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 things 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 may 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 solutions (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 deafult. This means that if you compare your code with the results from a given library, the results may differ. Tracing back the differences may often lead to an increased confusion.
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, here 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
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.
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
np.random.seed(2021)
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
clf = LinearRegression(fit_intercept=False).fit(X, y)
print(f"True beta: {true_beta}")
print(f"Fitted beta: {beta}")
print(f"Sklearn fitted beta: {clf.coef_}")
plt.figure()
plt.scatter(x, y, label="Data")
plt.plot(x, X @ beta, label="Fit")
plt.plot(x, clf.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
clf = 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 (sans intercept): {beta}")
print(f"Sklearn intercept: {clf.intercept_}")
print(f"Sklearn fitted beta (sans intercept): {clf.coef_}")
plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
plt.plot(x, clf.predict(X), "--", label="Sklearn (fit_intercept=True)")
plt.grid()
plt.legend()
plt.show()
Let us try to understand what this may imply mathematically when we subtract the mean values, also known as zero centering. To catch many birds with just one stone, we will focus on Ridge regression.
The cost/loss function for Ridge regression is $$ C(\beta_0, \beta_1, ... , \beta_{p-1}) = \sum_{i=0}^{n} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij}\beta_j\right)^2 + \lambda \sum_{j=1}^{p-1} \beta_i^2. $$
Note that the intercept term $\beta_0$is left out of the \( L_2 \) regularization term. The design matrix \( X \) does in this case not contain any intercept column. We want $$ \frac{\partial C}{\partial \beta_j} = 0, $$
for all \( j \), so let us start with \( \beta_0 \). This means that we have $$ \frac{\partial C}{\partial \beta_0} = -2\sum_{i=0}^{n-1} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij} \beta_j\right), $$
which gives $$ \sum_{i=0}^{n-1} \beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} \sum_{j=1}^{p-1} X_{ij} \beta_j. $$
If we assume that every column of \( \boldsymbol{X} \) is centered, which we can do by subtracting the mean,
X = X - np.mean(X,axis=0)
the sum \( \sum_{i=0}^{n-1} X_{ij} \) can be rewritten as $$ \sum_{i=0}^{n-1} \left(X_{ij} - \frac{1}{n}\sum_{i=0}^{n-1} X_{ij}\right) = \sum_{i=0}^{n-1} X_{ij} - \sum_{i=0}^{n-1} \frac{1}{n} \sum_{i=0}^{n-1}X_{ij}, $$ resulting in $$ \sum_{i=0}^{n-1} X_{ij} - n \frac{1}{n} \sum_{i=0}^{n-1}X_{ij} = 0. $$
Finally we have $$ n\beta_0 = \sum_{i=0}^{n-1} y_i - \sum_{j=1}^{p-1}\beta_j \sum_{i=0}^{n-1} X_{ij}, $$ or $$ \beta_0 = \frac{1}{n}\sum_{i=0}^{n-1} y_i = \overline{\boldsymbol{y}}, $$ the average value of \( \boldsymbol{y} \).
Replacing \( y_i \) with \( y_i - \beta_0 = y_i - \overline{\boldsymbol{y}} \) in the cost function will give us (in vector-matrix disguise) $$ C(\boldsymbol{\beta}) = (\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta})^T(\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta}) + \lambda \boldsymbol{\beta}^T\boldsymbol{\beta}, $$
which has the solution
\( \beta = (\tilde{X}^T\tilde{X} + \lambda I)^{-1}\tilde{X}^T\boldsymbol{\tilde{y}} \). where \( \boldsymbol{\tilde{y}} = \boldsymbol{y} - \overline{\boldsymbol{y}} \) and \( \tilde{X}_{ij} = X_{ij} - \frac{1}{n}\sum_{k=1}^{n-1}X_{kj} \).
Armed with this wisdom, we attempt first simply set the intercept eqault to False in our implementation of Ridge regression for a 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
X[:,0] = 1.0
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 = 4
MSEOwnRidgePredict = np.zeros(nlambdas)
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 4, 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
# include lasso using Scikit-Learn
# 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_)
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'r', label = 'MSE 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. The problem however is that can easily lead to a larger mean-squared error!
Let us see how we can change this code by zero centering.
