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).
$$
\begin{align*}
\boldsymbol{\beta}_{-i}(\lambda) & = ( \boldsymbol{X}_{-i, \ast}^{T}
\boldsymbol{X}_{-i, \ast} + \lambda \boldsymbol{I}_{pp})^{-1}
\boldsymbol{X}_{-i, \ast}^{T} \boldsymbol{y}_{-i}
\end{align*}
$$
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()
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
$$
\begin{equation}
\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{\beta} + \boldsymbol{\epsilon},
\tag{1}
\end{equation}
$$
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.
In project 1, when using Ridge and Lasso regression, we end up searching for the optimal parameter \( \lambda \) which minimizes our selected scores (MSE or \( R2 \) values for example). The brute force approach, as discussed in the code here for Ridge regression, consists in evaluating the MSE as function of different \( \lambda \) values. Based on these calculations, one tries then to determine the value of the hyperparameter \( \lambda \) which results in optimal scores (for example the smallest MSE or an \( R2=1 \)).
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(2021)
n = 100
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
Maxpolydegree = 5
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)
# Decide which values of lambda to use
nlambdas = 500
MSERidgePredict = np.zeros(nlambdas)
lambdas = np.logspace(-4, 2, nlambdas)
for i in range(nlambdas):
lmb = lambdas[i]
RegRidge = linear_model.Ridge(lmb)
RegRidge.fit(X_train,y_train)
ypredictRidge = RegRidge.predict(X_test)
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
Here we have performed a rather data greedy calculation as function of the regularization parameter \( \lambda \). There is no resampling here. The latter can easily be added by employing the function RidgeCV instead of just calling the Ridge function. For RidgeCV we need to pass the array of \( \lambda \) values. By inspecting the figure we can in turn determine which is the optimal regularization parameter. This becomes however less functional in the long run.
An alternative is to use the so-called grid search functionality included with the library Scikit-Learn, as demonstrated for the same example here.
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Ridge
from sklearn.model_selection import GridSearchCV
def R2(y_data, y_model):
return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
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(2021)
n = 100
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
Maxpolydegree = 5
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)
# Decide which values of lambda to use
nlambdas = 10
lambdas = np.logspace(-4, 2, nlambdas)
# create and fit a ridge regression model, testing each alpha
model = Ridge()
gridsearch = GridSearchCV(estimator=model, param_grid=dict(alpha=lambdas))
gridsearch.fit(X_train, y_train)
print(gridsearch)
ypredictRidge = gridsearch.predict(X_test)
# summarize the results of the grid search
print(f"Best estimated lambda-value: {gridsearch.best_estimator_.alpha}")
print(f"MSE score: {MSE(y_test,ypredictRidge)}")
print(f"R2 score: {R2(y_test,ypredictRidge)}")
By default the grid search function includes cross validation with five folds. The Scikit-Learn documentation contains more information on how to set the different parameters.
If we take out the random noise, running the above codes results in \( \lambda=0 \) yielding the best fit.
An alternative to the above manual grid set up, is to use a random search where the parameters are tuned from a random distribution (uniform below) for a fixed number of iterations. A model is constructed and evaluated for each combination of chosen parameters. We repeat the previous example but now with a random search. Note that values of \( \lambda \) are now limited to be within \( x\in [0,1] \). This domain may not be the most relevant one for the specific case under study.
import numpy as np
from sklearn.model_selection import train_test_split
from sklearn.linear_model import Ridge
from sklearn.model_selection import GridSearchCV
from scipy.stats import uniform as randuniform
from sklearn.model_selection import RandomizedSearchCV
def R2(y_data, y_model):
return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
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(2021)
n = 100
x = np.random.rand(n)
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.randn(n)
Maxpolydegree = 5
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)
param_grid = {'alpha': randuniform()}
# create and fit a ridge regression model, testing each alpha
model = Ridge()
gridsearch = RandomizedSearchCV(estimator=model, param_distributions=param_grid, n_iter=100)
gridsearch.fit(X_train, y_train)
print(gridsearch)
ypredictRidge = gridsearch.predict(X_test)
# summarize the results of the grid search
print(f"Best estimated lambda-value: {gridsearch.best_estimator_.alpha}")
print(f"MSE score: {MSE(y_test,ypredictRidge)}")
print(f"R2 score: {R2(y_test,ypredictRidge)}")
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)))
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)
from sklearn.preprocessing import LabelEncoder
from sklearn.model_selection import cross_validate
#Cross validation
accuracy = cross_validate(logreg,X_test,y_test,cv=10)['test_score']
print(accuracy)
print("Test set accuracy with Logistic Regression: {:.2f}".format(logreg.score(X_test,y_test)))
import scikitplot as skplt
y_pred = logreg.predict(X_test)
skplt.metrics.plot_confusion_matrix(y_test, y_pred, normalize=True)
plt.show()
y_probas = logreg.predict_proba(X_test)
skplt.metrics.plot_roc(y_test, y_probas)
plt.show()
skplt.metrics.plot_cumulative_gain(y_test, y_probas)
plt.show()
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
$$
f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots.
\tag{2}
$$
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), to be discussed next week.
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.
Using the definition of convexity, try to show that a function satisfying the properties above is convex (the third condition is not needed to show this).
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:
We revisit an example similar to what we had in the first homework set. We had a function of the type
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.
The Hessian matrix of \( C(\beta) \) is given by
$$
\boldsymbol{H} \equiv \begin{bmatrix}
\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\
\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\
\end{bmatrix} = \frac{2}{n}X^T X.
$$
This result implies that \( C(\beta) \) is a convex function since the matrix \( X^T X \) always is positive semi-definite.
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(f"Eigenvalues of Hessian Matrix:{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 adjust the gradient as follows
$$
\nabla_\beta C_{\text{ridge}}(\beta) = \frac{2}{n}\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\
\sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\
\end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (\frac{1}{n}X^T(X\beta - \mathbf{y})+\lambda \beta).
$$
We can easily extend our program to minimize \( C_{\text{ridge}}(\beta) \) using gradient descent and compare with the analytical solution given by
$$
\beta_{\text{ridge}} = \left(X^T X + n\lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}.
$$
The Hessian matrix of Ridge Regression for our simple example is given by
$$
\boldsymbol{H} \equiv \begin{bmatrix}
\frac{\partial^2 C(\beta)}{\partial \beta_0^2} & \frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} \\
\frac{\partial^2 C(\beta)}{\partial \beta_0 \partial \beta_1} & \frac{\partial^2 C(\beta)}{\partial \beta_1^2} & \\
\end{bmatrix} = \frac{2}{n}X^T X+2\lambda\boldsymbol{I}.
$$
This implies that the Hessian matrix is positive definite, hence the stationary point is a minimum. Note that the Ridge cost function is convex being a sum of two convex functions. Therefore, the stationary point is a global minimum of this function.
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 = n*lmbda* np.eye(XT_X.shape[0])
# Hessian matrix
H = (2.0/n)* XT_X+2*lmbda* np.eye(XT_X.shape[0])
# Get the eigenvalues
EigValues, EigVectors = np.linalg.eig(H)
print(f"Eigenvalues of Hessian Matrix:{EigValues}")
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 = 1.0/np.max(EigValues)
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()
Write a code which implements gradient descent for a logistic regression example.