# Logistic regression with $$\ell_1$$ regularization¶

In this example, we use CVXPY to train a logistic regression classifier with $$\ell_1$$ regularization. We are given data $$(x_i,y_i)$$, $$i=1,\ldots, m$$. The $$x_i \in {\bf R}^n$$ are feature vectors, while the $$y_i \in \{0, 1\}$$ are associated boolean classes.

Our goal is to construct a linear classifier $$\hat y = \mathbb{1}[\beta^T x > 0]$$, which is $$1$$ when $$\beta^T x$$ is positive and $$0$$ otherwise. We model the posterior probabilities of the classes given the data linearly, with

$\log \frac{\mathrm{Pr} (Y=1 \mid X = x)}{\mathrm{Pr} (Y=0 \mid X = x)} = \beta^T x.$

This implies that

$\mathrm{Pr} (Y=1 \mid X = x) = \frac{\exp(\beta^T x)}{1 + \exp(\beta^T x)}, \quad \mathrm{Pr} (Y=0 \mid X = x) = \frac{1}{1 + \exp(\beta^T x)}.$

We fit $$\beta$$ by maximizing the log-likelihood of the data, plus a regularization term $$\lambda \|\beta\|_1$$ with $$\lambda > 0$$:

$\ell(\beta) = \sum_{i=1}^{m} y_i \beta^T x_i - \log(1 + \exp (\beta^T x_i)) - \lambda \|\beta\|_1.$

Because $$\ell$$ is a concave function of $$\beta$$, this is a convex optimization problem.

import cvxpy as cp
import numpy as np
import matplotlib.pyplot as plt


In the following code we generate data with $$n=50$$ features by randomly choosing $$x_i$$ and supplying a sparse $$\beta_{\mathrm{true}} \in {\bf R}^n$$. We then set $$y_i = \mathbb{1}[\beta_{\mathrm{true}}^T x_i + z_i > 0]$$, where the $$z_i$$ are i.i.d. normal random variables. We divide the data into training and test sets with $$m=50$$ examples each.

np.random.seed(1)
n = 50
m = 50
def sigmoid(z):
return 1/(1 + np.exp(-z))

beta_true = np.array([1, 0.5, -0.5] + [0]*(n - 3))
X = (np.random.random((m, n)) - 0.5)*10
Y = np.round(sigmoid(X @ beta_true + np.random.randn(m)*0.5))

X_test = (np.random.random((2*m, n)) - 0.5)*10
Y_test = np.round(sigmoid(X_test @ beta_true + np.random.randn(2*m)*0.5))


We next formulate the optimization problem using CVXPY.

beta = cp.Variable(n)
lambd = cp.Parameter(nonneg=True)
log_likelihood = cp.sum(
cp.multiply(Y, X @ beta) - cp.logistic(X @ beta)
)
problem = cp.Problem(cp.Maximize(log_likelihood/n - lambd * cp.norm(beta, 1)))


We solve the optimization problem for a range of $$\lambda$$ to compute a trade-off curve. We then plot the train and test error over the trade-off curve. A reasonable choice of $$\lambda$$ is the value that minimizes the test error.

def error(scores, labels):
scores[scores > 0] = 1
scores[scores <= 0] = 0
return np.sum(np.abs(scores - labels)) / float(np.size(labels))

trials = 100
train_error = np.zeros(trials)
test_error = np.zeros(trials)
lambda_vals = np.logspace(-2, 0, trials)
beta_vals = []
for i in range(trials):
lambd.value = lambda_vals[i]
problem.solve()
train_error[i] = error( (X @ beta).value, Y)
test_error[i] = error( (X_test @ beta).value, Y_test)
beta_vals.append(beta.value)

%matplotlib inline
%config InlineBackend.figure_format = "svg"

plt.plot(lambda_vals, train_error, label="Train error")
plt.plot(lambda_vals, test_error, label="Test error")
plt.xscale("log")
plt.legend(loc="upper left")
plt.xlabel(r"$\lambda$", fontsize=16)
plt.show()


We also plot the regularization path, or the $$\beta_i$$ versus $$\lambda$$. Notice that a few features remain non-zero longer for larger $$\lambda$$ than the rest, which suggests that these features are the most important.

for i in range(n):
plt.plot(lambda_vals, [wi for wi in beta_vals])
plt.xlabel(r"$\lambda$", fontsize=16)
plt.xscale("log")


We plot the true $$\beta$$ versus reconstructed $$\beta$$, as chosen to minimize error on the test set. The non-zero coefficients are reconstructed with good accuracy. There are a few values in the reconstructed $$\beta$$ that are non-zero but should be zero.

idx = np.argmin(test_error)
plt.plot(beta_true, label=r"True $\beta$")
plt.plot(beta_vals[idx], label=r"Reconstructed $\beta$")
plt.xlabel(r"$i$", fontsize=16)
plt.ylabel(r"$\beta_i$", fontsize=16)
plt.legend(loc="upper right")

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