Warning: Converting non-floating point data to single - matlab

I've been working with feature extraction and epipolar geometry. However, I'm constantly coming across the the following error:
Warning: Converting non-floating point data to single.
In pdist2 (line 219)
In extractFeaturesU (line 93)
The code line that is returning a warning message is:
[distance, position] = sort(pdist2(double(repmat(featuresA, size(xPoints, 1))), featuresB), 2, 'ascend');
The part of the code containing the above line is displayed below.
%% extract features
corresponding = [];
rightBound = size(sharpImageB, 2);
xPoints = 3 : 3 : rightBound - 3;
for index = 1 : size(realWantedPoints, 2)
%extract features from wanted points
disp('extracting features from wanted points...');
if strcmp(desc, 'hog')
[featuresA, pointsA] = extractHOGFeatures(sharpImageA, realWantedPoints(:, index)', ...
'CellSize', [8 8], 'BlockSize', [2 2], 'NumBins', 9, 'UseSignedOrientation', true);
elseif strcmp(desc, 'block')
[featuresA, pointsA] = extractFeatures(sharpImageA, realWantedPoints(:, index)', ...
'Method', 'Block', 'BlockSize', 21, 'Upright', true);
elseif strcmp(desc, 'surf')
[featuresA, pointsA] = extractFeatures(sharpImageA, realWantedPoints(:, index)', ...
'Method', 'SURF', 'SURFSize', 64, 'Upright', true);
end
% generate epipolar line points
liner = star([1 0 0]) * [realWantedPoints(:, index); 1];
yPoints = -(liner(3) + (liner(1) * xPoints)) / liner(2);
matrixB = [xPoints', yPoints'];
% extract features from epipolar line points
disp('extracting features from epipolar line points...');
if strcmp('hog', desc)
[featuresB, pointsB] = extractHOGFeatures(sharpImageB, matrixB, ...
'CellSize', [8 8], 'BlockSize', [2 2], 'NumBins', 9, 'UseSignedOrientation', true);
elseif strcmp('block', desc)
[featuresB, pointsB] = extractFeatures(sharpImageB, matrixB, ...
'Method', 'Block', 'BlockSize', 21, 'Upright', true);
elseif strcmp('surf', desc)
[featuresB, pointsB] = extractFeatures(greyB, matrixB, ...
'Method', 'SURF', 'SURFSize', 64, 'Upright', true);
end
% calculate similarity
[distance, position] = sort(pdist2(double(repmat(featuresA, size(xPoints, 1))), featuresB), 2, 'ascend');
corresponding = [corresponding; pointsB(position(1), :)];
end
xB = corresponding(:, 1);
yB = corresponding(:, 2);

I can generate that error by calling pdist2 on two variables: one of type double and one of type single. Eg.
x = ones(5,1,'single');
y = ones(5,1,'double');
pdist2(x,y);
My guess is that your featuresB variable is single precision floating point and hence doesn't match the type of your first argument to pdist2 (which is double because you explicitly convert it).

Related

Unable to perform assignment because the size of the left side is 1-by-7-by-7 and the size of the right side is 6-by-6 [closed]

