no phase transition.
We will look at a system of \( L = 40 \) spins with a coupling constant of \( J = 1 \). To get enough training data we will generate 10000 states with their respective energies.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.axes_grid1 import make_axes_locatable
import seaborn as sns
import scipy.linalg as scl
from sklearn.model_selection import train_test_split
import sklearn.linear_model as skl
import tqdm
sns.set(color_codes=True)
cmap_args=dict(vmin=-1., vmax=1., cmap='seismic')
L = 40
n = int(1e4)
spins = np.random.choice([-1, 1], size=(n, L))
J = 1.0
energies = np.zeros(n)
for i in range(n):
energies[i] = - J * np.dot(spins[i], np.roll(spins[i], 1))
A more general form for the one-dimensional Ising model is $$ \begin{align} H = - \sum_j^L \sum_k^L s_j s_k J_{jk}. \label{_auto8} \end{align} $$
Here we allow for interactions beyond the nearest neighbors and a more adaptive coupling matrix. This latter expression can be formulated as a matrix-product on the form $$ \begin{align} H = X J, \label{_auto9} \end{align} $$
where \( X_{jk} = s_j s_k \) and \( J \) is the matrix consisting of the elements \( -J_{jk} \). This form of writing the energy fits perfectly with the form utilized in linear regression, viz. $$ \begin{align} \boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta} + \boldsymbol{\epsilon}. \label{_auto10} \end{align} $$ We organize the data as we did above
X = np.zeros((n, L ** 2))
for i in range(n):
X[i] = np.outer(spins[i], spins[i]).ravel()
y = energies
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.96)
X_train_own = np.concatenate(
(np.ones(len(X_train))[:, np.newaxis], X_train),
axis=1
)
X_test_own = np.concatenate(
(np.ones(len(X_test))[:, np.newaxis], X_test),
axis=1
)
We will do all fitting with Scikit-Learn,
clf = skl.LinearRegression().fit(X_train, y_train)
When extracting the \( J \)-matrix we make sure to remove the intercept
J_sk = clf.coef_.reshape(L, L)
And then we plot the results
fig = plt.figure(figsize=(20, 14))
im = plt.imshow(J_sk, **cmap_args)
plt.title("LinearRegression from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
plt.show()
The results perfectly with our previous discussion where we used our own code.
Having explored the ordinary least squares we move on to ridge regression. In ridge regression we include a regularizer. This involves a new cost function which leads to a new estimate for the weights \( \boldsymbol{\beta} \). This results in a penalized regression problem. The cost function is given by $$ \begin{align} C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}. \label{_auto11} \end{align} $$
_lambda = 0.1
clf_ridge = skl.Ridge(alpha=_lambda).fit(X_train, y_train)
J_ridge_sk = clf_ridge.coef_.reshape(L, L)
fig = plt.figure(figsize=(20, 14))
im = plt.imshow(J_ridge_sk, **cmap_args)
plt.title("Ridge from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
plt.show()
In the Least Absolute Shrinkage and Selection Operator (LASSO)-method we get a third cost function. $$ \begin{align} C(\boldsymbol{X}, \boldsymbol{\beta}; \lambda) = (\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y})^T(\boldsymbol{X}\boldsymbol{\beta} - \boldsymbol{y}) + \lambda \sqrt{\boldsymbol{\beta}^T\boldsymbol{\beta}}. \label{_auto12} \end{align} $$
Finding the extremal point of this cost function is not so straight-forward as in least squares and ridge. We will therefore rely solely on the function ``Lasso`` from Scikit-Learn.
clf_lasso = skl.Lasso(alpha=_lambda).fit(X_train, y_train)
J_lasso_sk = clf_lasso.coef_.reshape(L, L)
fig = plt.figure(figsize=(20, 14))
im = plt.imshow(J_lasso_sk, **cmap_args)
plt.title("Lasso from Scikit-learn", fontsize=18)
plt.xticks(fontsize=18)
plt.yticks(fontsize=18)
cb = fig.colorbar(im)
cb.ax.set_yticklabels(cb.ax.get_yticklabels(), fontsize=18)
plt.show()
It is quite striking how LASSO breaks the symmetry of the coupling constant as opposed to ridge and OLS. We get a sparse solution with \( J_{j, j + 1} = -1 \).
We see how the different models perform for a different set of values for \( \lambda \).
lambdas = np.logspace(-4, 5, 10)
train_errors = {
"ols_sk": np.zeros(lambdas.size),
"ridge_sk": np.zeros(lambdas.size),
"lasso_sk": np.zeros(lambdas.size)
}
test_errors = {
"ols_sk": np.zeros(lambdas.size),
"ridge_sk": np.zeros(lambdas.size),
"lasso_sk": np.zeros(lambdas.size)
}
plot_counter = 1
fig = plt.figure(figsize=(32, 54))
for i, _lambda in enumerate(tqdm.tqdm(lambdas)):
for key, method in zip(
["ols_sk", "ridge_sk", "lasso_sk"],
[skl.LinearRegression(), skl.Ridge(alpha=_lambda), skl.Lasso(alpha=_lambda)]
):
method = method.fit(X_train, y_train)
train_errors[key][i] = method.score(X_train, y_train)
test_errors[key][i] = method.score(X_test, y_test)
omega = method.coef_.reshape(L, L)
plt.subplot(10, 5, plot_counter)
plt.imshow(omega, **cmap_args)
plt.title(r"%s, $\lambda = %.4f$" % (key, _lambda))
plot_counter += 1
plt.show()
We see that LASSO reaches a good solution for low values of \( \lambda \), but will "wither" when we increase \( \lambda \) too much. Ridge is more stable over a larger range of values for \( \lambda \), but eventually also fades away.
