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

enter image description here

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.

enter image description here

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

dat = [dat1 dat2]'; % Y

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

Computing Cost, Gradient and plotting

%  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);

    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)));

    grid on;
    plot3(theta(1), theta(2), J_history(iter),'o')
    title(sprintf('Tita0 = %g, Tita1=%g', theta(1), theta(2)))
    hold on;

    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;


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

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.


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

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

    grad = zeros(size(theta));

    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) );



As per the below answer by @Esmailian, now I have something like this:

[m, n] = size(x);

x1_class = [ones(m, 1) x1' dat1'];
x2_class = [ones(m, 1) x2' dat2'];

x = [x1_class ; x2_class]

3 Answers 3


Regarding the code

  1. You should plot the decision boundary after training is finished, not inside the training loop, parameters are constantly changing there; unless you are tracking the change of decision boundary.

  2. x1 (x2) is the first feature and dat1 (dat2) is the second feature for the first (second) class, so the extended feature space x for both classes should be the union of (1, x1, dat1) and (1, x2, dat2).

Decision boundary

Assuming that data is $\boldsymbol{x}=(x_1, x_2)$ ((x, dat) or (plot_x, plot_y) in the code), and parameter is $\boldsymbol{\theta}=(\theta_0, \theta_1,\theta_2)$ ((theta(1), theta(2), theta(3)) in the code), here is the line that should be drawn as decision boundary: $$x_2 = -\frac{\theta_1}{\theta_2} x_1 - \frac{\theta_0}{\theta_2}$$ which can be drawn as a segment by connecting two points $(0, - \frac{\theta_0}{\theta_2})$ and $(- \frac{\theta_0}{\theta_1}, 0)$. However, if $\theta_2=0$, the line would be $x_1=-\frac{\theta_0}{\theta_1}$.

Where this comes from?

Decision boundary of Logistic regression is the set of all points $\boldsymbol{x}$ that satisfy $${\Bbb P}(y=1|\boldsymbol{x})={\Bbb P}(y=0|\boldsymbol{x}) = \frac{1}{2}.$$ Given $${\Bbb P}(y=1|\boldsymbol{x})=\frac{1}{1+e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}}}$$ where $\boldsymbol{\theta}=(\theta_0, \theta_1,\cdots,\theta_d)$, and $\boldsymbol{x}$ is extended to $\boldsymbol{x_+}=(1, x_1, \cdots, x_d)$ for the sake of readability to have$$\boldsymbol{\theta}^t\boldsymbol{x_+}=\theta_0 + \theta_1 x_1+\cdots+\theta_d x_d,$$ decision boundary can be derived as follows $$\begin{align*} &\frac{1}{1+e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}}} = \frac{1}{2} \\ &\Rightarrow \boldsymbol{\theta}^t\boldsymbol{x_+} = 0\\ &\Rightarrow \theta_0 + \theta_1 x_1+\cdots+\theta_d x_d = 0 \end{align*}$$ For two dimensional data $\boldsymbol{x}=(x_1, x_2)$ we have $$\begin{align*} & \theta_0 + \theta_1 x_1+\theta_2 x_2 = 0 \\ & \Rightarrow x_2 = -\frac{\theta_1}{\theta_2} x_1 - \frac{\theta_0}{\theta_2} \end{align*}$$ which is the separation line that should be drawn in $(x_1, x_2)$ plane.

Weighted decision boundary

If we want to weight the positive class ($y = 1$) more or less using $w$, here is the general decision boundary: $$w{\Bbb P}(y=1|\boldsymbol{x}) = {\Bbb P}(y=0|\boldsymbol{x}) = \frac{w}{w+1}$$

For example, $w=2$ means point $\boldsymbol{x}$ will be assigned to positive class if ${\Bbb P}(y=1|\boldsymbol{x}) > 0.33$ (or equivalently if ${\Bbb P}(y=0|\boldsymbol{x}) < 0.66$), which implies favoring the positive class (increasing the true positive rate).