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I am looking for a way to find same eigenvectors for 2 given matrices, this way I would make a joint diagonalisation. For this, I found out and tried to use qndiag (from https://github.com/pierreablin/qndiag.git ) from the following function :
function [D, B, infos] = qndiag(C, varargin)
% Joint diagonalization of matrices using the quasi-Newton method
%
% The algorithm is detailed in:
%
% P. Ablin, J.F. Cardoso and A. Gramfort. Beyond Pham’s algorithm
% for joint diagonalization. Proc. ESANN 2019.
% https://www.elen.ucl.ac.be/Proceedings/esann/esannpdf/es2019-119.pdf
% https://hal.archives-ouvertes.fr/hal-01936887v1
% https://arxiv.org/abs/1811.11433
%
% The function takes as input a set of matrices of size `(p, p)`, stored as
% a `(n, p, p)` array, `C`. It outputs a `(p, p)` array, `B`, such that the
% matrices `B * C(i,:,:) * B'` are as diagonal as possible.
%
% There are several optional parameters which can be provided in the
% varargin variable.
%
% Optional parameters:
% --------------------
% 'B0' Initial point for the algorithm.
% If absent, a whitener is used.
% 'weights' Weights for each matrix in the loss:
% L = sum(weights * KL(C, C')).
% No weighting (weights = 1) by default.
% 'maxiter' (int) Maximum number of iterations to perform.
% Default : 1000
%
% 'tol' (float) A positive scalar giving the tolerance at
% which the algorithm is considered to have converged.
% The algorithm stops when |gradient| < tol.
% Default : 1e-6
%
% lambda_min (float) A positive regularization scalar. Each
% eigenvalue of the Hessian approximation below
% lambda_min is set to lambda_min.
%
% max_ls_tries (int), Maximum number of line-search tries to
% perform.
%
% return_B_list (bool) Chooses whether or not to return the list
% of iterates.
%
% verbose (bool) Prints informations about the state of the
% algorithm if True.
%
% Returns
% -------
% D : Set of matrices jointly diagonalized
% B : Estimated joint diagonalizer matrix.
% infos : structure containing monitoring informations, containing the times,
% gradient norms and objective values.
%
% Example:
% --------
%
% [D, B] = qndiag(C, 'maxiter', 100, 'tol', 1e-5)
%
% Authors: Pierre Ablin <pierre.ablin#inria.fr>
% Alexandre Gramfort <alexandre.gramfort#inria.fr>
%
% License: MIT
% First tests
if nargin == 0,
error('No signal provided');
end
if length(size(C)) ~= 3,
error('Input C should be 3 dimensional');
end
if ~isa (C, 'double'),
fprintf ('Converting input data to double...');
X = double(X);
end
% Default parameters
C_mean = squeeze(mean(C, 1));
[p, d] = eigs(C_mean);
p = fliplr(p);
d = flip(diag(d));
B = p' ./ repmat(sqrt(d), 1, size(p, 1));
max_iter = 1000;
tol = 1e-6;
lambda_min = 1e-4;
max_ls_tries = 10;
return_B_list = false;
verbose = false;
weights = [];
% Read varargin
if mod(length(varargin), 2) == 1,
error('There should be an even number of optional parameters');
end
for i = 1:2:length(varargin)
param = lower(varargin{i});
value = varargin{i + 1};
switch param
case 'B0'
B0 = value;
case 'max_iter'
max_iter = value;
case 'tol'
tol = value;
case 'weights'
weights = value / mean(value(:));
case 'lambda_min'
lambda_min = value;
case 'max_ls_tries'
max_ls_tries = value;
case 'return_B_list'
return_B_list = value;
case 'verbose'
verbose = value;
otherwise
