FYS-STK4155/doc/LectureNotes/_build/jupyter_execute/week38.ipynb

Week 38: Logistic Regression and Optimization

Morten Hjorth-Jensen, Department of Physics and Center for Computing in Science Education, University of Oslo and Department of Physics and Astronomy and Facility for Rare Isotope Beams, Michigan State University

Date: September 18-22

Plans for week 38

Material for the active learning sessions on Tuesday and Wednesday.

• Lecture from last week on the bias-variance tradeoff

• Resampling techniques, cross-validation examples included here, see also the lectures from last week on the bootstrap method

• Exercise for week 38, see also the whiteboard notes from week 37 at https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/2023/NotesSep14.pdf

• Work on project 1, in particular resampling methods like cross-validation and bootstrap.

Material for the lecture on Thursday September 21.

• Logistic regression as our first encounter of classification methods. From binary cases to several categories.

• Start gradient and optimization methods

• Readings and Videos:

  • Hastie et al 4.1, 4.2 and 4.3 on logistic regression

  • For a good discussion on gradient methods, see Goodfellow et al section 4.3-4.5 and chapter 8. We will come back to the latter chapter in our discussion of Neural networks as well.

  • See also the whiteboard notes from week 37 at https://github.com/CompPhysics/MachineLearning/blob/master/doc/HandWrittenNotes/2023/NotesSep14.pdf for a discussion and derivation of the bias-variance tradeoff.

  • Video on Logistic regression

  • Yet another video on logistic regression

  • Video on gradient descent

Material from last week and relevant for the first project

Various steps in cross-validation

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).

How to set up the cross-validation for Ridge and/or Lasso

• Define a range of interest for the penalty parameter.

• Divide the data set into training and test set comprising samples \{1, \ldots, n\} \setminus i and \{ i \}, respectively.

• Fit the linear regression model by means of for example Ridge or Lasso regression for each \lambda in the grid using the training set, and the corresponding estimate of the error variance \boldsymbol{\sigma}_{-i}^2(\lambda), as

\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*}

• Evaluate the prediction performance of these models on the test set by C[y_i, \boldsymbol{X}_{i, \ast}; \boldsymbol{\beta}_{-i}(\lambda), \boldsymbol{\sigma}_{-i}^2(\lambda)]. Or, by the prediction error |y_i - \boldsymbol{X}_{i, \ast} \boldsymbol{\beta}_{-i}(\lambda)|, the relative error, the error squared or the R2 score function.

• Repeat the first three steps such that each sample plays the role of the test set once.

• Average the prediction performances of the test sets at each grid point of the penalty bias/parameter. It is an estimate of the prediction performance of the model corresponding to this value of the penalty parameter on novel data.

Cross-validation in brief

For the various values of k

1. shuffle the dataset randomly.

2. Split the dataset into k groups.

3. For each unique group:

a. Decide which group to use as set for test data

b. Take the remaining groups as a training data set

c. Fit a model on the training set and evaluate it on the test set

d. Retain the evaluation score and discard the model

5. Summarize the model using the sample of model evaluation scores

Code Example for Cross-validation and $k$-fold Cross-validation

The code here uses Ridge regression with cross-validation (CV) resampling and $k$-fold CV in order to fit a specific polynomial.

In [1]:

%matplotlib inline

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()

Material for lecture Thursday September 21

Logistic Regression

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

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.

Optimization and Deep learning

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.

Basics

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}.

Linear classifier

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{_auto1} \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.

Some selected properties

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.

Simple example

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.

In [2]:

# 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()

	ID	Age	Agegroup	CHD
0	1	21	1	0
1	2	23	1	0
2	3	25	1	1
3	4	29	1	0
4	5	21	1	0
...	...	...	...	...
95	96	61	8	1
96	97	69	8	1
97	98	65	8	1
98	99	64	8	1
99	100	63	8	0

100 rows × 4 columns

Plotting the mean value for each group

What we could attempt however is to plot the mean value for each group.

In [3]:

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.

The logistic function

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).

Examples of likelihood functions used in logistic regression and nueral networks

The following code plots the logistic function, the step function and other functions we will encounter from here and on.

In [4]:

"""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()

Two parameters

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}).

Maximum likelihood

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).

The cost function rewritten

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.

Minimizing the cross entropy

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).

A more compact expression

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}.

Extending to more predictors

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)}}.

Including more classes

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.

More classes

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.

Friday September 23

Searching for Optimal Regularization Parameters \lambda

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).

In [5]:

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.

Grid Search

An alternative is to use the so-called grid search functionality included with the library Scikit-Learn, as demonstrated for the same example here.

In [6]:

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)}")

GridSearchCV(estimator=Ridge(),
             param_grid={'alpha': array([1.00000000e-04, 4.64158883e-04, 2.15443469e-03, 1.00000000e-02,
       4.64158883e-02, 2.15443469e-01, 1.00000000e+00, 4.64158883e+00,
       2.15443469e+01, 1.00000000e+02])})
Best estimated lambda-value: 100.0
MSE score: 1.0892144853354966
R2 score: -0.0038332550504751595

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.

Randomized Grid Search

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.

In [7]:

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)}")

RandomizedSearchCV(estimator=Ridge(), n_iter=100,
                   param_distributions={'alpha': <scipy.stats._distn_infrastructure.rv_frozen object at 0x13ef4e1c0>})
Best estimated lambda-value: 0.9849967686928113
MSE score: 1.0853136633465326
R2 score: -0.0002382102844775691

Wisconsin Cancer Data

We show here how we can use a simple regression case on the breast cancer data using Logistic regression as our algorithm for classification.

In [8]:

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)))

(426, 30)
(143, 30)
Test set accuracy with Logistic Regression: 0.94

/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(

Using the correlation matrix

In addition to the above scores, we could also study the covariance (and the correlation matrix). We use Pandas to compute the correlation matrix.

In [9]:

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()

Discussing the correlation data

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

In [10]:

cancerpd = pd.DataFrame(cancer.data, columns=cancer.feature_names)

and then

In [11]:

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).

Other measures in classification studies: Cancer Data again

In [12]:

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()

(426, 30)
(143, 30)
[1.         0.86666667 1.         0.92857143 1.         0.85714286
 1.         0.92857143 0.92857143 1.        ]
Test set accuracy with Logistic Regression: 0.94

/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(
/Users/mhjensen/miniforge3/envs/myenv/lib/python3.9/site-packages/sklearn/linear_model/_logistic.py:814: ConvergenceWarning: lbfgs failed to converge (status=1):
STOP: TOTAL NO. of ITERATIONS REACHED LIMIT.

Increase the number of iterations (max_iter) or scale the data as shown in:
    https://scikit-learn.org/stable/modules/preprocessing.html
Please also refer to the documentation for alternative solver options:
    https://scikit-learn.org/stable/modules/linear_model.html#logistic-regression
  n_iter_i = _check_optimize_result(

Optimization, the central part of any Machine Learning algortithm

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.

Revisiting our Logistic Regression case

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.

The equations to solve

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.

Solving using Newton-Raphson's method

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

Warning:

Output truncated. This notebook contains too many cells to display efficiently.