11.2 Logistic regression

Logistic regression is an example of a binary classifier, where the output takes one two values 0 or 1 for each data point. We call the two values classes.

Formulation as an optimization problem

Define the sigmoid function


Next, given an observation \(x\in\real^d\) and a weights \(\theta\in\real^d\) we set


The weights vector \(\theta\) is part of the setup of the classifier. The expression \(h_\theta(x)\) is interpreted as the probability that \(x\) belongs to class 1. When asked to classify \(x\) the returned answer is

\[\begin{split}x\mapsto \begin{cases}\begin{array}{ll}1 & h_\theta(x)\geq 1/2, \\ 0 & h_\theta(x)<1/2.\end{array}\end{cases}\end{split}\]

When training a logistic regression algorithm we are given a sequence of training examples \(x_i\), each labelled with its class \(y_i\in \{0,1\}\) and we seek to find the weights \(\theta\) which maximize the likelihood function

\[\prod_i h_\theta(x_i)^{y_i}(1-h_\theta(x_i))^{1-y_i}.\]

Of course every single \(y_i\) equals 0 or 1, so just one factor appears in the product for each training data point. By taking logarithms we can define the logistic loss function:

\[J(\theta) = -\sum_{i:y_i=1} \log(h_\theta(x_i))-\sum_{i:y_i=0}\log(1-h_\theta(x_i)).\]

The training problem with regularization (a standard technique to prevent overfitting) is now equivalent to

\[\min_\theta J(\theta) + \lambda\|\theta\|_2.\]

This can equivalently be phrased as

(11.13)\[\begin{split}\begin{array}{lrllr} \minimize & \sum_i t_i +\lambda r & & & \\ \st & t_i & \geq - \log(h_\theta(x)) & = \log(1+\exp(-\theta^Tx_i)) & \mathrm{if}\ y_i=1, \\ & t_i & \geq - \log(1-h_\theta(x)) & = \log(1+\exp(\theta^Tx_i)) & \mathrm{if}\ y_i=0, \\ & r & \geq \|\theta\|_2. & & \end{array}\end{split}\]


As can be seen from (11.13) the key point is to implement the softplus bound \(t\geq \log(1+e^u)\), which is the simplest example of a log-sum-exp constraint for two terms. Here \(t\) is a scalar variable and \(u\) will be the affine expression of the form \(\pm \theta^Tx_i\). This is equivalent to

\[\exp(u-t) + \exp(-t)\leq 1\]

and further to

(11.14)\[\begin{split}\begin{array}{rclr} (z_1, 1, u-t) & \in & \EXP & (z_1\geq \exp(u-t)), \\ (z_2, 1, -t) & \in & \EXP & (z_2\geq \exp(-t)), \\ z_1+z_2 & \leq & 1. & \end{array}\end{split}\]

This formulation can be entered using affine conic constraints (see Sec. 6.2 (From Linear to Conic Optimization)).

Listing 11.8 Implementation of \(t\geq \log(1+e^u)\) as in (11.14). Click here to download.
fn softplus(task : & mut TaskCB, d : i32, n : i32, theta : i32, t : i32, X : &[f64], Y : &[bool]) -> Result<(),String> {
    let nvar = task.get_num_var()?;
    let ncon = task.get_num_con()?;
    let nafe = task.get_num_afe()?;
    task.append_vars(2*n)?;   // z1, z2
    task.append_cons(n)?;     // z1 + z2 = 1
    task.append_afes(4*n as i64)?;   //theta * X[i] - t[i], -t[i], z1[i], z2[i]
    let z1 = nvar;
    let z2 = nvar+n;
    let zcon = ncon;
    let thetaafe = nafe;
    let tafe = nafe+n   as i64;
    let z1afe = tafe+n  as i64;
    let z2afe = z1afe+n as i64;

    // Linear constraints
        let mut subi = vec![0i32; 2*n as usize];
        let mut subj = vec![0i32; 2*n as usize];
        let mut aval = vec![0.0;  2*n as usize];

        for i in 0..n {
        for ((i,&zx),si, sj, av) in izip!(iproduct!(0..n,&[z1,z2]),
                                          aval.iter_mut()) {
            *si = zcon+i;
            *sj = zx+i as i32;
            *av = 1.0;

        task.put_aij_list(subi.as_slice(), subj.as_slice(), aval.as_slice())?;
        task.put_con_bound_slice_const(zcon, zcon+n, Boundkey::FX, 1.0, 1.0)?;
        task.put_var_bound_slice_const(nvar, nvar+2*n, Boundkey::FR, -INF, INF)?;
    // Affine conic expressions
    let mut afeidx = vec![0i64; (d*n+4*n) as usize];
    let mut varidx = vec![0i32; (d*n+4*n) as usize];
    let mut fval   = vec![0.0;  (d*n+4*n) as usize];