To determine which value of \( \lambda \) is best we plot the accuracy of the models when predicting the training and the testing set. We expect the accuracy of the training set to be quite good, but if the accuracy of the testing set is much lower this tells us that we might be subject to an overfit model. The ideal scenario is an accuracy on the testing set that is close to the accuracy of the training set.
fig = plt.figure(figsize=(20, 14))
colors = {
"ols_sk": "r",
"ridge_sk": "y",
"lasso_sk": "c"
}
for key in train_errors:
plt.semilogx(
lambdas,
train_errors[key],
colors[key],
label="Train {0}".format(key),
linewidth=4.0
)
for key in test_errors:
plt.semilogx(
lambdas,
test_errors[key],
colors[key] + "--",
label="Test {0}".format(key),
linewidth=4.0
)
plt.legend(loc="best", fontsize=18)
plt.xlabel(r"$\lambda$", fontsize=18)
plt.ylabel(r"$R^2$", fontsize=18)
plt.tick_params(labelsize=18)
plt.show()
From the above figure we can see that LASSO with \( \lambda = 10^{-2} \) achieves a very good accuracy on the test set. This by far surpasses the other models for all values of \( \lambda \).
In linear regression our main interest was centered on learning the coefficients of a functional fit (say a polynomial) in order to be able to predict the response of a continuous variable on some unseen data. The fit to the continuous variable \( y_i \) is based on some independent variables \( \boldsymbol{x}_i \). Linear regression resulted in analytical expressions for standard ordinary Least Squares or Ridge regression (in terms of matrices to invert) for several quantities, ranging from the variance and thereby the confidence intervals of the parameters \( \boldsymbol{\beta} \) to the mean squared error. If we can invert the product of the design matrices, linear regression gives then a simple recipe for fitting our data.
Classification problems, however, are concerned with outcomes taking the form of discrete variables (i.e. categories). We may for example, on the basis of DNA sequencing for a number of patients, like to find out which mutations are important for a certain disease; or based on scans of various patients' brains, figure out if there is a tumor or not; or given a specific physical system, we'd like to identify its state, say whether it is an ordered or disordered system (typical situation in solid state physics); or classify the status of a patient, whether she/he has a stroke or not and many other similar situations.
The most common situation we encounter when we apply logistic regression is that of two possible outcomes, normally denoted as a binary outcome, true or false, positive or negative, success or failure etc.
Logistic regression will also serve as our stepping stone towards neural network algorithms and supervised deep learning. For logistic learning, the minimization of the cost function leads to a non-linear equation in the parameters \( \boldsymbol{\beta} \). The optimization of the problem calls therefore for minimization algorithms. This forms the bottle neck of all machine learning algorithms, namely how to find reliable minima of a multi-variable function. This leads us to the family of gradient descent methods. The latter are the working horses of basically all modern machine learning algorithms.
We note also that many of the topics discussed here on logistic regression are also commonly used in modern supervised Deep Learning models, as we will see later.
We consider the case where the dependent variables, also called the responses or the outcomes, \( y_i \) are discrete and only take values from \( k=0,\dots,K-1 \) (i.e. \( K \) classes).
The goal is to predict the output classes from the design matrix \( \boldsymbol{X}\in\mathbb{R}^{n\times p} \) made of \( n \) samples, each of which carries \( p \) features or predictors. The primary goal is to identify the classes to which new unseen samples belong.
Let us specialize to the case of two classes only, with outputs \( y_i=0 \) and \( y_i=1 \). Our outcomes could represent the status of a credit card user that could default or not on her/his credit card debt. That is $$ y_i = \begin{bmatrix} 0 & \mathrm{no}\\ 1 & \mathrm{yes} \end{bmatrix}. $$
Before moving to the logistic model, let us try to use our linear regression model to classify these two outcomes. We could for example fit a linear model to the default case if \( y_i > 0.5 \) and the no default case \( y_i \leq 0.5 \).
We would then have our weighted linear combination, namely $$ \begin{equation} \boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{\beta} + \boldsymbol{\epsilon}, \label{_auto13} \end{equation} $$ where \( \boldsymbol{y} \) is a vector representing the possible outcomes, \( \boldsymbol{X} \) is our \( n\times p \) design matrix and \( \boldsymbol{\beta} \) represents our estimators/predictors.
The main problem with our function is that it takes values on the entire real axis. In the case of logistic regression, however, the labels \( y_i \) are discrete variables. A typical example is the credit card data discussed below here, where we can set the state of defaulting the debt to \( y_i=1 \) and not to \( y_i=0 \) for one the persons in the data set (see the full example below).
One simple way to get a discrete output is to have sign functions that map the output of a linear regressor to values \( \{0,1\} \), \( f(s_i)=sign(s_i)=1 \) if \( s_i\ge 0 \) and 0 if otherwise. We will encounter this model in our first demonstration of neural networks. Historically it is called the ``perceptron" model in the machine learning literature. This model is extremely simple. However, in many cases it is more favorable to use a ``soft" classifier that outputs the probability of a given category. This leads us to the logistic function.