Here is the line for this general case: $$\begin{align*} &\frac{1}{1+e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}}} = \frac{1}{w+1} \\ &\Rightarrow e^{-\boldsymbol{\theta}^t\boldsymbol{x_+}} = w\\ &\Rightarrow \boldsymbol{\theta}^t\boldsymbol{x_+} = -\text{ln}w\\ &\Rightarrow \theta_0 + \theta_1 x_1+\cdots+\theta_d x_d = -\text{ln}w \end{align*}$$

  • $\begingroup$ Thanks for your insight, I understand why I need three thetas and the weighted decision boundary you explain. I am still having trouble implementing this in code however. ABove you say: "Assuming that input x=(x1, x2) ... Isn't this what I present in the code above? $\endgroup$
    – rrz0
    Apr 20, 2019 at 8:53
  • $\begingroup$ Gradient descent is implemented correctly, but I am having issue with mismatching matrix dimensions when it comes to the input X and classdata. Shouldn't x be nx3 ? [x1, x2, 1] $\endgroup$
    – rrz0
    Apr 20, 2019 at 9:06
  • $\begingroup$ @Rrz0 that is the extended 3D version as I explained, which is actually X+ = [1, x, dat], you do not need this for drawing the line. Please check my last update regarding 'plot_x'. $\endgroup$
    – Esmailian
    Apr 20, 2019 at 9:11
  • $\begingroup$ Thanks for the update. I edited the question to show the cost function.. When computing the sigmoid we get X * theta. I believe this results in a dimension error if X is not [n x 3] $\endgroup$
    – rrz0
    Apr 20, 2019 at 9:15
  • 1
    $\begingroup$ @Rrz0 I spotted the problem. Check out the update. $\endgroup$
    – Esmailian
    Apr 20, 2019 at 9:29

Your decision boundary is a surface in 3D as your points are in 2D.

With Wolfram Language

Create the data sets.

mqtrue = 5;
cqtrue = 30;
With[{x = Subdivide[0, 3, 50]},
  dat1 = Transpose@{x, mqtrue x + 5 RandomReal[1, Length@x]};
With[{x = Subdivide[7, 10, 50]},
  dat2 = Transpose@{x, mqtrue x + cqtrue + 5 RandomReal[1, Length@x]};

View in 2D (ListPlot) and the 3D (ListPointPlot3D) regression space.

ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"]

Mathematica graphics

I Append the response variable to the data.

datPlot =
  MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}

enter image description here

Perform a Logistic regression (LogitModelFit). You could use GeneralizedLinearModelFit with ExponentialFamily set to "Binomial" as well.

With[{dat = Join[dat1, dat2]},
 model =
   MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
   {x, y}, {x, y}]

Mathematica graphics

From the FittedModel "Properties" we need "Function".


{AdjustedLikelihoodRatioIndex, DevianceTableDeviances, ParameterConfidenceIntervalTableEntries, AIC, DevianceTableEntries, ParameterConfidenceRegion, AnscombeResiduals, DevianceTableResidualDegreesOfFreedom, ParameterErrors, BasisFunctions, DevianceTableResidualDeviances, ParameterPValues, BestFit, EfronPseudoRSquared, ParameterTable, BestFitParameters, EstimatedDispersion, ParameterTableEntries, BIC, FitResiduals, ParameterZStatistics, CookDistances, Function, PearsonChiSquare, CorrelationMatrix, HatDiagonal, PearsonResiduals, CovarianceMatrix, LikelihoodRatioIndex, PredictedResponse, CoxSnellPseudoRSquared, LikelihoodRatioStatistic, Properties, CraggUhlerPseudoRSquared, LikelihoodResiduals, ResidualDeviance, Data, LinearPredictor, ResidualDegreesOfFreedom, DesignMatrix, LogLikelihood, Response, DevianceResiduals, NullDeviance, StandardizedDevianceResiduals, Deviances, NullDegreesOfFreedom, StandardizedPearsonResiduals, DevianceTable, ParameterConfidenceIntervals, WorkingResiduals, DevianceTableDegreesOfFreedom, ParameterConfidenceIntervalTable}


Mathematica graphics

Use this for prediction

model["Function"][8, 54]

and plot the decision boundary surface in 3D along with the data (datPlot) using Show and Plot3D

modelPlot =
   model["Function"][x, y],
    Sequence @@ 
     MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
   Mesh -> None,
   PlotStyle -> Opacity[.25, Green],
   PlotPoints -> 30

enter image description here

With ParametricPlot3D and Manipulate you can examine decision boundary curves for values of the variables. For example, keeping x fixed and letting y vary or vice versa.

   {x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
   {u, y, model["Function"][u, y]}, {u, 0, 10}, PlotStyle -> Purple],
  PlotLabel -> 
   StringTemplate["model[`1`, `2`] = `3`"] @@ {x, y, model["Function"][x, y]}
 {{x, 6, Style["x", Orange, Bold]}, 0, 10, Appearance -> "Labeled"},
 {{y, 40, Style["y", Purple, Bold]}, 0, 80, Appearance -> "Labeled"}

enter image description here


You can also plot contours of the probability in 2D.

plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];

 db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
  Plot[db, {x, 0, 10}, PlotStyle -> Red]
 {{p, .5}, 0, 1, Appearance -> "Labeled"}

enter image description here

Hope this helps.