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 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 = 4
MSEOwnRidgePredict = np.zeros(nlambdas)
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 1, 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 @ OwnRidgeBeta + intercept_
#EQUIVALENT PREDICTION:
#Add intercept to prediction
ypredictOwnRidge = X_test_scaled @ OwnRidgeBeta + y_scaler
print("Values for own Ridge prediction")
print(ypredictOwnRidge)
RegRidge = linear_model.Ridge(lmb)
RegRidge.fit(X_train,y_train)
ypredictRidge = RegRidge.predict(X_test)
print("Values for SL Ridge prediction")
print(ypredictRidge)
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_)
# 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()
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.
Ideally we want our cost/loss function to be convex(concave).
First we give the definition of a convex set: A set \( C \) in \( \mathbb{R}^n \) is said to be convex if, for all \( x \) and \( y \) in \( C \) and all \( t \in (0,1) \) , the point \( (1 − t)x + ty \) also belongs to C. Geometrically this means that every point on the line segment connecting \( x \) and \( y \) is in \( C \) as discussed below.
The convex subsets of \( \mathbb{R} \) are the intervals of \( \mathbb{R} \). Examples of convex sets of \( \mathbb{R}^2 \) are the regular polygons (triangles, rectangles, pentagons, etc...).
Convex function: Let \( X \subset \mathbb{R}^n \) be a convex set. Assume that the function \( f: X \rightarrow \mathbb{R} \) is continuous, then \( f \) is said to be convex if $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$ for all \( x_1, x_2 \in X \) and for all \( t \in [0,1] \). If \( \leq \) is replaced with a strict inequaltiy in the definition, we demand \( x_1 \neq x_2 \) and \( t\in(0,1) \) then \( f \) is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting \( f(x_1) \) and \( f(x_2) \), the value of the function on the interval \( [x_1,x_2] \) is always below the line as illustrated below.
In the following we state first and second-order conditions which ensures convexity of a function \( f \). We write \( D_f \) to denote the domain of \( f \), i.e the subset of \( R^n \) where \( f \) is defined. For more details and proofs we refer to: S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press.
Suppose \( f \) is differentiable (i.e \( \nabla f(x) \) is well defined for all \( x \) in the domain of \( f \)). Then \( f \) is convex if and only if \( D_f \) is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holds for all \( x,y \in D_f \). This condition means that for a convex function the first order Taylor expansion (right hand side above) at any point a global under estimator of the function. To convince yourself you can make a drawing of \( f(x) = x^2+1 \) and draw the tangent line to \( f(x) \) and note that it is always below the graph.
Assume that \( f \) is twice differentiable, i.e the Hessian matrix exists at each point in \( D_f \). Then \( f \) is convex if and only if \( D_f \) is a convex set and its Hessian is positive semi-definite for all \( x\in D_f \). For a single-variable function this reduces to \( f''(x) \geq 0 \). Geometrically this means that \( f \) has nonnegative curvature everywhere.
This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
The next result is of great importance to us and the reason why we are going on about convex functions. In machine learning we frequently have to minimize a loss/cost function in order to find the best parameters for the model we are considering.
Ideally we want the global minimum (for high-dimensional models it is hard to know if we have local or global minimum). However, if the cost/loss function is convex the following result provides invaluable information:
Consider the problem of finding \( x \in \mathbb{R}^n \) such that \( f(x) \) is minimal, where \( f \) is convex and differentiable. Then, any point \( x^* \) that satisfies \( \nabla f(x^*) = 0 \) is a global minimum.
This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
Video of Lecture and link to handwritten notes.
Before we proceed, we would like to discuss the approach called the standard Steepest descent (different from the above steepest descent discussion), which again leads to us having to be able to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
The success of the CG method for finding solutions of non-linear problems is based on the theory of conjugate gradients for linear systems of equations. It belongs to the class of iterative methods for solving problems from linear algebra of the type $$ \begin{equation*} \boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}. \end{equation*} $$
In the iterative process we end up with a problem like $$ \begin{equation*} \boldsymbol{r}= \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}, \end{equation*} $$ where \( \boldsymbol{r} \) is the so-called residual or error in the iterative process.
When we have found the exact solution, \( \boldsymbol{r}=0 \).
The residual is zero when we reach the minimum of the quadratic equation $$ \begin{equation*} P(\boldsymbol{x})=\frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T\boldsymbol{b}, \end{equation*} $$
with the constraint that the matrix \( \boldsymbol{A} \) is positive definite and symmetric. This defines also the Hessian and we want it to be positive definite.
We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \). We can assume without loss of generality that $$ \begin{equation*} \boldsymbol{x}_0=0, \end{equation*} $$ or consider the system $$ \begin{equation*} \boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0, \end{equation*} $$ instead.