error(['Parameter ''' param ''' unknown'])
end
end
[n_samples, n_features, ~] = size(C);
D = transform_set(B, C, false);
current_loss = NaN;
% Monitoring
if return_B_list
B_list = []
end
t_list = [];
gradient_list = [];
loss_list = [];
if verbose
print('Running quasi-Newton for joint diagonalization');
print('iter | obj | gradient');
end
for t=1:max_iter
if return_B_list
B_list(k) = B;
end
diagonals = zeros(n_samples, n_features);
for k=1:n_samples
diagonals(k, :) = diag(squeeze(D(k, :, :)));
end
% Gradient
if isempty(weights)
G = squeeze(mean(bsxfun(#rdivide, D, ...
reshape(diagonals, n_samples, n_features, 1)), ...
1)) - eye(n_features);
else
G = squeeze(mean(...
bsxfun(#times, ...
reshape(weights, n_samples, 1, 1), ...
bsxfun(#rdivide, D, ...
reshape(diagonals, n_samples, n_features, 1))), ...
1)) - eye(n_features);
end
g_norm = norm(G);
if g_norm < tol
break
end
% Hessian coefficients
if isempty(weights)
h = mean(bsxfun(#rdivide, ...
reshape(diagonals, n_samples, 1, n_features), ...
reshape(diagonals, n_samples, n_features, 1)), 1);
else
h = mean(bsxfun(#times, ...
reshape(weights, n_samples, 1, 1), ...
bsxfun(#rdivide, ...
reshape(diagonals, n_samples, 1, n_features), ...
reshape(diagonals, n_samples, n_features, 1))), ...
1);
end
h = squeeze(h);
% Quasi-Newton's direction
dt = h .* h' - 1.;
dt(dt < lambda_min) = lambda_min; % Regularize
direction = -(G .* h' - G') ./ dt;
% Line search
[success, new_D, new_B, new_loss, direction] = ...
linesearch(D, B, direction, current_loss, max_ls_tries, weights);
D = new_D;
B = new_B;
current_loss = new_loss;
% Monitoring
gradient_list(t) = g_norm;
loss_list(t) = current_loss;
if verbose
print(sprintf('%d - %.2e - %.2e', t, current_loss, g_norm))
end
end
infos = struct();
infos.t_list = t_list;
infos.gradient_list = gradient_list;
infos.loss_list = loss_list;
if return_B_list
infos.B_list = B_list
end
end
function [op] = transform_set(M, D, diag_only)
[n, p, ~] = size(D);
if ~diag_only
op = zeros(n, p, p);
for k=1:length(D)
op(k, :, :) = M * squeeze(D(k, :, :)) * M';
end
else
op = zeros(n, p);
for k=1:length(D)
op(k, :) = sum(M .* (squeeze(D(k, :, :)) * M'), 1);
end
end
end
function [v] = slogdet(A)
v = log(abs(det(A)));
end
function [out] = loss(B, D, is_diag, weights)
[n, p, ~] = size(D);
if ~is_diag
diagonals = zeros(n, p);
for k=1:n
diagonals(k, :) = diag(squeeze(D(k, :, :)));
end
else
diagonals = D;
end
logdet = -slogdet(B);
if ~isempty(weights)
diagonals = bsxfun(#times, diagonals, reshape(weights, n, 1));
end
out = logdet + 0.5 * sum(log(diagonals(:))) / n;
end
function [success, new_D, new_B, new_loss, delta] = linesearch(D, B, direction, current_loss, n_ls_tries, weights)
[n, p, ~] = size(D);
step = 1.;
if current_loss == NaN
current_loss = loss(B, D, false);
end
success = false;
for n=1:n_ls_tries
M = eye(p) + step * direction;
new_D = transform_set(M, D, true);
new_B = M * B;
new_loss = loss(new_B, new_D, true, weights);
if new_loss < current_loss
success = true;
break
end
step = step / 2;
end
new_D = transform_set(M, D, false);
delta = step * direction;
end
I use it with the following script that you can test with downloading the 2 input matrices at the bottom of this post :
clc; clear
m=7 % dimension
n=2 % number of matrices
% Load spectro and WL+GCph+XC
FISH_GCsp = load('Fisher_GCsp_flat.txt');
FISH_XC = load('Fisher_XC_GCph_WL_flat.txt');