The following example on data for coronary heart disease (CHD) as function of age may serve as an illustration. In the code here we read and plot whether a person has had CHD (output = 1) or not (output = 0). This ouput is plotted the person's against age. Clearly, the figure shows that attempting to make a standard linear regression fit may not be very meaningful.
# Common imports
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from sklearn.model_selection import train_test_split
from sklearn.utils import resample
from sklearn.metrics import mean_squared_error
from IPython.display import display
from pylab import plt, mpl
plt.style.use('seaborn')
mpl.rcParams['font.family'] = 'serif'
# Where to save the figures and data files
PROJECT_ROOT_DIR = "Results"
FIGURE_ID = "Results/FigureFiles"
DATA_ID = "DataFiles/"
if not os.path.exists(PROJECT_ROOT_DIR):
os.mkdir(PROJECT_ROOT_DIR)
if not os.path.exists(FIGURE_ID):
os.makedirs(FIGURE_ID)
if not os.path.exists(DATA_ID):
os.makedirs(DATA_ID)
def image_path(fig_id):
return os.path.join(FIGURE_ID, fig_id)
def data_path(dat_id):
return os.path.join(DATA_ID, dat_id)
def save_fig(fig_id):
plt.savefig(image_path(fig_id) + ".png", format='png')
infile = open(data_path("chddata.csv"),'r')
# Read the chd data as csv file and organize the data into arrays with age group, age, and chd
chd = pd.read_csv(infile, names=('ID', 'Age', 'Agegroup', 'CHD'))
chd.columns = ['ID', 'Age', 'Agegroup', 'CHD']
output = chd['CHD']
age = chd['Age']
agegroup = chd['Agegroup']
numberID = chd['ID']
display(chd)
plt.scatter(age, output, marker='o')
plt.axis([18,70.0,-0.1, 1.2])
plt.xlabel(r'Age')
plt.ylabel(r'CHD')
plt.title(r'Age distribution and Coronary heart disease')
plt.show()
What we could attempt however is to plot the mean value for each group.
agegroupmean = np.array([0.1, 0.133, 0.250, 0.333, 0.462, 0.625, 0.765, 0.800])
group = np.array([1, 2, 3, 4, 5, 6, 7, 8])
plt.plot(group, agegroupmean, "r-")
plt.axis([0,9,0, 1.0])
plt.xlabel(r'Age group')
plt.ylabel(r'CHD mean values')
plt.title(r'Mean values for each age group')
plt.show()
We are now trying to find a function \( f(y\vert x) \), that is a function which gives us an expected value for the output \( y \) with a given input \( x \). In standard linear regression with a linear dependence on \( x \), we would write this in terms of our model $$ f(y_i\vert x_i)=\beta_0+\beta_1 x_i. $$
This expression implies however that \( f(y_i\vert x_i) \) could take any value from minus infinity to plus infinity. If we however let \( f(y\vert y) \) be represented by the mean value, the above example shows us that we can constrain the function to take values between zero and one, that is we have \( 0 \le f(y_i\vert x_i) \le 1 \). Looking at our last curve we see also that it has an S-shaped form. This leads us to a very popular model for the function \( f \), namely the so-called Sigmoid function or logistic model. We will consider this function as representing the probability for finding a value of \( y_i \) with a given \( x_i \).
Another widely studied model, is the so-called perceptron model, which is an example of a "hard classification" model. We will encounter this model when we discuss neural networks as well. Each datapoint is deterministically assigned to a category (i.e \( y_i=0 \) or \( y_i=1 \)). In many cases, and the coronary heart disease data forms one of many such examples, it is favorable to have a "soft" classifier that outputs the probability of a given category rather than a single value. For example, given \( x_i \), the classifier outputs the probability of being in a category \( k \). Logistic regression is the most common example of a so-called soft classifier. In logistic regression, the probability that a data point \( x_i \) belongs to a category \( y_i=\{0,1\} \) is given by the so-called logit function (or Sigmoid) which is meant to represent the likelihood for a given event, $$ p(t) = \frac{1}{1+\mathrm \exp{-t}}=\frac{\exp{t}}{1+\mathrm \exp{t}}. $$ Note that \( 1-p(t)= p(-t) \).
The following code plots the logistic function, the step function and other functions we will encounter from here and on.
"""The sigmoid function (or the logistic curve) is a
function that takes any real number, z, and outputs a number (0,1).
It is useful in neural networks for assigning weights on a relative scale.
The value z is the weighted sum of parameters involved in the learning algorithm."""