One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form $$ \begin{equation*} f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n. \end{equation*} $$ This suggests taking the first basis vector \( \boldsymbol{r}_1 \) (see below for definition) to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \), which equals $$ \begin{equation*} \boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b}, \end{equation*} $$ and \( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \).
We can compute the residual iteratively as $$ \begin{equation*} \boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1}, \end{equation*} $$ which equals $$ \begin{equation*} \boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{r}_k), \end{equation*} $$ or $$ \begin{equation*} (\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{r}_k, \end{equation*} $$ which gives $$ \alpha_k = \frac{\boldsymbol{r}_k^T\boldsymbol{r}_k}{\boldsymbol{r}_k^T\boldsymbol{A}\boldsymbol{r}_k} $$ leading to the iterative scheme $$ \begin{equation*} \boldsymbol{x}_{k+1}=\boldsymbol{x}_k-\alpha_k\boldsymbol{r}_{k}, \end{equation*} $$
import numpy as np
import numpy.linalg as la
import scipy.optimize as sopt
import matplotlib.pyplot as pt
from mpl_toolkits.mplot3d import axes3d
def f(x):
return 0.5*x[0]**2 + 2.5*x[1]**2
def df(x):
return np.array([x[0], 5*x[1]])
fig = pt.figure()
ax = fig.gca(projection="3d")
xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
fmesh = f(np.array([xmesh, ymesh]))
ax.plot_surface(xmesh, ymesh, fmesh)
And then as countor plot
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh)
guesses = [np.array([2, 2./5])]
Find guesses
x = guesses[-1]
s = -df(x)
Run it!
def f1d(alpha):
return f(x + alpha*s)
alpha_opt = sopt.golden(f1d)
next_guess = x + alpha_opt * s
guesses.append(next_guess)
print(next_guess)
What happened?
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
it_array = np.array(guesses)
pt.plot(it_array.T[0], it_array.T[1], "x-")
In the CG method we define so-called conjugate directions and two vectors \( \boldsymbol{s} \) and \( \boldsymbol{t} \) are said to be conjugate if $$ \begin{equation*} \boldsymbol{s}^T\boldsymbol{A}\boldsymbol{t}= 0. \end{equation*} $$ The philosophy of the CG method is to perform searches in various conjugate directions of our vectors \( \boldsymbol{x}_i \) obeying the above criterion, namely $$ \begin{equation*} \boldsymbol{x}_i^T\boldsymbol{A}\boldsymbol{x}_j= 0. \end{equation*} $$ Two vectors are conjugate if they are orthogonal with respect to this inner product. Being conjugate is a symmetric relation: if \( \boldsymbol{s} \) is conjugate to \( \boldsymbol{t} \), then \( \boldsymbol{t} \) is conjugate to \( \boldsymbol{s} \).
An example is given by the eigenvectors of the matrix $$ \begin{equation*} \boldsymbol{v}_i^T\boldsymbol{A}\boldsymbol{v}_j= \lambda\boldsymbol{v}_i^T\boldsymbol{v}_j, \end{equation*} $$ which is zero unless \( i=j \).
Assume now that we have a symmetric positive-definite matrix \( \boldsymbol{A} \) of size \( n\times n \). At each iteration \( i+1 \) we obtain the conjugate direction of a vector $$ \begin{equation*} \boldsymbol{x}_{i+1}=\boldsymbol{x}_{i}+\alpha_i\boldsymbol{p}_{i}. \end{equation*} $$ We assume that \( \boldsymbol{p}_{i} \) is a sequence of \( n \) mutually conjugate directions. Then the \( \boldsymbol{p}_{i} \) form a basis of \( R^n \) and we can expand the solution $ \boldsymbol{A}\boldsymbol{x} = \boldsymbol{b}$ in this basis, namely $$ \begin{equation*} \boldsymbol{x} = \sum^{n}_{i=1} \alpha_i \boldsymbol{p}_i. \end{equation*} $$
The coefficients are given by $$ \begin{equation*} \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}. \end{equation*} $$ Multiplying with \( \boldsymbol{p}_k^T \) from the left gives $$ \begin{equation*} \boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{x} = \sum^{n}_{i=1} \alpha_i\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{p}_i= \boldsymbol{p}_k^T \boldsymbol{b}, \end{equation*} $$ and we can define the coefficients \( \alpha_k \) as $$ \begin{equation*} \alpha_k = \frac{\boldsymbol{p}_k^T \boldsymbol{b}}{\boldsymbol{p}_k^T \boldsymbol{A} \boldsymbol{p}_k} \end{equation*} $$
If we choose the conjugate vectors \( \boldsymbol{p}_k \) carefully, then we may not need all of them to obtain a good approximation to the solution \( \boldsymbol{x} \). We want to regard the conjugate gradient method as an iterative method. This will us to solve systems where \( n \) is so large that the direct method would take too much time.