% Marginalizing over uncommon parameters between the two matrices
COV_GCsp_first = inv(FISH_GCsp);
COV_XC_first = inv(FISH_XC);
COV_GCsp = COV_GCsp_first(1:m,1:m);
COV_XC = COV_XC_first(1:m,1:m);
% Invert to get Fisher matrix
FISH_sp = inv(COV_GCsp);
FISH_xc = inv(COV_XC);
% Drawing a random set of commuting matrices
C=zeros(n,m,m);
B0=zeros(n,m,m);
C(1,:,:) = FISH_sp
C(2,:,:) = FISH_xc
%[D, B] = qndiag(C, 'max_iter', 1e6, 'tol', 1e-6);
[D, B] = qndiag(C);
% Print diagonal matrices
B*C(1,:,:)*B'
B*C(2,:,:)*B'
But unfortunately, I get the following error :
Unable to perform assignment because the size of the left side is 1-by-7-by-7 and the size of the
right side is 6-by-6.
Error in qndiag>transform_set (line 224)
op(k, :, :) = M * squeeze(D(k, :, :)) * M';
Error in qndiag (line 128)
D = transform_set(B, C, false);
Error in compute_joint_diagonalization (line 32)
[D, B] = qndiag(C);
I don't understand the utility of function squeeze the most important : why the function eigs returns only 6 values and not 7 like in my data (the input matrices has 7x7 size).
What might be wrong with this issue of array dimension and how can I fix it ?
I put the 2 input files available here :
Matrix Fisher_GCsp_flat.txt
Matrix Fisher_XC_GCph_WL_flat.txt
You can test the above code that calls qndiag for these 2 matrices.
Update 1
To allow people interested to test quickly the code, I put a link of the archive:
Archive_Matlab_StackOver.tar
You just have to untar and execute under Matlab the script compute_joint_diagonalization.m and you will see normally the above error (regarding the use of eigs and squeeze functions).
It should help you understand the origin of this issue.
Update 2
If I replace [p, d] = eigs(C_mean) by [p, d] = eigs(C_mean,7) , I get another error :
Index in position 1 exceeds array bounds (must not exceed 2).
Error in qndiag>transform_set (line 224)
op(k, :, :) = M * squeeze(D(k, :, :)) * M';
Error in qndiag (line 128)
D = transform_set(B, C, false);
Error in compute_joint_diagonalization (line 27)
[D, B] = qndiag(C);
However, the dimensions of the 2 matrices used are 7x7 and should be correctly processed with eigs(C_mean,7).
Update 3
The size of op, D, M and k are equal to (including after the error message) :
size(D) =
2 7 7
length(D) =
7
size(M) =
7 7
size(op) =
2 7 7
Index in position 1 exceeds array bounds (must not exceed 2).
Error in qndiag>transform_set (line 231)
op(k, :, :) = M * squeeze(D(k, :, :)) * M';
Error in qndiag (line 128)
D = transform_set(B, C, false);
Error in compute_joint_diagonalization (line 27)
[D, B] = qndiag(C);
Notice that k varies from 1 to length(D)=7.
Is there an issue which could appear with these dimensions ?
From the documentation for eigs:
d = eigs(A) returns a vector of the six largest magnitude eigenvalues of matrix A.
If you want all seven, you need to call d = eigs(A,7) or d = eig(A). For a small matrix (e.g. < 1000 x 1000) it's usually easier to just get all the eigenvalues with eig, rather than get a subset with eigs.
Edit: Responding to your "Update 3"
for k=1:length(D) should be replaced by for k=1:n. This needs to be changed on two lines. Judging from your error message they are lines 231 and 236.
L = length(X) returns the length of the largest array dimension in X, which in your case is 7, i.e. too high for the first dimension.