import numpy
import matplotlib.pyplot as plt
import math as mt
z = numpy.arange(-5, 5, .1)
sigma_fn = numpy.vectorize(lambda z: 1/(1+numpy.exp(-z)))
sigma = sigma_fn(z)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(z, sigma)
ax.set_ylim([-0.1, 1.1])
ax.set_xlim([-5,5])
ax.grid(True)
ax.set_xlabel('z')
ax.set_title('sigmoid function')
plt.show()
"""Step Function"""
z = numpy.arange(-5, 5, .02)
step_fn = numpy.vectorize(lambda z: 1.0 if z >= 0.0 else 0.0)
step = step_fn(z)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(z, step)
ax.set_ylim([-0.5, 1.5])
ax.set_xlim([-5,5])
ax.grid(True)
ax.set_xlabel('z')
ax.set_title('step function')
plt.show()
"""tanh Function"""
z = numpy.arange(-2*mt.pi, 2*mt.pi, 0.1)
t = numpy.tanh(z)
fig = plt.figure()
ax = fig.add_subplot(111)
ax.plot(z, t)
ax.set_ylim([-1.0, 1.0])
ax.set_xlim([-2*mt.pi,2*mt.pi])
ax.grid(True)
ax.set_xlabel('z')
ax.set_title('tanh function')
plt.show()
We assume now that we have two classes with \( y_i \) either \( 0 \) or \( 1 \). Furthermore we assume also that we have only two parameters \( \beta \) in our fitting of the Sigmoid function, that is we define probabilities $$ \begin{align*} p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\ p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}), \end{align*} $$ where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
Note that we used $$ p(y_i=0\vert x_i, \boldsymbol{\beta}) = 1-p(y_i=1\vert x_i, \boldsymbol{\beta}). $$
In order to define the total likelihood for all possible outcomes from a dataset \( \mathcal{D}=\{(y_i,x_i)\} \), with the binary labels \( y_i\in\{0,1\} \) and where the data points are drawn independently, we use the so-called Maximum Likelihood Estimation (MLE) principle. We aim thus at maximizing the probability of seeing the observed data. We can then approximate the likelihood in terms of the product of the individual probabilities of a specific outcome \( y_i \), that is $$ \begin{align*} P(\mathcal{D}|\boldsymbol{\beta})& = \prod_{i=1}^n \left[p(y_i=1|x_i,\boldsymbol{\beta})\right]^{y_i}\left[1-p(y_i=1|x_i,\boldsymbol{\beta}))\right]^{1-y_i}\nonumber \\ \end{align*} $$ from which we obtain the log-likelihood and our cost/loss function $$ \mathcal{C}(\boldsymbol{\beta}) = \sum_{i=1}^n \left( y_i\log{p(y_i=1|x_i,\boldsymbol{\beta})} + (1-y_i)\log\left[1-p(y_i=1|x_i,\boldsymbol{\beta}))\right]\right). $$
Reordering the logarithms, we can rewrite the cost/loss function as $$ \mathcal{C}(\boldsymbol{\beta}) = \sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right). $$
The maximum likelihood estimator is defined as the set of parameters that maximize the log-likelihood where we maximize with respect to \( \beta \). Since the cost (error) function is just the negative log-likelihood, for logistic regression we have that $$ \mathcal{C}(\boldsymbol{\beta})=-\sum_{i=1}^n \left(y_i(\beta_0+\beta_1x_i) -\log{(1+\exp{(\beta_0+\beta_1x_i)})}\right). $$ This equation is known in statistics as the cross entropy. Finally, we note that just as in linear regression, in practice we often supplement the cross-entropy with additional regularization terms, usually \( L_1 \) and \( L_2 \) regularization as we did for Ridge and Lasso regression.
The cross entropy is a convex function of the weights \( \boldsymbol{\beta} \) and, therefore, any local minimizer is a global minimizer.
Minimizing this cost function with respect to the two parameters \( \beta_0 \) and \( \beta_1 \) we obtain $$ \frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \beta_0} = -\sum_{i=1}^n \left(y_i -\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right), $$ and $$ \frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \beta_1} = -\sum_{i=1}^n \left(y_ix_i -x_i\frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}}\right). $$
Let us now define a vector \( \boldsymbol{y} \) with \( n \) elements \( y_i \), an \( n\times p \) matrix \( \boldsymbol{X} \) which contains the \( x_i \) values and a vector \( \boldsymbol{p} \) of fitted probabilities \( p(y_i\vert x_i,\boldsymbol{\beta}) \). We can rewrite in a more compact form the first derivative of cost function as $$ \frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right). $$
If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements \( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as $$ \frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}. $$
Within a binary classification problem, we can easily expand our model to include multiple predictors. Our ratio between likelihoods is then with \( p \) predictors $$ \log{ \frac{p(\boldsymbol{\beta}\boldsymbol{x})}{1-p(\boldsymbol{\beta}\boldsymbol{x})}} = \beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p. $$ Here we defined \( \boldsymbol{x}=[1,x_1,x_2,\dots,x_p] \) and \( \boldsymbol{\beta}=[\beta_0, \beta_1, \dots, \beta_p] \) leading to $$ p(\boldsymbol{\beta}\boldsymbol{x})=\frac{ \exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}{1+\exp{(\beta_0+\beta_1x_1+\beta_2x_2+\dots+\beta_px_p)}}. $$
Till now we have mainly focused on two classes, the so-called binary system. Suppose we wish to extend to \( K \) classes. Let us for the sake of simplicity assume we have only two predictors. We have then following model $$ \log{\frac{p(C=1\vert x)}{p(K\vert x)}} = \beta_{10}+\beta_{11}x_1, $$ and $$ \log{\frac{p(C=2\vert x)}{p(K\vert x)}} = \beta_{20}+\beta_{21}x_1, $$ and so on till the class \( C=K-1 \) class $$ \log{\frac{p(C=K-1\vert x)}{p(K\vert x)}} = \beta_{(K-1)0}+\beta_{(K-1)1}x_1, $$
and the model is specified in term of \( K-1 \) so-called log-odds or logit transformations.
In our discussion of neural networks we will encounter the above again in terms of a slightly modified function, the so-called Softmax function.