We denote the initial guess for \( \boldsymbol{x} \) as \( \boldsymbol{x}_0 \). We can assume without loss of generality that $$ \begin{equation*} \boldsymbol{x}_0=0, \end{equation*} $$ or consider the system $$ \begin{equation*} \boldsymbol{A}\boldsymbol{z} = \boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_0, \end{equation*} $$ instead.
One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form $$ \begin{equation*} f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n. \end{equation*} $$ This suggests taking the first basis vector \( \boldsymbol{p}_1 \) to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \), which equals $$ \begin{equation*} \boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b}, \end{equation*} $$ and \( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \). The other vectors in the basis will be conjugate to the gradient, hence the name conjugate gradient method.
Let \( \boldsymbol{r}_k \) be the residual at the \( k \)-th step: $$ \begin{equation*} \boldsymbol{r}_k=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k. \end{equation*} $$ Note that \( \boldsymbol{r}_k \) is the negative gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_k \), so the gradient descent method would be to move in the direction \( \boldsymbol{r}_k \). Here, we insist that the directions \( \boldsymbol{p}_k \) are conjugate to each other, so we take the direction closest to the gradient \( \boldsymbol{r}_k \) under the conjugacy constraint. This gives the following expression $$ \begin{equation*} \boldsymbol{p}_{k+1}=\boldsymbol{r}_k-\frac{\boldsymbol{p}_k^T \boldsymbol{A}\boldsymbol{r}_k}{\boldsymbol{p}_k^T\boldsymbol{A}\boldsymbol{p}_k} \boldsymbol{p}_k. \end{equation*} $$
We can also compute the residual iteratively as $$ \begin{equation*} \boldsymbol{r}_{k+1}=\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_{k+1}, \end{equation*} $$ which equals $$ \begin{equation*} \boldsymbol{b}-\boldsymbol{A}(\boldsymbol{x}_k+\alpha_k\boldsymbol{p}_k), \end{equation*} $$ or $$ \begin{equation*} (\boldsymbol{b}-\boldsymbol{A}\boldsymbol{x}_k)-\alpha_k\boldsymbol{A}\boldsymbol{p}_k, \end{equation*} $$ which gives $$ \begin{equation*} \boldsymbol{r}_{k+1}=\boldsymbol{r}_k-\boldsymbol{A}\boldsymbol{p}_{k}, \end{equation*} $$
We will use linear regression as a case study for the gradient descent methods. Linear regression is a great test case for the gradient descent methods discussed in the lectures since it has several desirable properties such as:
x = 2*np.random.rand(m,1)
y = 4+3*x+np.random.randn(m,1)
with \( x_i \in [0,1] \) is chosen randomly using a uniform distribution. Additionally we have a stochastic noise chosen according to a normal distribution \( \cal {N}(0,1) \). The linear regression model is given by $$ h_\beta(x) = \boldsymbol{y} = \beta_0 + \beta_1 x, $$ such that $$ \boldsymbol{y}_i = \beta_0 + \beta_1 x_i. $$
Let \( \mathbf{y} = (y_1,\cdots,y_n)^T \), \( \mathbf{\boldsymbol{y}} = (\boldsymbol{y}_1,\cdots,\boldsymbol{y}_n)^T \) and \( \beta = (\beta_0, \beta_1)^T \)
It is convenient to write \( \mathbf{\boldsymbol{y}} = X\beta \) where \( X \in \mathbb{R}^{100 \times 2} \) is the design matrix given by (we keep the intercept here) $$ X \equiv \begin{bmatrix} 1 & x_1 \\ \vdots & \vdots \\ 1 & x_{100} & \\ \end{bmatrix}. $$ The cost/loss/risk function is given by ( $$ C(\beta) = \frac{1}{n}||X\beta-\mathbf{y}||_{2}^{2} = \frac{1}{n}\sum_{i=1}^{100}\left[ (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2\right] $$ and we want to find \( \beta \) such that \( C(\beta) \) is minimized.
Computing \( \partial C(\beta) / \partial \beta_0 \) and \( \partial C(\beta) / \partial \beta_1 \) we can show that the gradient can be written as $$ \nabla_{\beta} C(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ \sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ \end{bmatrix} = \frac{2}{n}X^T(X\beta - \mathbf{y}), $$ where \( X \) is the design matrix defined above.