    // Thetas
    let mut k : usize = 0;
    for ((i,j),afei,vari) in izip!(iproduct!(0..n,0..d),
                                   & mut afeidx[k..k+(n*d) as usize],
                                   & mut varidx[k..k+(n*d) as usize]) {
        *afei = thetaafe + i as i64;
        *vari = theta    + j;

    for ((&yi,_j),&xij,fv) in izip!(iproduct!(Y,0..d), X, & mut fval[k..k+(n*d) as usize]) {
        *fv = (if yi {-1.0} else {1.0}) * xij;
    k += (n*d) as usize;

    for fv in fval[k..k+(2*n) as usize].iter_mut() { *fv = -1.0; }
    for (i,afei,vari) in izip!(0..n,
                               &mut afeidx[k..k+(2*n) as usize].iter_mut().step_by(2),
                               &mut varidx[k..k+(2*n) as usize].iter_mut().step_by(2)) {
        *afei = thetaafe+i as i64; *vari = t+i;
    for (i,afei,vari) in izip!(0..n,
                               &mut afeidx[k+1..k+(2*n) as usize].iter_mut().step_by(2),
                               &mut varidx[k+1..k+(2*n) as usize].iter_mut().step_by(2)) {
        *afei = tafe+i as i64; *vari = t+i;
    k += (n*2) as usize;

    for fv in fval[k..k+(2*n) as usize].iter_mut() { *fv = 1.0; }
    for (i,afei,vari) in izip!(0..n,
                               &mut afeidx[k..k+(2*n) as usize].iter_mut().step_by(2),
                               &mut varidx[k..k+(2*n) as usize].iter_mut().step_by(2)) {
        *afei = z1afe+i as i64; *vari = z1+i;
    for (i,afei,vari) in izip!(0..n,
                               &mut afeidx[k+1..k+(2*n) as usize].iter_mut().step_by(2),
                               &mut varidx[k+1..k+(2*n) as usize].iter_mut().step_by(2)) {
        *afei = z2afe+i as i64; *vari = z2+i;

    // Add the expressions
    task.put_afe_f_entry_list(afeidx.as_slice(), varidx.as_slice(), fval.as_slice())?;

        // Add a single row with the constant expression "1.0"
        let oneafe = task.get_num_afe()?;
        task.put_afe_g(oneafe, 1.0)?;

        // Add an exponential cone domain
        let dom = task.append_primal_exp_cone_domain()?;

        // Conic constraints
        let zeros = &[0.0,0.0,0.0];
        let mut acci = task.get_num_acc()?;
        for i in 0..n as i64 {
            task.append_acc(dom, &[z1afe+i, oneafe, thetaafe+i], zeros)?;
            task.append_acc(dom, &[z2afe+i, oneafe, tafe+i],     zeros)?;

            acci += 2;


Once we have this subroutine, it is easy to implement a function that builds the regularized loss function model (11.13).

Listing 11.9 Implementation of (11.13). Click here to download.
fn logistic_regression(X : &[f64],
                       Y : &[bool],
                       lamb : f64) -> Result<Vec<f64>,String> {
    let n = Y.len() as i32;
    let d = (X.len()/Y.len()) as i32;  // num samples, dimension

    /* Create the optimization task. */
    let mut task = match Task::new() {
        Some(e) => e,
        None => return Err("Failed to create task".to_string()),
    task.put_stream_callback(Streamtype::LOG, |msg| print!("{}",msg))?;

    // Variables [r; theta; t]
    let nvar : i32 = 1+d+n;
    task.put_var_bound_slice_const(0, nvar, Boundkey::FR, -INF, INF)?;
    let (r,theta,t) = (0i32,1i32,1+d);
    for j in 0..d { task.put_var_name(theta+j,format!("theta[{}]",j).as_str())?; }
    for j in 0..n { task.put_var_name(t+j,format!("t[{}]",j).as_str())?; }

    // Objective lambda*r + sum(t)
    task.put_c_j(r, lamb)?;
    for i in 0..n { task.put_c_j(t+i, 1.0)?; }

    // Softplus function constraints
    softplus(& mut task, d, n, theta, t, X, Y)?;

    // Regularization
    // Append a sequence of linear expressions (r, theta) to F
    let numafe = task.get_num_afe()?;
    task.append_afes(1+d as i64)?;
    task.put_afe_f_entry(numafe, r, 1.0)?;
    for i in 0..d {
        task.put_afe_f_entry(numafe + i as i64 + 1, theta + i, 1.0)?;

    // Add the constraint
        let dom = task.append_quadratic_cone_domain((1+d) as i64)?;
                            vec![0.0; 1+d as usize].as_slice())?;
    // Solution

    let mut xx = vec![0.0; d as usize];
    task.get_xx_slice(Soltype::ITR, theta, theta+d as i32,xx.as_mut_slice())?;

Example: 2D dataset fitting

In the next figure we apply logistic regression to the training set of 2D points taken from the example ex2data2.txt . The two-dimensional dataset was converted into a feature vector \(x\in\real^{28}\) using monomial coordinates of degrees at most 6.


Fig. 11.2 Logistic regression example with none, medium and strong regularization (small, medium, large \(\lambda\)). Without regularization we get obvious overfitting.