# How to plot logistic decision boundary?

I am running logistic regression on a small dataset which looks like this:

After implementing gradient descent and the cost function, I am getting a 100% accuracy in the prediction stage, However I want to be sure that everything is in order so I am trying to plot the decision boundary line which separates the two datasets.

Below I present plots showing the cost function and theta parameters. As can be seen, currently I am printing the decision boundary line incorrectly.

Extracting data

clear all; close all; clc;

alpha = 0.01;
num_iters = 1000;

%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);

x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

x = [x1 x2]'; % X

subplot(2,2,1);
dat = [dat1 dat2]'; % Y

scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);

%  Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);

% Add intercept term to x and X_test
x = [ones(m, 1) x];

% Initialize fitting parameters
theta = zeros(n + 1, 1);
%initial_theta = [0.2; 0.2];

J_history = zeros(num_iters, 1);

plot_x = [min(x(:,2))-2,  max(x(:,2))+2]

for iter = 1:num_iters
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(theta, x, classdata);
theta = theta - alpha * grad;
J_history(iter) = cost;

fprintf('Iteration #%d - Cost = %d... \r\n',iter, cost);

subplot(2,2,2);
hold on; grid on;
plot(iter, J_history(iter), '.r');  title(sprintf('Plot of cost against number of iterations. Cost is %g',J_history(iter)));
xlabel('Iterations')
ylabel('MSE')
drawnow

subplot(2,2,3);
grid on;
plot3(theta(1), theta(2), J_history(iter),'o')
title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
xlabel('Tita0')
ylabel('Tita1')
zlabel('Cost')
hold on;
drawnow

subplot(2,2,1);
grid on;
% Calculate the decision boundary line
plot_y = theta(2).*plot_x + theta(1);  % <--- Boundary line
% Plot, and adjust axes for better viewing
plot(plot_x, plot_y)
hold on;
drawnow

end

fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Gradient at initial theta (zeros): \n');

The above code is implementing gradient descent correctly (I think) but I am still unable to show the boundary line plot. Any suggestions would be appreciated.

logistic_costFunction

function [J, grad] = logistic_costFunction(theta, X, y)

% Initialize some useful values
m = length(y); % number of training examples

h = sigmoid(X * theta);
J = -(1 / m) * sum( (y .* log(h)) + ((1 - y) .* log(1 - h)) );

for i = 1 : size(theta, 1)
grad(i) = (1 / m) * sum( (h - y) .* X(:, i) );
end

end
• Your dataset is 2D, so you'll have three theta's actually. Can you share the part where you trained the logistic regression? – gunes Apr 19 at 9:43
• Could you kindly explain why I should have three thetas? It seems that I am missing something fundamental here. As can be seen I am initializing theta according to: theta = zeros(n + 1, 1); – Rrz0 Apr 19 at 9:53
• Cross posted: datascience.stackexchange.com/questions/49573/… – naive Apr 19 at 11:06

You have a set of $$(x_1,x_2)$$ points which belong to two different classes. You'll enumerate the two classes as $$0$$ and $$1$$, which will be your target outputs, i.e. $$y$$'s. In your implementation, you use $$y=x_2$$, which isn't correct. For example, the cross entropy you use assumes that $$y$$ is either $$0$$ or $$1$$. Therefore, since you have a 2D dataset, your boundary line will be in the form $$\theta_0+\theta_1x_1+\theta_2x_2=0$$, where you use both coordinates. While classifying you'll compare your logistic regression output, $$h=\sigma(\theta^Tx)$$ , with some threshold, $$\tau$$, commonly (but not necessarily) chosen as $$\tau=0$$ in balanced datasets like this, and decide class $$1$$ for $$h>1/2$$, $$0$$ otherwise. This is what a logistic regression is in a nutshell.

Note: In your dataset, the samples can be perfectly classified via only $$x_1$$, with some threshold, e.g. $$x_1>\tau\rightarrow$$ class $$1$$, but assume you don't have this information and you want to use all the features for simplicity. This is a specific situatuon to your dataset. However, even in this case, your $$y$$'s will be again $$0$$ and $$1$$, but nothing else.

Edit: Here, I'm making some modifications to make your code work. First of all, you do the training with only $$x_1$$ coordinate, in which you get estimates for only $$\theta_1,\theta_0$$. Since your data is 2D, as I've explained above, you need three variables, i.e. $$\theta_2,\theta_1,\theta_0$$, because you actually seek for a boundary equation in 2D plane. The reason you see decrease in your cost is the fact that your data is actually separable only with $$x_1$$ coordinate. This isn't wrong, but contradicts your aim to find a separating boundary line in x-y plane. If you use only $$x_1$$, you'll get a boundary equation in the form of a vertical line: $$\theta_1x_1+\theta_0=0$$, which is why your plot_y variable is oscillating around $$0$$, because you seem to calculate the RHS estimate of this equation in each iteration. The correct thing to do is using both $$x_1$$ and $$x_2$$ as your features, and predicting $$y$$, via estimates of $$\theta_{2-0}$$.

Another practical difficulty here is the scales of your features. $$x_2$$ has a considerably larger scale than $$x_1$$ and if you use SGD with constant learning rate as here, you need to choose a very small one to suit all, or use different learning rates for each feature. Better, use feature scaling. I've implemented it in your code:

clear all; close all; clc;

alpha = 0.1;
num_iters = 1000;

%% Plotting data
x1 = linspace(0,3,50);
mqtrue = 5;
cqtrue = 30;
dat1 = mqtrue*x1+5*randn(1,50);

x2 = linspace(7,10,50);
dat2 = mqtrue*x2 + (cqtrue + 5*randn(1,50));

x = [x1 x2]'; % X

dat = [dat1 dat2]'; % Y
x = [x dat];

scatter(x1, dat1); hold on;
scatter(x2, dat2, '*'); hold on;
classdata = (dat>40);

%  Setup the data matrix appropriately, and add ones for the intercept term
[m, n] = size(x);

minX = min(x);
maxX = max(x);

x = (x - repmat(minX,m,1)) ./ repmat(maxX-minX,m,1);
plot_x = [min(x(:,2)),  max(x(:,2))];

% Add intercept term to x and X_test
x = [ones(m, 1) x];

% Initialize fitting parameters
theta = zeros(n + 1, 1);
%initial_theta = [0.2; 0.2];

J_history = zeros(num_iters, 1);

% alpha = [0.1 0.01 0.001]';
for iter = 1:num_iters
% Compute and display initial cost and gradient
[cost, grad] = logistic_costFunction(theta, x, classdata);
theta = theta - alpha .* grad;
J_history(iter) = cost;

fprintf('Iteration #%d - Cost = %d... \r\n',iter, cost);
end

plot_y = -(theta(2)*plot_x + theta(1)) / theta(3);  % <--- Boundary line
plot((plot_x*(maxX(1)-minX(1))+minX(1)), plot_y*(maxX(2)-minX(2))+minX(2))

fprintf('Cost at initial theta (zeros): %f\n', cost);
fprintf('Gradient at initial theta (zeros): \n');
Note that, the correct boundary line is: $$x_2=-\frac{\theta_1x_1+\theta_0}{\theta_2}$$ And, the output is:
• your X matrix should be $n\times3$, where each line is $[x_{i1}, x_{i2}, 1]$ – gunes Apr 19 at 11:42