We can now write a program that minimizes \( C(\beta) \) using the gradient descent method with a constant learning rate \( \gamma \) according to $$ \beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots $$
We can use the expression we computed for the gradient and let use a \( \beta_0 \) be chosen randomly and let \( \gamma = 0.001 \). Stop iterating when \( ||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8} \). Note that the code below does not include the latter stop criterion.
And finally we can compare our solution for \( \beta \) with the analytic result given by \( \beta= (X^TX)^{-1} X^T \mathbf{y} \).
Here our simple example
# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys
# the number of datapoints
n = 100
x = 2*np.random.rand(n,1)
y = 4+3*x+np.random.randn(n,1)
X = np.c_[np.ones((n,1)), x]
# Hessian matrix
H = (2.0/n)* X.T @ X
# Get the eigenvalues
EigValues, EigVectors = np.linalg.eig(H)
print(EigValues)
beta_linreg = np.linalg.inv(X.T @ X) @ X.T @ y
print(beta_linreg)
beta = np.random.randn(2,1)
eta = 1.0/np.max(EigValues)
Niterations = 1000
for iter in range(Niterations):
gradient = (2.0/n)*X.T @ (X @ beta-y)
beta -= eta*gradient
print(beta)
xnew = np.array([[0],[2]])
xbnew = np.c_[np.ones((2,1)), xnew]
ypredict = xbnew.dot(beta)
ypredict2 = xbnew.dot(beta_linreg)
plt.plot(xnew, ypredict, "r-")
plt.plot(xnew, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Gradient descent example')
plt.show()
# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDRegressor
n = 100
x = 2*np.random.rand(n,1)
y = 4+3*x+np.random.randn(n,1)
X = np.c_[np.ones((n,1)), x]
beta_linreg = np.linalg.inv(X.T @ X) @ (X.T @ y)
print(beta_linreg)
sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)
sgdreg.fit(x,y.ravel())
print(sgdreg.intercept_, sgdreg.coef_)
We have also discussed Ridge regression where the loss function contains a regularized term given by the \( L_2 \) norm of \( \beta \), $$ C_{\text{ridge}}(\beta) = \frac{1}{n}||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0. $$
In order to minimize \( C_{\text{ridge}}(\beta) \) using GD we only have adjust the gradient as follows $$ \nabla_\beta C_{\text{ridge}}(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ \sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ \end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta). $$
We can easily extend our program to minimize \( C_{\text{ridge}}(\beta) \) using gradient descent and compare with the analytical solution given by $$ \beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}. $$
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys
# the number of datapoints
n = 100
x = 2*np.random.rand(n,1)
y = 4+3*x+np.random.randn(n,1)
X = np.c_[np.ones((n,1)), x]
XT_X = X.T @ X
#Ridge parameter lambda
lmbda = 0.001
Id = lmbda* np.eye(XT_X.shape[0])
beta_linreg = np.linalg.inv(XT_X+Id) @ X.T @ y
print(beta_linreg)
# Start plain gradient descent
beta = np.random.randn(2,1)
eta = 0.1
Niterations = 100
for iter in range(Niterations):
gradients = 2.0/n*X.T @ (X @ (beta)-y)+2*lmbda*beta
beta -= eta*gradients
print(beta)
ypredict = X @ beta
ypredict2 = X @ beta_linreg
plt.plot(x, ypredict, "r-")
plt.plot(x, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Gradient descent example for Ridge')
plt.show()


Local
The output directly from the webcam and manipulated on a canvas
ID
Video
Canvas
Remote
The incoming images received from other connected clients
ID
Feed


John Shiver
Software Engineer: Backend | Devops | DataWork Experience
Current2022-09
Founding Engineer
PlanetX Remote First- First engineering hire
- Designed and implemented backend API for Alpha and Beta products with Go, Postgres, and ElasticSearch
- Designed and implemented cloud infrastructure, using Terraform and AWS
- Onboarded and directed new backend engineers, directed contract engineers
2022-072022-02
Senior Backend Engineer
Lootrush Remote First- Built all backend web3 Rust services: buy / sell / trade NFTs
- Built all internal data sync Rust services: capture business relevant blockchain data and player game data
- Rust,Event Sourcing,CQRS,DDD
2022-022021-06
Staff Engineer
Bread (acquired by ADS in Dec 2020) New York, NY- Technical lead for newly formed core engineering team: 4 engineers
- Implemented core logic for business rule engine Go microservice, allows non-technical users to define rule sets which are executed by other services via gRPC