The softmax function is used in various multiclass classification methods, such as multinomial logistic regression (also known as softmax regression), multiclass linear discriminant analysis, naive Bayes classifiers, and artificial neural networks. Specifically, in multinomial logistic regression and linear discriminant analysis, the input to the function is the result of \( K \) distinct linear functions, and the predicted probability for the \( k \)-th class given a sample vector \( \boldsymbol{x} \) and a weighting vector \( \boldsymbol{\beta} \) is (with two predictors): $$ p(C=k\vert \mathbf {x} )=\frac{\exp{(\beta_{k0}+\beta_{k1}x_1)}}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}. $$ It is easy to extend to more predictors. The final class is $$ p(C=K\vert \mathbf {x} )=\frac{1}{1+\sum_{l=1}^{K-1}\exp{(\beta_{l0}+\beta_{l1}x_1)}}, $$
and they sum to one. Our earlier discussions were all specialized to the case with two classes only. It is easy to see from the above that what we derived earlier is compatible with these equations.
To find the optimal parameters we would typically use a gradient descent method. Newton's method and gradient descent methods are discussed in the material on optimization methods.
We show here how we can use a simple regression case on the breast cancer data using Logistic regression as our algorithm for classification.
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
# Load the data
cancer = load_breast_cancer()
X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
print(X_train.shape)
print(X_test.shape)
# Logistic Regression
logreg = LogisticRegression(solver='lbfgs')
logreg.fit(X_train, y_train)
print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
#now scale the data
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
# Logistic Regression
logreg.fit(X_train_scaled, y_train)
print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
In addition to the above scores, we could also study the covariance (and the correlation matrix). We use Pandas to compute the correlation matrix.
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
cancer = load_breast_cancer()
import pandas as pd
# Making a data frame
cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
fig, axes = plt.subplots(15,2,figsize=(10,20))
malignant = cancer.data[cancer.target == 0]
benign = cancer.data[cancer.target == 1]
ax = axes.ravel()
for i in range(30):
_, bins = np.histogram(cancer.data[:,i], bins =50)
ax[i].hist(malignant[:,i], bins = bins, alpha = 0.5)
ax[i].hist(benign[:,i], bins = bins, alpha = 0.5)
ax[i].set_title(cancer.feature_names[i])
ax[i].set_yticks(())
ax[0].set_xlabel("Feature magnitude")
ax[0].set_ylabel("Frequency")
ax[0].legend(["Malignant", "Benign"], loc ="best")
fig.tight_layout()
plt.show()
import seaborn as sns
correlation_matrix = cancerpd.corr().round(1)
# use the heatmap function from seaborn to plot the correlation matrix
# annot = True to print the values inside the square
plt.figure(figsize=(15,8))
sns.heatmap(data=correlation_matrix, annot=True)
plt.show()
In the above example we note two things. In the first plot we display the overlap of benign and malignant tumors as functions of the various features in the Wisconsing breast cancer data set. We see that for some of the features we can distinguish clearly the benign and malignant cases while for other features we cannot. This can point to us which features may be of greater interest when we wish to classify a benign or not benign tumour.
In the second figure we have computed the so-called correlation matrix, which in our case with thirty features becomes a \( 30\times 30 \) matrix.
We constructed this matrix using pandas via the statements
cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)
and then
correlation_matrix = cancerpd.corr().round(1)
Diagonalizing this matrix we can in turn say something about which features are of relevance and which are not. This leads us to the classical Principal Component Analysis (PCA) theorem with applications. This will be discussed later this semester (week 43).
import matplotlib.pyplot as plt
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.datasets import load_breast_cancer
from sklearn.linear_model import LogisticRegression
# Load the data
cancer = load_breast_cancer()
X_train, X_test, y_train, y_test = train_test_split(cancer.data,cancer.target,random_state=0)
print(X_train.shape)
print(X_test.shape)
# Logistic Regression
logreg = LogisticRegression(solver='lbfgs')
logreg.fit(X_train, y_train)
print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
#now scale the data
from sklearn.preprocessing import StandardScaler
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
# Logistic Regression
logreg.fit(X_train_scaled, y_train)
print("Test set accuracy Logistic Regression with scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
from sklearn.preprocessing import LabelEncoder
from sklearn.model_selection import cross_validate
#Cross validation
accuracy = cross_validate(logreg,X_test_scaled,y_test,cv=10)['test_score']
print(accuracy)
print("Test set accuracy with Logistic Regression and scaled data: {:.2f}".format(logreg.score(X_test_scaled,y_test)))
import scikitplot as skplt
y_pred = logreg.predict(X_test_scaled)
skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)
plt.show()
y_probas = logreg.predict_proba(X_test_scaled)
skplt.metrics.plot_roc(y_test, y_probas)
plt.show()
skplt.metrics.plot_cumulative_gain(y_test, y_probas)
plt.show()
Overview Video, why do we care about gradient methods?
Almost every problem in machine learning and data science starts with a dataset \( X \), a model \( g(\beta) \), which is a function of the parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows us to judge how well the model \( g(\beta) \) explains the observations \( X \). The model is fit by finding the values of \( \beta \) that minimize the cost function. Ideally we would be able to solve for \( \beta \) analytically, however this is not possible in general and we must use some approximative/numerical method to compute the minimum.