  • $\begingroup$ Beautiful plots. Some important notes: Logistic regression is used by OP for "classification" in 2D space, therefore "decision boundary" should be drawn in the same dimension $d$ as feature space (2D here) and it is a straight 2D line (unlike the last plot), which is also not the same as those animated lines (it must be parallel to that waterfall). However, "output of logistic regression", i.e. $(\boldsymbol{x},P(y=1|\boldsymbol{x}))$, as you have beautifully illustrated, needs $d+1$ for visualization. $\endgroup$
    – Esmailian
    Apr 19, 2019 at 19:38
  • 1
    $\begingroup$ @Esmailian See update. $\endgroup$
    – Edmund
    Apr 19, 2019 at 22:56
  • $\begingroup$ Thank you for your very helpful insight and excellent visualizations. $\endgroup$
    – rrz0
    Apr 20, 2019 at 8:54
  • $\begingroup$ That's a pretty strange-looking decision boundary, though. Normally you'd want the boundary to be perpendicular to the line connecting the clusters, I'd say. This is a great way to spot fitting artefacts like that. $\endgroup$ Oct 9, 2020 at 10:44
  • $\begingroup$ @SjoerdSmit The red line in the last plot is a contour of the green "decision" surface. It is a line along the ramp that joins values zero and one on the z-axis. $\endgroup$
    – Edmund
    Oct 10, 2020 at 13:12

I have generally found that to be quite easy to do in Python without having to solve the boundary equation. (Code mostly adapted from that question in SO)

Import packages :

import numpy as np
from matplotlib import pyplot as plt
from sklearn.linear_model import LogisticRegression

Generate data :

mu_vec1 = np.array([0,10])
cov_mat1 = np.array([[10,8],[8,10]])
x1_samples = np.random.multivariate_normal(mu_vec1, cov_mat1, 100)
mu_vec1 = mu_vec1.reshape(1,2).T # to 1-col vector

mu_vec2 = np.array([10,70])
cov_mat2 = np.array([[10,8],[8,10]])
x2_samples = np.random.multivariate_normal(mu_vec2, cov_mat2, 100)
mu_vec2 = mu_vec2.reshape(1,2).T

Plot generated data and save the associated figure :

fig = plt.figure()
plt.scatter(x1_samples[:,0],x1_samples[:,1], c= 'blue', marker='o')
plt.scatter(x2_samples[:,0],x2_samples[:,1], c= 'orange', marker='o')


simulated data

Fit the logistic regression :

X = np.concatenate((x1_samples,x2_samples), axis = 0)
y = np.array([0]*100 + [1]*100)

model_logistic = LogisticRegression()
model_logistic.fit(X, y)

Create a mesh, predict the regression on that mesh, plot the associated contour along simulated data and save the plot :

h = .02  # step size in the mesh
x_min, x_max = X[:, 0].min() - 1, X[:, 0].max() + 1
y_min, y_max = X[:, 1].min() - 1, X[:, 1].max() + 1
xx, yy = np.meshgrid(np.arange(x_min, x_max, h),
                     np.arange(y_min, y_max, h))

Z = model_logistic.predict(np.c_[xx.ravel(), yy.ravel()])

fig = plt.figure()
plt.scatter(x1_samples[:,0],x1_samples[:,1], c= 'blue', marker='o')
plt.scatter(x2_samples[:,0],x2_samples[:,1], c= 'orange', marker='o')

Z = Z.reshape(xx.shape)
plt.contour(xx, yy, Z, 1, colors='black')


data + decision boundary

This generally works for more complex dependencies (non linearities in variables) or more complex models (svm as per linked SO answer).

  • $\begingroup$ Don't you have to add the constant term; the intercept? # Intercept (a.k.a. bias) added to the decision function. (theta 0) parameter0 = log_reg.intercept_ That is, if you do not, then the line will be different than in the prior answer (see above and stackoverflow.com/a/55199096). $\endgroup$
    – Carl
    Apr 4 at 5:20
  • $\begingroup$ Yes you are right. The problem being separable, you'll need to handle the constants carefully. probably not a good toy exemple. Nowadays I would use this: scikit-learn.org/stable/modules/generated/… $\endgroup$ Apr 4 at 7:17
  • $\begingroup$ If you fix the code, I will upvote it. (I usually don't downvote.) $\endgroup$
    – Carl
    Apr 4 at 20:39
  • $\begingroup$ I honestly don't have time to get back in this, esp. on a degenerate toy problem. If it bother you so much you can suggests edits to other people answers. $\endgroup$ Apr 5 at 5:53

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