2021-062020-05
Senior Backend Engineer
Bread New York, NY- Implemented Go microservice responsible for generating and storing legal disclosures related to loan origination
- Refactored team's existing Go microservices, unit test coverage 10->70%, integration test coverage 0->80%
2020-032019-07
Senior Backend Engineer
Stocktwits New York, NY- Developed Go services useed to calculate and deliver user stock portfolio performance
- Implemented a distributed lock package using Redis
2019-072018-03
Data Engineer
Squarespace New York, NY- Responsible for various ETLs primarily using Kafka, HDFS, Postgres, and ElasticSearch
- Founded new data engineering team: built k8 infrastructure + all shared Airflow code for the team
2018-032016-09
Systems Engineer
Yext New York, NY- Develop and maintain MYSQL replication management program in Python used for disaster recovery
- Develop and maintain Python CLI tool to sync DNS records between multiple providers
2016-092016-03
Software Engineer
WayUp New York, NY- Designed and implemented various Django backend services, primarily email notifications and in app chat
2016-032014-10
Software Engineer
Offerup Seattle, WA- Designed and implemented various Django backend services at scale (3000+ requests / second)
Education
2019
Bridge Program: Computer Science
NYU Tandon Engineering New York, NYGraduated with Distinction: 4.0
20112008
J.D.
University of San Francisco School of Law San Francisco, CACALI Award: Highest Grade in Economic Regulatory Systems
20082004


MAGNESIUM CHLORIDE- magnesium chloride injection
Mylan Institutional LLC
Disclaimer: This drug has not been found by FDA to be safe and effective, and this labeling has not been approved by FDA. For further information about unapproved drugs, click here.
Magnesium Chloride Injection is a sterile solution of Magnesium Chloride Hexahydrate in Water for Injection q.s. Each mL contains Magnesium Chloride Hexahydrate 200 mg, Sodium Chloride 9 mg, Benzyl Alcohol 1% as a preservative, Water for Injection, q.s. pH adjusted with Hydrochloric Acid and/or Sodium Hydroxide. Total osmolarity equivalent to 2.951 mOsm/mL.
Contains 1.97 mEq of Mg++ and Cl- per mL.
The structural formula is MgCl2•6H2O
Magnesium is the second most plentiful cation within cellular fluids. It is an important activator of many enzyme systems and deficits are accompanied by a variety of functional disturbances.
Magnesium Chloride Injection should not be administered if there is renal impairment, marked myocardial disease or to comatose patients.
The usual precautions for parenteral administration should be observed. Administer with caution if flushing and sweating occurs. A preparation of a calcium salt should be readily available for intravenous injection to counteract potential serious signs of magnesium intoxication. As long as deep tendon reflexes are active it is probable that the patient will not develop respiratory paralysis. Respiration and blood pressure should be carefully observed during and after administration of Magnesium Chloride Injection.
Animal reproduction studies have not been conducted with magnesium chloride. It is also not known whether magnesium chloride can cause fetal harm when administered to a pregnant woman or can affect reproduction capacity. Magnesium Chloride should be given to a pregnant woman only if clearly needed.
Flushing, sweating, sharply lowered blood pressure,hypothermia, stupor and ultimately respiratory depression.
For intravenous infusion: 4 grams in 250 mL of 5%Dextrose Injection, at a rate not exceeding 3 mL per minute. Serum magnesium levels should serve as a guide to continued dosage.
1 to 40 grams daily.
Parenteral drug products should be inspected visually for particulate matter and discoloration prior to administration whenever solution and container permit.
Magnesium Chloride Injection 200 mg/mL (20% w/v).
NDC 67457-134-50
50 mL Multiple-Dose Vial. Individually boxed.
Store at 20° to 25°C (68° to 77°F). [See USP Controlled
Room Temperature.]
Manufactured for:
Mylan Institutional LLC
Rockford, IL 61103
Manufactured by:
Mylan Institutional
Galway, Ireland
0546L104
REVISED FEBRUARY 2013
MI:MAGNIJ:R2
PRINCIPAL DISPLAY PANEL - 200 mg/mL
NDC 67457-134-50 50 mL
Magnesium
Chloride
Injection
200 mg/mL
(20% w/v)
2.951 mOsm/mL
For Intravenous Use After
Dilution
Rx only Multiple-Dose Vial
Manufactured for:
Mylan Institutional LLC
Rockford, IL 61103 U.S.A.