In our discussion on Logistic Regression we studied the case of two classes, with \( y_i \) either \( 0 \) or \( 1 \). Furthermore we assumed also that we have only two parameters \( \beta \) in our fitting, that is we defined probabilities $$ \begin{align*} p(y_i=1|x_i,\boldsymbol{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\ p(y_i=0|x_i,\boldsymbol{\beta}) &= 1 - p(y_i=1|x_i,\boldsymbol{\beta}), \end{align*} $$ where \( \boldsymbol{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
Our compact equations used a definition of a vector \( \boldsymbol{y} \) with \( n \) elements \( y_i \), an \( n\times p \) matrix \( \boldsymbol{X} \) which contains the \( x_i \) values and a vector \( \boldsymbol{p} \) of fitted probabilities \( p(y_i\vert x_i,\boldsymbol{\beta}) \). We rewrote in a more compact form the first derivative of the cost function as $$ \frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = -\boldsymbol{X}^T\left(\boldsymbol{y}-\boldsymbol{p}\right). $$
If we in addition define a diagonal matrix \( \boldsymbol{W} \) with elements \( p(y_i\vert x_i,\boldsymbol{\beta})(1-p(y_i\vert x_i,\boldsymbol{\beta}) \), we can obtain a compact expression of the second derivative as $$ \frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T} = \boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X}. $$ This defines what is called the Hessian matrix.
If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
Our iterative scheme is then given by $$ \boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}\partial \boldsymbol{\beta}^T}\right)^{-1}_{\boldsymbol{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}}\right)_{\boldsymbol{\beta}^{\mathrm{old}}}, $$ or in matrix form as $$ \boldsymbol{\beta}^{\mathrm{new}} = \boldsymbol{\beta}^{\mathrm{old}}-\left(\boldsymbol{X}^T\boldsymbol{W}\boldsymbol{X} \right)^{-1}\times \left(-\boldsymbol{X}^T(\boldsymbol{y}-\boldsymbol{p}) \right)_{\boldsymbol{\beta}^{\mathrm{old}}}. $$ The right-hand side is computed with the old values of \( \beta \).
If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
Let us quickly remind ourselves how we derive the above method.
Perhaps the most celebrated of all one-dimensional root-finding routines is Newton's method, also called the Newton-Raphson method. This method requires the evaluation of both the function \( f \) and its derivative \( f' \) at arbitrary points. If you can only calculate the derivative numerically and/or your function is not of the smooth type, we normally discourage the use of this method.
The Newton-Raphson formula consists geometrically of extending the tangent line at a current point until it crosses zero, then setting the next guess to the abscissa of that zero-crossing. The mathematics behind this method is rather simple. Employing a Taylor expansion for \( x \) sufficiently close to the solution \( s \), we have $$ f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots. \label{eq:taylornr} $$
For small enough values of the function and for well-behaved functions, the terms beyond linear are unimportant, hence we obtain $$ f(x)+(s-x)f'(x)\approx 0, $$ yielding $$ s\approx x-\frac{f(x)}{f'(x)}. $$
Having in mind an iterative procedure, it is natural to start iterating with $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. $$
The above is Newton-Raphson's method. It has a simple geometric interpretation, namely \( x_{n+1} \) is the point where the tangent from \( (x_n,f(x_n)) \) crosses the \( x \)-axis. Close to the solution, Newton-Raphson converges fast to the desired result. However, if we are far from a root, where the higher-order terms in the series are important, the Newton-Raphson formula can give grossly inaccurate results. For instance, the initial guess for the root might be so far from the true root as to let the search interval include a local maximum or minimum of the function. If an iteration places a trial guess near such a local extremum, so that the first derivative nearly vanishes, then Newton-Raphson may fail totally
Newton's method can be generalized to systems of several non-linear equations and variables. Consider the case with two equations $$ \begin{array}{cc} f_1(x_1,x_2) &=0\\ f_2(x_1,x_2) &=0,\end{array} $$ which we Taylor expand to obtain $$ \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1 \partial f_1/\partial x_1+h_2 \partial f_1/\partial x_2+\dots\\ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1 \partial f_2/\partial x_1+h_2 \partial f_2/\partial x_2+\dots \end{array}. $$ Defining the Jacobian matrix \( {\bf \boldsymbol{J}} \) we have $$ {\bf \boldsymbol{J}}=\left( \begin{array}{cc} \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\ \partial f_2/\partial x_1 &\partial f_2/\partial x_2 \end{array} \right), $$ we can rephrase Newton's method as $$ \left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)= \left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right), $$ where we have defined $$ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)= -{\bf \boldsymbol{J}}^{-1} \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right). $$ We need thus to compute the inverse of the Jacobian matrix and it is to understand that difficulties may arise in case \( {\bf \boldsymbol{J}} \) is nearly singular.
It is rather straightforward to extend the above scheme to systems of more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
The basic idea of gradient descent is that a function \( F(\mathbf{x}) \), \( \mathbf{x} \equiv (x_1,\cdots,x_n) \), decreases fastest if one goes from \( \bf {x} \) in the direction of the negative gradient \( -\nabla F(\mathbf{x}) \).
It can be shown that if $$ \mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), $$ with \( \gamma_k > 0 \).
For \( \gamma_k \) small enough, then \( F(\mathbf{x}_{k+1}) \leq F(\mathbf{x}_k) \). This means that for a sufficiently small \( \gamma_k \) we are always moving towards smaller function values, i.e a minimum.
The previous observation is the basis of the method of steepest descent, which is also referred to as just gradient descent (GD). One starts with an initial guess \( \mathbf{x}_0 \) for a minimum of \( F \) and computes new approximations according to $$ \mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0. $$
The parameter \( \gamma_k \) is often referred to as the step length or the learning rate within the context of Machine Learning.
