How to plot logistic regression decision boundary?
$begingroup$
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);
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... rn',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): %fn', 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.
machine-learning logistic-regression
$endgroup$
add a comment |
$begingroup$
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);
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... rn',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): %fn', 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.
machine-learning logistic-regression
$endgroup$
add a comment |
$begingroup$
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);
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... rn',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): %fn', 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.
machine-learning logistic-regression
$endgroup$
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);
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... rn',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): %fn', 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.
machine-learning logistic-regression
machine-learning logistic-regression
asked 21 hours ago
Rrz0Rrz0
1738
1738
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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"]
I Append
the response variable to the data.
datPlot =
ListPointPlot3D[
MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
]
Perform a Logistic regression (LogitModelFit
). You could use GeneralizedLinearModelFit
with ExponentialFamily
set to "Binomial"
as well.
With[{dat = Join[dat1, dat2]},
model =
LogitModelFit[
MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
{x, y}, {x, y}]
]
From the FittedModel
"Properties"
we need "Function"
.
model["Properties"]
{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}
model["Function"]
Use this for prediction
model["Function"][8, 54]
0.0196842
and plot the decision boundary surface in 3D along with the data (datPlot
) using Show
and Plot3D
modelPlot =
Show[
datPlot,
Plot3D[
model["Function"][x, y],
Evaluate[
Sequence @@
MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
Mesh -> None,
PlotStyle -> Opacity[.25, Green],
PlotPoints -> 30
]
]
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.
Manipulate[
Show[
modelPlot,
ParametricPlot3D[
{x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
ParametricPlot3D[
{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"}
]
You can also project contours of this surface into 2D (Plot
). For example, keeping y
fixed and letting x
vary.
yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
Manipulate[
Show[
ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
PlotTheme -> "Detailed"],
Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
Exclusions -> None]
],
{{y, 40}, 0, yMax, Appearance -> "Labeled"}
]
Update
You can also plot contours of the probability in 2D.
plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];
Manipulate[
db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
Show[
plot,
Plot[db, {x, 0, 10}, PlotStyle -> Red]
],
{{p, .5}, 0, 1, Appearance -> "Labeled"}
]
Hope this helps.
$endgroup$
$begingroup$
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
10 hours ago
1
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
add a comment |
$begingroup$
Regarding the code
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.
Decision boundary
Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat)
or (x, 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
$${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
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$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
decision boundary can be derived as follows
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
end{align*}$$
For two dimensional input $boldsymbol{x}=(x_1, x_2)$ 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:
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
&Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
end{align*}$$
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
oldest
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active
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$begingroup$
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"]
I Append
the response variable to the data.
datPlot =
ListPointPlot3D[
MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
]
Perform a Logistic regression (LogitModelFit
). You could use GeneralizedLinearModelFit
with ExponentialFamily
set to "Binomial"
as well.
With[{dat = Join[dat1, dat2]},
model =
LogitModelFit[
MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
{x, y}, {x, y}]
]
From the FittedModel
"Properties"
we need "Function"
.
model["Properties"]
{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}
model["Function"]
Use this for prediction
model["Function"][8, 54]
0.0196842
and plot the decision boundary surface in 3D along with the data (datPlot
) using Show
and Plot3D
modelPlot =
Show[
datPlot,
Plot3D[
model["Function"][x, y],
Evaluate[
Sequence @@
MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
Mesh -> None,
PlotStyle -> Opacity[.25, Green],
PlotPoints -> 30
]
]
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.
Manipulate[
Show[
modelPlot,
ParametricPlot3D[
{x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
ParametricPlot3D[
{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"}
]
You can also project contours of this surface into 2D (Plot
). For example, keeping y
fixed and letting x
vary.
yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
Manipulate[
Show[
ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
PlotTheme -> "Detailed"],
Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
Exclusions -> None]
],
{{y, 40}, 0, yMax, Appearance -> "Labeled"}
]
Update
You can also plot contours of the probability in 2D.
plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];
Manipulate[
db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
Show[
plot,
Plot[db, {x, 0, 10}, PlotStyle -> Red]
],
{{p, .5}, 0, 1, Appearance -> "Labeled"}
]
Hope this helps.