connected points and smallest value

I would like to know how I can to take each D(Points) and look at its connexed points (in 8 connex) but only on the side of the limit (ie the diagonal points top and bottom right and on the same line to the right, ie 3 points Connect on 8) and select the coordinates of the connexed point which has the smallest value of D. And I would like to repeat this until I obtain that the smallest value of D equal 0
% Creation of matrix example
c=zeros(500,500);
c(1:100,250)=1;c(100:300,200)=1;c(300:400,270)=1; c(400:500,250)=1;
c(100,200:250)=1;c(300,200:270)=1;c(400,250:270)=1;
figure, imagesc(c)
Points= [211,388;64,200;160,437;237,478;110,270;100,34];
hold on, plot(Points(:,1),Points(:,2),'ro'), hold off
%Distance map
D = bwdist(cumsum(c, 2) > 0, 'euclidean');
figure, imagesc(D)
The key function here is sub2ind which converts subscripts to linear indices. It is very handy when you need to work on specific points inside an array.
% Let's prepare the 8 shifts needed (i add a no-move shift in first place to simplify the algorithm)
delta_x = [0, -1, -1, -1, 0, 0, 1, 1, 1];
delta_y = [0, -1, 0, 1, -1, 1, -1, 0, 1];
sz_D = size(D);
n_points = size(Points, 1);
is_running = true;
while is_running
% All the shift combinaisons
new_ind_x = bsxfun(#plus, Points(:,1), delta_x);
new_ind_y = bsxfun(#plus, Points(:,2), delta_y);
% Saturation to stay in the image
new_ind_x = min(max(new_ind_x, 1), sz_D(2));
new_ind_y = min(max(new_ind_y, 1), sz_D(1));
% Get the values in D and find the index of the minimum points
points_lin_ind = sub2ind(sz_D, new_ind_y, new_ind_x);
points_val = D(points_lin_ind);
[min_values, min_ind] = min(points_val, [], 2);
% Select subscripts in new_ind_x and new_ind_y
min_lin_ind = sub2ind([n_points, 9], (1:n_points).', min_ind);
Points = [new_ind_x(min_lin_ind), new_ind_y(min_lin_ind)];
% Exit condition
if all(min_values == 0)
is_running = false;
end
end
PS : Not tested.

Why is my Matlab hgtransform matrix invalid?