Ideally the sequence \( \{\mathbf{x}_k \}_{k=0} \) converges to a global minimum of the function \( F \). In general we do not know if we are in a global or local minimum. In the special case when \( F \) is a convex function, all local minima are also global minima, so in this case gradient descent can converge to the global solution. The advantage of this scheme is that it is conceptually simple and straightforward to implement. However the method in this form has some severe limitations:
In machine learing we are often faced with non-convex high dimensional cost functions with many local minima. Since GD is deterministic we will get stuck in a local minimum, if the method converges, unless we have a very good intial guess. This also implies that the scheme is sensitive to the chosen initial condition.
Note that the gradient is a function of \( \mathbf{x} = (x_1,\cdots,x_n) \) which makes it expensive to compute numerically.
The gradient descent method is sensitive to the choice of learning rate \( \gamma_k \). This is due to the fact that we are only guaranteed that \( F(\mathbf{x}_{k+1}) \leq F(\mathbf{x}_k) \) for sufficiently small \( \gamma_k \). The problem is to determine an optimal learning rate. If the learning rate is chosen too small the method will take a long time to converge and if it is too large we can experience erratic behavior.
Many of these shortcomings can be alleviated by introducing randomness. One such method is that of Stochastic Gradient Descent (SGD), see below.


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Ruby's Marshal module allows serialization and deserialization of many standard and arbitrary Ruby objects in a compact binary format. It is relatively fast, available in stdlibs standard and allows conservation of language-specific properties (such as symbols or encoding-aware strings).
Feature-wise, it is comparable to other language-specific implementations, such as:
From internal perspective, serialized stream consists of a simple magic header and a record.
This page hosts a formal specification of ruby_marshal using Kaitai Struct. This specification can be automatically translated into a variety of programming languages to get a parsing library.
meta:
id: ruby_marshal
license: CC0-1.0
endian: le
doc: |
Ruby's Marshal module allows serialization and deserialization of
many standard and arbitrary Ruby objects in a compact binary
format. It is relatively fast, available in stdlibs standard and
allows conservation of language-specific properties (such as symbols
or encoding-aware strings).
Feature-wise, it is comparable to other language-specific
implementations, such as:
* Java's
[Serializable](https://docs.oracle.com/javase/8/docs/api/java/io/Serializable.html)
* .NET
[BinaryFormatter](https://learn.microsoft.com/en-us/dotnet/api/system.runtime.serialization.formatters.binary.binaryformatter?view=net-7.0)
* Python's
[marshal](https://docs.python.org/3/library/marshal.html),
[pickle](https://docs.python.org/3/library/pickle.html) and
[shelve](https://docs.python.org/3/library/shelve.html)
From internal perspective, serialized stream consists of a simple
magic header and a record.
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Stream+Format'
seq:
- id: version
contents: [4, 8]
- id: records
type: record
types:
record:
doc: |
Each record starts with a single byte that determines its type
(`code`) and contents. If necessary, additional info as parsed
as `body`, to be determined by `code`.
seq:
- id: code
type: u1
enum: codes
- id: body
type:
switch-on: code
cases:
'codes::bignum': bignum
'codes::instance_var': instance_var
'codes::packed_int': packed_int
'codes::ruby_array': ruby_array
'codes::ruby_hash': ruby_hash
'codes::ruby_string': ruby_string
'codes::ruby_struct': ruby_struct
'codes::ruby_symbol': ruby_symbol
'codes::ruby_symbol_link': packed_int
'codes::ruby_object_link': packed_int
packed_int:
doc: |
Ruby uses sophisticated system to pack integers: first `code`
byte either determines packing scheme or carries encoded
immediate value (thus allowing smaller values from -123 to 122
(inclusive) to take only one byte. There are 11 encoding schemes
in total:
* 0 is encoded specially (as 0)
* 1..122 are encoded as immediate value with a shift
* 123..255 are encoded with code of 0x01 and 1 extra byte
* 0x100..0xffff are encoded with code of 0x02 and 2 extra bytes
* 0x10000..0xffffff are encoded with code of 0x03 and 3 extra
bytes
* 0x1000000..0xffffffff are encoded with code of 0x04 and 4
extra bytes
* -123..-1 are encoded as immediate value with another shift
* -256..-124 are encoded with code of 0xff and 1 extra byte
* -0x10000..-257 are encoded with code of 0xfe and 2 extra bytes
* -0x1000000..0x10001 are encoded with code of 0xfd and 3 extra
bytes
* -0x40000000..-0x1000001 are encoded with code of 0xfc and 4
extra bytes
Values beyond that are serialized as bignum (even if they
technically might be not Bignum class in Ruby implementation,
i.e. if they fit into 64 bits on a 64-bit platform).