$endgroup$
$begingroup$
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
10 hours ago
1
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
add a comment |
$begingroup$
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"]
I Append
the response variable to the data.
datPlot =
ListPointPlot3D[
MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
]
Perform a Logistic regression (LogitModelFit
). You could use GeneralizedLinearModelFit
with ExponentialFamily
set to "Binomial"
as well.
With[{dat = Join[dat1, dat2]},
model =
LogitModelFit[
MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
{x, y}, {x, y}]
]
From the FittedModel
"Properties"
we need "Function"
.
model["Properties"]
{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}
model["Function"]
Use this for prediction
model["Function"][8, 54]
0.0196842
and plot the decision boundary surface in 3D along with the data (datPlot
) using Show
and Plot3D
modelPlot =
Show[
datPlot,
Plot3D[
model["Function"][x, y],
Evaluate[
Sequence @@
MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
Mesh -> None,
PlotStyle -> Opacity[.25, Green],
PlotPoints -> 30
]
]
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.
Manipulate[
Show[
modelPlot,
ParametricPlot3D[
{x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
ParametricPlot3D[
{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"}
]
You can also project contours of this surface into 2D (Plot
). For example, keeping y
fixed and letting x
vary.
yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
Manipulate[
Show[
ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
PlotTheme -> "Detailed"],
Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
Exclusions -> None]
],
{{y, 40}, 0, yMax, Appearance -> "Labeled"}
]
Update
You can also plot contours of the probability in 2D.
plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];
Manipulate[
db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
Show[
plot,
Plot[db, {x, 0, 10}, PlotStyle -> Red]
],
{{p, .5}, 0, 1, Appearance -> "Labeled"}
]
Hope this helps.
$endgroup$
$begingroup$
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
10 hours ago
1
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
add a comment |
$begingroup$
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"]
I Append
the response variable to the data.
datPlot =
ListPointPlot3D[
MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
]
Perform a Logistic regression (LogitModelFit
). You could use GeneralizedLinearModelFit
with ExponentialFamily
set to "Binomial"
as well.
With[{dat = Join[dat1, dat2]},
model =
LogitModelFit[
MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
{x, y}, {x, y}]
]
From the FittedModel
"Properties"
we need "Function"
.
model["Properties"]
{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}
model["Function"]
Use this for prediction
model["Function"][8, 54]
0.0196842
and plot the decision boundary surface in 3D along with the data (datPlot
) using Show
and Plot3D
modelPlot =
Show[
datPlot,
Plot3D[
model["Function"][x, y],
Evaluate[
Sequence @@
MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
Mesh -> None,
PlotStyle -> Opacity[.25, Green],
PlotPoints -> 30
]
]
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.
Manipulate[
Show[
modelPlot,
ParametricPlot3D[
{x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
ParametricPlot3D[
{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"}
]
You can also project contours of this surface into 2D (Plot
). For example, keeping y
fixed and letting x
vary.
yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
Manipulate[
Show[
ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
PlotTheme -> "Detailed"],
Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
Exclusions -> None]
],
{{y, 40}, 0, yMax, Appearance -> "Labeled"}
]
Update
You can also plot contours of the probability in 2D.
plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];
Manipulate[
db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
Show[
plot,
Plot[db, {x, 0, 10}, PlotStyle -> Red]
],
{{p, .5}, 0, 1, Appearance -> "Labeled"}
]
Hope this helps.
$endgroup$
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"]
I Append
the response variable to the data.
datPlot =
ListPointPlot3D[
MapThread[Append, {#, Boole@Thread[#[[All, 2]] > 40]}] & /@ {dat1, dat2}
]
Perform a Logistic regression (LogitModelFit
). You could use GeneralizedLinearModelFit
with ExponentialFamily
set to "Binomial"
as well.