I am attempting to apply a transform matrix by setting the 'Matrix' property of a matlab hgtransform object. The transform matrix is below:
866.0254e-003 500.0000e-003 0.0000e+000 500.0000e-003
500.0000e-003 -866.0254e-003 0.0000e+000 500.0000e-003
0.0000e+000 0.0000e+000 1.0000e+000 0.0000e+000
0.0000e+000 0.0000e+000 0.0000e+000 1.0000e+000
This particular matrix is intended to represent a translation
(0.5, 0.5, 0)
and rotation around the Z axis of pi/6.
When I try to do this:
% make a unit box
sx = 1;
sy = 1;
sz = 1;
shapeData.Vertices = [ -sx/2, -sy/2, -sz/2;
sx/2, -sy/2, -sz/2;
sx/2, sy/2, -sz/2;
-sx/2, sy/2, -sz/2;
-sx/2, -sy/2, sz/2;
sx/2, -sy/2, sz/2;
sx/2, sy/2, sz/2;
-sx/2, sy/2, sz/2; ];
shapeData.Faces = [ 1, 4, 3, 2;
1, 5, 6, 2;
2, 6, 7, 3;
7, 8, 4, 3;
8, 5, 1, 4;
8, 7, 6, 5 ];
figure;
axes;
transformObject = hgtransform (gca);
patchObject = patch (gca, ...
'Faces', shapeData.Faces, ...
'Vertices', shapeData.Vertices, ...
'FaceColor', 'red', ...
'FaceAlpha', 1.0, ...
'EdgeColor', 'none', ...
'FaceLighting', 'gouraud', ...
'AmbientStrength', 0.15, ...
'Parent', transformObject);
M = [ ...
866.0254e-003 500.0000e-003 0.0000e+000 500.0000e-003; ...
500.0000e-003 -866.0254e-003 0.0000e+000 500.0000e-003; ...
0.0000e+000 0.0000e+000 1.0000e+000 0.0000e+000; ...
0.0000e+000 0.0000e+000 0.0000e+000 1.0000e+000; ...
];
set ( transformObject, 'Matrix', M );
I get the error:
Error using matlab.graphics.primitive.Transform/set
Invalid value for Matrix property
Why?
EDIT
The code that generated the transform matrix. First you need the following class which constructs orientation (rotation) matrices:
classdef orientmat
properties (GetAccess = public, SetAccess = protected)
orientationMatrix;
end
methods
function this = orientmat (spectype, spec)
% orentmat constructor
%
% Syntax
%
% om = orientmat (spectype, spec)
%
% Input
%
%
switch spectype
case 'orientation'
this.orientationMatrix = spec;
case 'euler'
this.orientationMatrix = SpinCalc('EA123toDCM', rad2deg (spec), eps (), 1);
case 'euler123'
this.orientationMatrix = SpinCalc('EA123toDCM', rad2deg (spec), eps (), 1);
case 'euler321'
this.orientationMatrix = SpinCalc('EA321toDCM', rad2deg (spec), eps (), 1);
case 'vector'
% axis and angle (angle in rad = norm of matrix)
wcrs = [ 0 spec(3) -spec(2)
-spec(3) 0 spec(1)
spec(2) -spec(1) 0] ;
this.orientationMatrix = expm (wcrs);
case '2vectors'
% normalise the fisr vector
spec.vec1 = this.unit (spec.vec1);
spec.vec2 = this.unit (spec.vec2);
spec.vec3 = cross (spec.vec1, spec.vec2);
spec.vec2 = this.unit (cross (this.unit (spec.vec3), spec.vec1));
switch spec.vec1axis
case 1
X = spec.vec1;
if spec.vec2axis == 2
Y = spec.vec2;
Z = spec.vec3;
elseif spec.vec2axis == 3
Y = spec.vec3;
Z = spec.vec2;
end
case 2
Y = spec.vec1;
if spec.vec2axis == 1
X = spec.vec2;
Z = spec.vec3;
elseif spec.vec2axis == 3
X = spec.vec3;
Z = spec.vec2;
end
case 3
Z = spec.vec1;
if spec.vec2axis == 2
X = spec.vec2;
Y = spec.vec3;
elseif spec.vec2axis == 3
X = spec.vec3;
Y = spec.vec2;
end
end
this.orientationMatrix = [ X, Y, Z ];
end
end
end
% operator overloading
methods
function om = plus (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix + om2.orientationMatrix);
end
function om = minus (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix - om2.orientationMatrix);
end
function om = times (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix .* om2.orientationMatrix);
end
function om = mtimes (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix * om2.orientationMatrix);
end
function om = double (om1)
om = om1.orientationMatrix;
end
function om = uminus (om1)
om = mbdyn.pre.orientmat ('orientation', -om1.orientationMatrix);
end
function om = uplus (om1)
om = mbdyn.pre.orientmat ('orientation', +om1.orientationMatrix);
end
function om = transpose (om1)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix.');
end
function om = ctranspose (om1)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix');
end
end
methods (Access = private)
function out = unit (self, vec)
out = vec ./ norm (vec);
end
end
end
Then do:
om = orientmat ('2vectors', struct ('vec1axis', 1, 'vec1', [cos(pi/6);sin(pi/6);0], 'vec2axis', 3, 'vec2', [0;0;1]));
M = [ om.orientationMatrix, [0.5; 0.5; 0]; 0, 0, 0, 1 ];
Now there may be an issue with the rotation not actually being what I intend, but as far as I can see it is still a valid transformation matrix?
The answer was that Matlab only accepts non-negative scaling terms, see Transforms Supported by hgtransform

On a a plotted graph I will like to know the area with maximum triangular intersection