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Fixnum+and+long'
seq:
- id: code
type: u1
- id: encoded
type:
switch-on: code
cases:
# 0x00: none
0x01: u1
0x02: u2
0x03: u2
0x04: u4
0xff: u1
0xfe: u2
0xfd: u2
0xfc: u4
- id: encoded2
type:
switch-on: code
cases:
0x03: u1
0xfd: u1
doc: |
One extra byte for 3-byte integers (0x03 and 0xfd), as
there is no standard `u3` type in KS.
instances:
is_immediate:
value: code > 4 and code < 0xfc
value:
value: >
is_immediate ? (code < 0x80 ? code - 5 : (4 - (~code & 0x7f))) :
code == 0 ? 0 :
code == 0xff ? (encoded - 0x100) :
code == 0xfe ? (encoded - 0x10000) :
code == 0xfd ? ((encoded2 << 16 | encoded) - 0x1000000) :
code == 0x03 ? (encoded2 << 16 | encoded) :
encoded
ruby_symbol:
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Symbols+and+Byte+Sequence'
seq:
- id: len
type: packed_int
- id: name
size: len.value
type: str
encoding: UTF-8
ruby_string:
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-String'
seq:
- id: len
type: packed_int
- id: body
size: len.value
ruby_array:
seq:
- id: num_elements
type: packed_int
- id: elements
type: record
repeat: expr
repeat-expr: num_elements.value
ruby_hash:
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Hash+and+Hash+with+Default+Value'
seq:
- id: num_pairs
type: packed_int
- id: pairs
type: pair
repeat: expr
repeat-expr: num_pairs.value
bignum:
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Bignum'
seq:
- id: sign
type: u1
doc: A single byte containing `+` for a positive value or `-` for a negative value.
- id: len_div_2
type: packed_int
doc: Length of bignum body, divided by 2.
- id: body
size: len_div_2.value * 2
doc: Bytes that represent the number, see ruby-lang.org docs for reconstruction algorithm.
ruby_struct:
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Struct'
seq:
- id: name
type: record
doc: Symbol containing the name of the struct.
- id: num_members
type: packed_int
doc: Number of members in a struct
- id: members
type: pair
repeat: expr
repeat-expr: num_members.value
instance_var:
doc-ref: 'https://docs.ruby-lang.org/en/2.4.0/marshal_rdoc.html#label-Instance+Variables'
seq:
- id: obj
type: record
- id: num_vars
type: packed_int
- id: vars
type: pair
repeat: expr
repeat-expr: num_vars.value
pair:
seq:
- id: key
type: record
- id: value
type: record
enums:
codes:
0x22: ruby_string
0x30: const_nil
0x3a: ruby_symbol
0x3b: ruby_symbol_link
0x40: ruby_object_link
0x46: const_false
0x49: instance_var
0x53: ruby_struct
0x54: const_true
0x5b: ruby_array
0x69: packed_int
0x6c: bignum
0x7b: ruby_hash


igx-date-picker component.
You can read more about configuring the igx-date-picker component in the README or the official documentation.


Lyrics of Sunshine and Shadow
by Paul Laurence Dunbar
At Night
Additional Information
- Year Published: 1913
- Language: English
- Country of Origin: United States of America
- Source: Dunbar, P.L. (1913). The Complete Poems of Paul Laurence Dunbar. New York: Dodd, Mead, and Company.
-
Readability:
- Flesch–Kincaid Level: 6.5
- Word Count: 467
- Genre: Poetry
- Keywords: night, sleep, tired
- ✎ Cite This
Downloads
Whut time ‘d dat clock strike?
Nine? No—eight;
I didn’t think hit was so late.
Aer chew! I must ‘a’ got a cough,
I raally b’lieve I did doze off—
Hit’s mighty soothin’ to de tiah,
A–dozin’ dis way by de fiah;
Oo oom—hit feels so good to stretch
I sutny is one weary wretch!

He des ain’t wo’th his boa’d an’ keep;
I des don’t b’lieve he’d bat his eyes
If Gab’el called him fo’m de skies!
But sleepin’s good dey ain’t no doubt—
Dis pipe o’ mine is done gone out.
Don’t bu’n a minute, bless my soul,
Des please to han’ me dat ah coal.
You ‘Lias git up now, my son,
Seems lak my nap is des begun;
You sutny mus’ ma’k down de day
Wen I treats comp’ny dis away!
W’y, Brother Jones, dat drowse come on,
An’ laws! I dremp dat you was gone!
You ‘Lias, whaih yo’ mannahs, suh,
To hyeah me call an’ nevah stuh!

Dis wood fiah is invitin’ dho’,
Hit seems to wa’m de ve’y flo’—
An’ nuffin’ ain’t a whit ez sweet,
Ez settin’ toastin’ of yo’ feet.
Hit mek you drowsy, too, but La!
Hyeah, ‘Lias, don’t you hyeah yo’ ma?
Ef I gits sta’ted f’om dis cheah
I’ lay, you scamp, I’ll mek you heah!
To–morrer mo’nin’ I kin bawl
Twell all de neighbohs hyeah me call;
An’ you’ll be snoozin’ des ez deep
Ez if de day was made fu’ sleep;
Hit’s funny when you got a cough
Somehow yo’ voice seems too fu’ off—
Can’t wake dat boy fu’ all I say,
I reckon he’ll sleep daih twell day!


@(products: List[models.Product], categories: List[models.Category], user: models.users.User) @main("Items") {
Info on Products
Eye Makeup Products include products that are used around the eye to enhance the appearance of the eyes and to emphazise the beauty of the eyes. They include such products as eye-shadow, eye-liner, eye-brow products, and other products that can help to enhance and accent the eyes.
Eye Makeup Products include products that are used around the eye to enhance the appearance of the eyes and to emphazise the beauty of the eyes. They include such products as eye-shadow, eye-liner, eye-brow products, and other products that can help to enhance and accent the eyes.
Eye Makeup Products include products that are used around the eye to enhance the appearance of the eyes and to emphazise the beauty of the eyes. They include such products as eye-shadow, eye-liner, eye-brow products, and other products that can help to enhance and accent the eyes.