Manufactured by:
Mylan Institutional
Galway, Ireland
MI:134:1C:R2
Mylan.com
Each mL contains: Magnesium
Chloride Hexahydrate 200 mg,
Sodium Chloride 9 mg, Benzyl
Alcohol 1% as a preservative,
Water for Injection q.s. pH adjusted
with Hydrochloric Acid and/or
Sodium Hydroxide.
Total osmolarity equivalent to
2.951 mOsm/mL.
Contains 1.97 mEq of Mg++ and CI¯
per mL.
WARNING: DO NOT USE IF A
PRECIPITATE IS PRESENT.
Store at 20° to 25°C (68° to 77°F).
[See USP Controlled Room
Temperature.]
Usual Dosage: See accompanying
prescribing information.
|
MAGNESIUM CHLORIDE
magnesium chloride injection | ||||||||||||||||||||
| ||||||||||||||||||||
| ||||||||||||||||||||
| ||||||||||||||||||||
| ||||||||||||||||||||
|
| Labeler - Mylan Institutional LLC (790384502) |


Employee Moon
View Employees Tution Applications and Update Status
Log Out
Employee Info
Form Information - Table 2
Status Information - Table 3
Balance Information - Table 4
Employee Application Approval
Form Id:
Empoyee Id:
Request Additional Info:
Notify Employee:
Yes
No
Grade Status(Direct Supervisor)
Passed
Failed
Marked Urgent
Escalation Email Direct Supervisor
Automatic Approval Direct Supervisor
Direct Supervisor Aprroval
Approved
Denied


Click Here to Download the CSV file of the Full Metrics Result.
Top 100 Least Maintainable New Methods
Most of these Methods are in:
{WorstNewMethodsAreInModules}
{TableBodyOfWorstNewMethods}
Top 100 Least Maintainable Methods
Most of these Methods are in:
{WorstMethodsAreInModules}
{TableBodyOfWorstMethods}
Top 100 Least Maintainable Classes
Most of these Classes are in :
{WorstClassesAreInModules}
{TableBodyOfWorstClasses}
Assemblies


Creating Machine
Typically, machines are created in a controller or in your routes file.
Use the Ussd Facade and call the function machine. That's all it takes.
Ussd::machine();
Session id
Machine Requires a session id to work. You can set it on the machine call like this.
Ussd::machine()->setSessionId($request->session_id);
Session id from request
You may want to set the session id straight for the request with having to use a request object. To achieve this, call setSessionIdFromRequest and specify the request key.
Ussd::machine()->setSessionIdFromRequest('sesssio_id');
UserInput
Another important variable is the Input from the user. That too can be set like the session id.
Ussd::machine()->setInput($request->user_input);
// or
Ussd::machine()->setInputFromRequest('user_input');
Other parameters
Other parameters like phoneNumbe and network are also available to set when needed.
Ussd::machine()->setPhoneNumber($request->phone_number);
Ussd::machine()->setNetwork($request->network);
// or
Ussd::machine()->setPhoneNumberFromRequest('phone_number');
Ussd::machine()->setNetworkFromRequest('network');
Parameters can be chained
Ussd::machine()->setPhoneNumber($request->phone_number)->setNetworkFromRequest('network');
Rarameters can be set at once
Ussd:machine()->set([
'phone_number' => $request->phone_number,
'network' => $request->operatorNetwork,
'session_id' => $request->sessionID,
'input' => $request->user_input,
]);
// or
/*
When key from request can be converted to the same camel
as phoneNumber, network, sessionId, input, then
there is no need to use associative array.
on the value is enough.
*/
Ussd:machine()->setFromRequest([
'phone_number',
'input' => 'user_input',
'network',
'session_id',
]);
Choose cache store
When you don't specify the cache store, the store variable in config/ussd.php is used. When specified the machine uses the specified one. But you can override that with setStore.
Ussd:machine()->setStore('redis');
Initial State
The machine can not run without tell it where to start from. When you do, don't worry. You don't need to tell it of the other states, it will automatically know and work accordingly.
Ussd:machine()->setInitialState(Welcome::class);
Output of the machine
The machine will generate an array as the out put with two property message
and code
. The message is the string to be responsed to the mobile operator and the code show waether to provide an input or not to the user.