With[{dat = Join[dat1, dat2]},
model =
LogitModelFit[
MapThread[Append, {dat, Boole@Thread[dat[[All, 2]] > 40]}],
{x, y}, {x, y}]
]
From the FittedModel
"Properties"
we need "Function"
.
model["Properties"]
{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}
model["Function"]
Use this for prediction
model["Function"][8, 54]
0.0196842
and plot the decision boundary surface in 3D along with the data (datPlot
) using Show
and Plot3D
modelPlot =
Show[
datPlot,
Plot3D[
model["Function"][x, y],
Evaluate[
Sequence @@
MapThread[Prepend, {MinMax /@ Transpose@Join[dat1, dat2], {x, y}}]],
Mesh -> None,
PlotStyle -> Opacity[.25, Green],
PlotPoints -> 30
]
]
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.
Manipulate[
Show[
modelPlot,
ParametricPlot3D[
{x, u, model["Function"][x, u]}, {u, 0, 80}, PlotStyle -> Orange],
ParametricPlot3D[
{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"}
]
You can also project contours of this surface into 2D (Plot
). For example, keeping y
fixed and letting x
vary.
yMax = Ceiling@*Max@Join[dat1, dat2][[All, 2]];
Manipulate[
Show[
ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers",
PlotTheme -> "Detailed"],
Plot[yMax model["Function"][x, y], {x, 0, 10}, PlotStyle -> Purple,
Exclusions -> None]
],
{{y, 40}, 0, yMax, Appearance -> "Labeled"}
]
Update
You can also plot contours of the probability in 2D.
plot = ListPlot[{dat1, dat2}, PlotMarkers -> "OpenMarkers", PlotTheme -> "Detailed"];
Manipulate[
db = y /. First@Quiet@Solve[model["Function"][x, y] == p, y];
Show[
plot,
Plot[db, {x, 0, 10}, PlotStyle -> Red]
],
{{p, .5}, 0, 1, Appearance -> "Labeled"}
]
Hope this helps.
edited 7 hours ago
answered 16 hours ago
EdmundEdmund
250311
250311
$begingroup$
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
10 hours ago
1
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
add a comment |
$begingroup$
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
10 hours ago
1
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
$begingroup$
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
10 hours ago
$begingroup$
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
10 hours ago
1
1
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
$begingroup$
@Esmailian See update.
$endgroup$
– Edmund
7 hours ago
add a comment |
$begingroup$
Regarding the code
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.
Decision boundary
Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat)
or (x, 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
$${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
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$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
decision boundary can be derived as follows
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
end{align*}$$
For two dimensional input $boldsymbol{x}=(x_1, x_2)$ 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:
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
&Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
end{align*}$$
$endgroup$
add a comment |
$begingroup$
Regarding the code
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.
Decision boundary
Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat)
or (x, 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
$${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
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$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
decision boundary can be derived as follows
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
end{align*}$$
For two dimensional input $boldsymbol{x}=(x_1, x_2)$ 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:
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
&Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
end{align*}$$
$endgroup$
add a comment |
$begingroup$
Regarding the code
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.
Decision boundary
Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat)
or (x, 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
$${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
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$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
decision boundary can be derived as follows
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
end{align*}$$
For two dimensional input $boldsymbol{x}=(x_1, x_2)$ 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:
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
&Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
end{align*}$$
$endgroup$
Regarding the code
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.
Decision boundary
Assuming that input is $boldsymbol{x}=(x_1, x_2)$ ((x, dat)
or (x, 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
$${Bbb P}(y=1|boldsymbol{x})=frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}}$$
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$$boldsymbol{theta}^tboldsymbol{x_+}=theta_0 + theta_1 x_1+cdots+theta_d x_d,$$
decision boundary can be derived as follows
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{2} \
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = 0\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = 0
end{align*}$$
For two dimensional input $boldsymbol{x}=(x_1, x_2)$ 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:
$$begin{align*}
&frac{1}{1+e^{-boldsymbol{theta}^tboldsymbol{x_+}}} = frac{1}{w+1} \
&Rightarrow e^{-boldsymbol{theta}^tboldsymbol{x_+}} = w\
&Rightarrow boldsymbol{theta}^tboldsymbol{x_+} = -text{ln}w\
&Rightarrow theta_0 + theta_1 x_1+cdots+theta_d x_d = -text{ln}w
end{align*}$$
edited 12 hours ago
answered 19 hours ago
EsmailianEsmailian
3,486420
3,486420
add a comment |
add a comment |
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