I have the code below to link nine different coordinates in the plotted graph:
A1={[1, 1; 1, 5; 3, 9; 4, 2; 4, 6; 6, 2; 7, 6; 6, 9; 9, 9]};
A = cell2mat(A1);
figure
plot(A(:,1),A(:,2),'oc','LineWidth',2,'MarkerSize',5);
axis([0 10 0 10]);
xlabel('X-Coordinates')
ylabel('Y-Coordinates')
grid on
hold on
for ii = 1:size(A, 1) - 1
for jj = ii + 1:size(A, 1)
line([A(ii, 1), A(jj, 1)], [A(ii, 2), A(jj, 2)])
end
end
The linkages form different triangles. My problem is I will like to know the region with the highest number of triangular intersections.
Can anyone please help with this problem? Thanks
The code below answered my question:
A1={[1, 1; 1, 5; 3, 9; 4, 2; 4, 6; 6, 2; 7, 6; 6, 9; 9, 9]};
A = cell2mat(A1);
k = boundary(A);
hold on;
plot(A(:,1),A(:,2),'oc','LineWidth',2,'MarkerSize',5);
axis([0 10 0 10]);
xlabel('X-Coordinates')
ylabel('Y-Coordinates')
grid on
for ii = 1:size(A, 1) - 1
for jj = ii + 1:size(A, 1)
line([A(ii, 1), A(jj, 1)], [A(ii, 2), A(jj, 2)])
end
end
% Create the combination of all points that make the triangles
% This could be used to plot the lines as well
N = size(A,1);
comb = [];
for i = 1:N-2
for j = i+1:N-1
comb = [comb; repmat([i j], N-j,1) (j+1:N)']; %#ok<AGROW>
end
end
nComb = size(comb,1);
% Create a mesh grid
dg = 0.01; % Resolution - tune this!
gridEdge = [min(A);max(A)];
[X, Y] = meshgrid(gridEdge(1,1):dg:gridEdge(2,1), gridEdge(1,2):dg:gridEdge(2,2));
% Check if a point is inside each triangle
[isInside, onEdge] = deal(zeros(numel(X),nComb));
for i = 1:nComb
[isInside(:,i), onEdge(:,i)] = inpolygon(X(:),Y(:),A(comb(i,:),1),A(comb(i,:),2));
end
% Remove points on edge
isInside = isInside - onEdge;
% Get index of points with most intersection
inTri = sum(isInside,2);
idx = find(inTri == max(inTri));
% Plot result
hold on
plot(X(idx),Y(idx),'.')
text(mean(X(idx)),mean(Y(:)),num2str(max(inTri)),'FontSize',20)
A big thank you to nilZ0r.

Finding an equation of linear classifier for two separable sets of points using perceptron learning

I would like to write a matlab function to find an equation of a linear classifier for 2 separable sets of points using one single-layer perceptron. I have got 2 files:
script file - run.m:
x_1 = [3, 3, 2, 4, 5];
y_1 = [3, 4, 5, 2, 2];
x_2 = [6, 7, 5, 9, 8];
y_2 = [3, 3, 4, 2, 5];
target_array = [0 0 0 0 0 1 1 1 1 1];
[ func ] = classify_perceptron([x_1 x_2; y_1 y_2], target_array);
x = -2:10;
y = arrayfun(func, x);
plot(x_1, y_1, 'o', x_2, y_2, 'X', x, y);
axis([-2, 10, -2, 10]);
classify_perceptron.m
function [ func ] = classify_perceptron( points, target )
% points - matrix of x,y coordinates
% target - array of expected results
% func - function handler which appropriately classifies a point
% given by x, y arguments supplied to this function
target_arr = target;
weights = rand(1, 2);
translation = rand();
for i=1:size(points, 2)
flag = true;
while flag
result = weights * points(:, i) + translation;
y = result > 0;
e = target_arr(1, i) - y;
if e ~= 0
weights = weights + (e * points(:, i))';
translation = translation + e;
else
flag = false;
end
end
end
func = #(x)(-(translation + (weights(1, 1) * x)) / weights(1, 2));
return
end
The problem is that I don't know where I am making the mistake that leads to incorrect result. It looks like the slope of the line is right, however translation should be a bit bigger. I would be really thankful for pointing me in the right direction. The result I get is presented in the picture below:
Ok, so I have made a significant progress. In case someone runs into the same problem I present to you the solution. The problem has been solved by adding a variable learning_rate = 0.1 and packing the loop iterating over points into another loop iterating as many times as specified in the variable epochs (e.g. 300) .