<?php
use Illuminate\Support\Facades\Route;
use Sparors\Ussd\Facades\Ussd;
use App\Http\Ussd\Welcome;
Route::get('/', function () {
$ussd = Ussd::machine()
->setFromRequest([
'phone_number',
'input' => 'user_input',
'network', 'session_id',
])
->setInitialState(Welcome::class);
return response()->json($ussd->run());
});
/*
output
{
"message": "Welcome class over here"
"code": 1
}
*/
NB: code 1 for input, 2 for no input.
what ! my operatory does not understand that response
Don't worry, we know you can not tell your ussd operator to change their structure just to make your application work. You can set how to responsed to your operator by using a callable. Below is an example.
<?php
use Illuminate\Support\Facades\Route;
use Sparors\Ussd\Facades\Ussd;
use App\Http\Ussd\Welcome;
Route::get('/', function () {
$ussd = Ussd::machine()
->setFromRequest([
'phone_number' => 'msisdn',
'input' => 'msg',
'network',
'session_id' => 'UserSessionID',
])
->setInitialState(Welcome::class)
->setResponse(function(string $message, int $code) {
return [
'USSDResp' => [
'action' => $code === 2 ? 'prompt' : 'input',
'menus' => '',
'title' => $message,
];
];
});
return response()->json($ussd->run());
});
/*
output
{
"USSDResp": {
"action": "input",
"menus": "",
"title": "Welcome class over here"
}
}
*/


Mellow Cobra's C++ Pascal Interpreter
Copyright (c) 2017 Grayson Dubois @mellowcobra
Created following the tutorial on
Ruslan's Blog
View the
README


I'm a consultant with Pampered Chef, proud member of the Berkshire Hathaway Family of Companies. Based in Port Perry, ON, Canada, I am thrilled to share with you superior quality kitchen tools and appliances along with recipe ideas and tips to make your time in the kitchen a little easier! I've loved Pampered Chef for years and bought my first stoneware piece from them decades ago. It still gets used several times a week and is very well seasoned!
Currently residing in Port Perry, ON, just an hour outside of Toronto, with the best of small town rural living, just a short drive away from the city!
As a single mom of a teenage girl, and working full-time remotely, I was looking for something fun where I could meet new people that aligned with my goal to cook from home more frequently. Enter Pampered Chef almost 3 years ago. Now I get to have fun and help my wonderful hosts earn amazing products for free or at discounts!
In addition to my full-time job and Pampered Chef, my daughter and I share a love for Starbucks and when she's not being a typical brooding teenager, we share lots of laughs together. I love spending time with family, crafting and baking and I'm a self-professed crazy cat lady with 5 kitties at home!


Introduction
In this project, we use MOVO robot to perform object detection and navigation. The goal for the robot is to identify the chairs and people in the environment and searching for empty chair location in the environment. Once the robot found the empty chair location, it registers it to the map. After it identifies all the empty chairs in the environment, users can ask the robot to direct to the closet empty chair.
Chair detection and localization
We use FasterRCNN with ResNet101 backbone trained on the VOC dataset, which has 20 classes including 'person' and 'chair'. We found directly running inference on pretrained model can already generate descent result in our testing environment. Therefore, no fine-tuning is performed.
We fetch the RGB values of the image that MOVO sees through the color channel and also fetch the depth image using the Kinect's sensor and converted to XYZ coordinates with respect to the robot. Based on the bounding box from the output of the fastRCNN, we use the centroid of the bounding box as the estimation for the chair location.
Here is a video demo of object detection and object registration on map. We use joy-stick to control the robot navigation.
Here is the result for all the chairs that robot detect and found in the environment. We see total of 13 out of 15 detection is correct and there are 2 false positives.
After detecting all the empty chairs, Movo can guide people to an empty chair by navigate autonomously in the environment.


Util
Util is a collection of Twitter's core JVM libraries including: concurrency, I/O, caching, stats, time and duration, conversions and so on.
Contributing
We feel that a welcoming community is important and we ask that you follow Twitter’s Open Source Code of Conduct in all interactions with the community.
Util is actively maintained by Twitter’s infrastructure team.
The master branch represents the most recent published release while active development happens on the develop branch.
Before endeavoring on large changes, please discuss them with the Finagle Google Group to receive feedback and suggestions.
For all patches, please review our contributing docs.


{% extends "base.html" %} {% load bootstrap3 %} {% block content %}
published March 14, 2019
This page explains what data we collect from you and what we use it for. Changes to the contents of this policy will be published here and sent to subscribers' emails. Past versions of this page are available on request.
Here is what data we will collect from you:
Here is what we will never do with your data:
Here is what we may do with your data: