I'm trying to make my own project, learning about linear regression.
For some reasone I got wrong linear regression.
I've tried changing alfa values and iterations but it didnt help.
I was checking also for other advices here but I couldnt make it.
GRADIENT DESCENT
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
%GRADIENTDESCENT Performs gradient descent to learn theta
% theta = GRADIENTDESCENT(X, y, theta, alpha, num_iters) updates theta by
% taking num_iters gradient steps with learning rate alpha
% Initialize some useful values
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
x = X(:,2);
h = theta(1) + (theta(2) * x);
theta_zero = theta(1) - alpha * (1/m) * sum(h-y); %a-step*gradient(1)
theta_one = theta(2) - alpha * (1/m) * sum((h - y) .* x); %b-step*gardient(2)
theta = [theta_zero; theta_one];
J_history(iter) = computeCost(X, y, theta);
end
end
MAIN CODE
clear ; close all; clc
%% ======================= Part 2: Plotting =======================
fprintf('Plotting Data ...\n')
data = csvread('Fish.txt');
X = data(:, 1); y = data(:, 2);
m = length(y); % number of training examples
figure;
plot(X, y, 'dp', 'MarkerSize', 5);
title('Wykres')
fprintf('Program paused. Press enter to continue.\n');
pause;
%% =================== Part 3: Gradient descent ===================
fprintf('Running Gradient Descent ...\n')
X = [ones(m, 1), data(:,1)]; % Add a column of ones to x
%disp(X)
theta = zeros(size(X, 2), 1); % initialize fitting parameters
% Some gradient descent settings
iterations = 1500;
alpha = 0.000001;
% compute and display initial cost
computeCost(X, y, theta)
% run gradient descent
theta = gradientDescent(X, y, theta, alpha, iterations);
% print theta to screen
fprintf('Theta found by gradient descent: ');
fprintf('%f %f \n', theta(1), theta(2));
% Plot the linear fit
hold on; % keep previous plot visible
plot(X(:,2), X*theta, '-')
legend('Training data', 'Linear regression')
hold off % don't overlay any more plots on this figure
FILE
Fish.txt
First column is weight, second is length
242,23.2
290,24
340,23.9
363,26.3
430,26.5
450,26.8
500,26.8
390,27.6
450,27.6
500,28.5
475,28.4
500,28.7
500,29.1
340,29.5
600,29.4
600,29.4
700,30.4
700,30.4
610,30.9
650,31
575,31.3
685,31.4
620,31.5
680,31.8
700,31.9
725,31.8
720,32
714,32.7
850,32.8
1000,33.5
920,35
955,35
925,36.2
975,37.4
950,38
40,12.9
69,16.5
78,17.5
87,18.2
120,18.6
0,19
110,19.1
120,19.4
150,20.4
145,20.5
160,20.5
140,21
160,21.1
169,22
161,22
200,22.1
180,23.6
290,24
272,25
390,29.5
270,23.6
270,24.1
306,25.6
540,28.5
800,33.7
1000,37.3
55,13.5
60,14.3
90,16.3
120,17.5
150,18.4
140,19
170,19
145,19.8
200,21.2
273,23
300,24
5.9,7.5
32,12.5
40,13.8
51.5,15
70,15.7
100,16.2
78,16.8
80,17.2
85,17.8
85,18.2
110,19
115,19
125,19
130,19.3
120,20
120,20
130,20
135,20
110,20
130,20.5
150,20.5
145,20.7
150,21
170,21.5
225,22
145,22
188,22.6
180,23
197,23.5
218,25
300,25.2
260,25.4
265,25.4
250,25.4
250,25.9
300,26.9
320,27.8
514,30.5
556,32
840,32.5
685,34
700,34
700,34.5
690,34.6
900,36.5
650,36.5
820,36.6
850,36.9
900,37
1015,37
820,37.1
1100,39
1000,39.8
1100,40.1
1000,40.2
1000,41.1
200,30
300,31.7
300,32.7
300,34.8
430,35.5
345,36
456,40
510,40
540,40.1
500,42
567,43.2
770,44.8
950,48.3
1250,52
1600,56
1550,56
1650,59
6.7,9.3
7.5,10
7,10.1
9.7,10.4
9.8,10.7
8.7,10.8
10,11.3
9.9,11.3
9.8,11.4
12.2,11.5
13.4,11.7
12.2,12.1
19.7,13.2
19.9,13.8
Thank you in advance!
I tried using standard deviation and Ive replaced
theta = zeros(2, 1);
for
theta = zeros(size(X, 2), 1);
I've also changed values of alpha
Related
I am trying to call to a script, and I am getting an error message. Please help.
The following is what I am trying to run in Octave:
%%% May 23, 2016
%%% Potential Fields for Robot Path Planning
%
%
% Initially proposed for real-time collision avoidance [Khatib 1986].
% Hundreds of papers published on APF
% A potential field is a scalar function over the free space.
% To navigate, the robot applies a force proportional to the
% negated gradient of the potential field.
% A navigation function is an ideal potential field
clc
close all
clear
%% Defining environment variables
startPos = [5,5];
goalPos = [90, 95];
obs1Pos = [50, 50];
obsRad = 10;
goalR = 0.2; % The radius of the goal
goalS = 20; % The spread of attraction of the goal
obsS = 30; % The spread of repulsion of the obstacle
alpha = 0.8; % Strength of attraction
beta = 0.6; % Strength of repulsion
%% Carry out the Potential Field Math as follows:
u = zeros(100, 100);
v = zeros(100, 100);
testu = zeros(100, 100);
testv = zeros(100, 100);
for x = 1:1:100
for y = 1:1:100
[uG, vG] = GoalDelta(x, y, goalPos(1), goalPos(2), goalR, goalS, alpha);
[uO, vO] = ObsDelta(x, y, obs1Pos(2), obs1Pos(1), obsRad, obsS, beta);
xnet = uG + uO;
ynet = vG + vO;
vspeed = sqrt(xnet^2 + ynet^2);
theta = atan2(ynet,xnet);
u(x,y) = vspeed*cos(theta);
v(x,y) = vspeed*sin(theta);
% hold on
end
end
%%
[X,Y] = meshgrid(1:1:100,1:1:100);
figure
quiver(X, Y, u, v, 3)
%% Defining the grid
% Plotting the obstacles
circles(obs1Pos(1),obs1Pos(2),obsRad, 'facecolor','red')
axis square
hold on % Plotting start position
circles(startPos(1),startPos(2),2, 'facecolor','green')
hold on % Plotting goal position
circles(goalPos(1),goalPos(2),2, 'facecolor','yellow')
%% Priting of the path
currentPos = startPos;
x = 0;
while sqrt((goalPos(1)-currentPos(1))^2 + (goalPos(2)-currentPos(2))^2) > 1
tempPos = currentPos + [u(currentPos(1),currentPos(2)), v(currentPos(1),currentPos(2))]
currentPos = round(tempPos)
hold on
plot(currentPos(1),currentPos(2),'-o', 'MarkerFaceColor', 'black')
pause(0.5)
end
The following is the error message I am getting:
error: invalid call to script C:\Users\MyComputer\Downloads\circles.m
error: called from
circles
ECE8743_PotentialFields_Obstacle_1 at line 59 column 1
>>
Here is circles.m:
function [ h ] = circles(x,y,r,varargin)
% h = circles(x,y,r,varargin) plots circles of radius r at points x and y.
% x, y, and r can be scalars or N-D arrays.
%
% Chad Greene, March 2014. Updated August 2014.
% University of Texas Institute for Geophysics.
%
%% Syntax
% circles(x,y,r)
% circles(...,'points',numberOfPoints)
% circles(...,'rotation',degreesRotation)
% circles(...,'ColorProperty',ColorValue)
% circles(...,'LineProperty',LineValue)
% h = circles(...)
%
%% Description
%
% circles(x,y,r) plots circle(s) of radius or radii r centered at points given by
% x and y. Inputs x, y, and r may be any combination of scalar,
% vector, or 2D matrix, but dimensions of all nonscalar inputs must agree.
%
% circles(...,'points',numberOfPoints) allows specification of how many points to use
% for the outline of each circle. Default value is 1000, but this may be
% increased to increase plotting resolution. Or you may specify a small
% number (e.g. 4 to plot a square, 5 to plot a pentagon, etc.).
%
% circles(...,'rotation',degreesRotation) rotates the shape by a given
% degreesRotation, which can be a scalar or a matrix. This is useless for
% circles, but may be desired for polygons with a discernible number of corner points.
%
% circles(...,'ColorProperty',ColorValue) allows declaration of
% 'facecolor' or 'facealpha'
% as name-value pairs. Try declaring any fill property as name-value pairs.
%
% circles(...,'LineProperty',LineValue) allows declaration of 'edgecolor',
% 'linewidth', etc.
%
% h = circles(...) returns the handle(s) h of the plotted object(s).
%
%
%% EXAMPLES:
%
% Example 1:
% circles(5,10,3)
%
% % Example 2:
% x = 2:7;
% y = [5,15,12,25,3,18];
% r = [3 4 5 5 7 3];
% figure
% circles(x,y,r)
%
% % Example 3:
% figure
% circles(1:10,5,2)
%
% % Example 4:
% figure
% circles(5,15,1:5,'facecolor','none')
%
% % Example 5:
% figure
% circles(5,10,3,'facecolor','green')
%
% % Example 6:
% figure
% h = circles(5,10,3,'edgecolor',[.5 .2 .9])
%
% % Example 7:
% lat = repmat((10:-1:1)',1,10);
% lon = repmat(1:10,10,1);
% r = .4;
% figure
% h1 = circles(lon,lat,r,'linewidth',4,'edgecolor','m','facecolor',[.6 .4 .8]);
% hold on;
% h2 = circles(1:.5:10,((1:.5:10).^2)/10,.12,'edgecolor','k','facecolor','none');
% axis equal
%
% % Example 8: Circles have corners
% This script approximates circles with 1000 points. If all those points
% are too complex for your Pentium-II, you can reduce the number of points
% used to make each circle. If 1000 points is not high enough resolution,
% you can increase the number of points. Or if you'd like to draw
% triangles or squares, or pentagons, you can significantly reduce the
% number of points. Let's try drawing a stop sign:
%
% figure
% h = circles(1,1,10,'points',8,'color','red');
% axis equal
% % and we see that our stop sign needs to be rotated a little bit, so we'll
% % delete the one we drew and try again:
% delete(h)
% h = circles(1,1,10,'points',8,'color','red','rot',45/2);
% text(1,1,'STOP','fontname','helvetica CY',...
% 'horizontalalignment','center','fontsize',140,...
% 'color','w','fontweight','bold')
%
% figure
% circles([1 3 5],2,1,'points',4,'rot',[0 45 35])
%
%
% TIPS:
% 1. Include the name-value pair 'facecolor','none' to draw outlines
% (non-filled) circles.
%
% 2. Follow the circles command with axis equal to fix distorted circles.
%
% See also: fill, patch, and scatter.
%% Check inputs:
assert(isnumeric(x),'Input x must be numeric.')
assert(isnumeric(y),'Input y must be numeric.')
assert(isnumeric(r),'Input r must be numeric.')
if ~isscalar(x) && ~isscalar(y)
assert(numel(x)==numel(y),'If neither x nor y is a scalar, their dimensions must match.')
end
if ~isscalar(x) && ~isscalar(r)
assert(numel(x)==numel(r),'If neither x nor r is a scalar, their dimensions must match.')
end
if ~isscalar(r) && ~isscalar(y)
assert(numel(r)==numel(y),'If neither y nor r is a scalar, their dimensions must match.')
end
%% Parse inputs:
% Define number of points per circle:
tmp = strcmpi(varargin,'points')|strcmpi(varargin,'NOP')|strcmpi(varargin,'corners')|...
strncmpi(varargin,'vert',4);
if any(tmp)
NOP = varargin{find(tmp)+1};
tmp(find(tmp)+1)=1;
varargin = varargin(~tmp);
else
NOP = 1000; % 1000 points on periphery by default
end
% Define rotation
tmp = strncmpi(varargin,'rot',3);
if any(tmp)
rotation = varargin{find(tmp)+1};
assert(isnumeric(rotation)==1,'Rotation must be numeric.')
rotation = rotation*pi/180; % converts to radians
tmp(find(tmp)+1)=1;
varargin = varargin(~tmp);
else
rotation = 0; % no rotation by default.
end
% Be forgiving if the user enters "color" instead of "facecolor"
tmp = strcmpi(varargin,'color');
if any(tmp)
varargin{tmp} = 'facecolor';
end
%% Begin operations:
% Make inputs column vectors:
x = x(:);
y = y(:);
r = r(:);
rotation = rotation(:);
% Determine how many circles to plot:
numcircles = max([length(x) length(y) length(r) length(rotation)]);
% Create redundant arrays to make the plotting loop easy:
if length(x)<numcircles
x(1:numcircles) = x;
end
if length(y)<numcircles
y(1:numcircles) = y;
end
if length(r)<numcircles
r(1:numcircles) = r;
end
if length(rotation)<numcircles
rotation(1:numcircles) = rotation;
end
% Define an independent variable for drawing circle(s):
t = 2*pi/NOP*(1:NOP);
% Query original hold state:
holdState = ishold;
hold on;
% Preallocate object handle:
h = NaN(size(x));
% Plot circles singly:
for n = 1:numcircles
h(n) = fill(x(n)+r(n).*cos(t+rotation(n)), y(n)+r(n).*sin(t+rotation(n)),'',varargin{:});
end
% Return to original hold state:
if ~holdState
hold off
end
% Delete object handles if not requested by user:
if nargout==0
clear h
end
end
What am I doing wrong here? How do I correct this error? I am rather new to Octave and Matlab, so any help is greatly appreciated.
this is the first post after pasting the error so maybe it'll help anyone later:
in my case pasting the script into a function worked out!
function retval = name_of_script(parametres)
%your script
endfunction
I am a beginner at Octave, and I have limited knowledge of Matlab. Here is what I am trying to do. I am trying to run this script:
%%% May 23, 2016
%%% Potential Fields for Robot Path Planning
%
%
% Initially proposed for real-time collision avoidance [Khatib 1986].
% Hundreds of papers published on APF
% A potential field is a scalar function over the free space.
% To navigate, the robot applies a force proportional to the
% negated gradient of the potential field.
% A navigation function is an ideal potential field
clc
close all
clear
%% Defining environment variables
startPos = [5,5];
goalPos = [90, 95];
obs1Pos = [50, 50];
obsRad = 10;
goalR = 0.2; % The radius of the goal
goalS = 20; % The spread of attraction of the goal
obsS = 30; % The spread of repulsion of the obstacle
alpha = 0.8; % Strength of attraction
beta = 0.6; % Strength of repulsion
%% Carry out the Potential Field Math as follows:
u = zeros(100, 100);
v = zeros(100, 100);
testu = zeros(100, 100);
testv = zeros(100, 100);
for x = 1:1:100
for y = 1:1:100
[uG, vG] = GoalDelta(x, y, goalPos(1), goalPos(2), goalR, goalS, alpha);
[uO, vO] = ObsDelta(x, y, obs1Pos(2), obs1Pos(1), obsRad, obsS, beta);
xnet = uG + uO;
ynet = vG + vO;
vspeed = sqrt(xnet^2 + ynet^2);
theta = atan2(ynet,xnet);
u(x,y) = vspeed*cos(theta);
v(x,y) = vspeed*sin(theta);
% hold on
end
end
%%
[X,Y] = meshgrid(1:1:100,1:1:100);
figure
quiver(X, Y, u, v, 3)
%% Defining the grid
% Plotting the obstacles
circles(obs1Pos(1),obs1Pos(2),obsRad, 'facecolor','red')
axis square
hold on % Plotting start position
circles(startPos(1),startPos(2),2, 'facecolor','green')
hold on % Plotting goal position
circles(goalPos(1),goalPos(2),2, 'facecolor','yellow')
%% Priting of the path
currentPos = startPos;
x = 0;
while sqrt((goalPos(1)-currentPos(1))^2 + (goalPos(2)-currentPos(2))^2) > 1
tempPos = currentPos + [u(currentPos(1),currentPos(2)), v(currentPos(1),currentPos(2))]
currentPos = round(tempPos)
hold on
plot(currentPos(1),currentPos(2),'-o', 'MarkerFaceColor', 'black')
pause(0.5)
end
Here is GoalDelta.m:
% This function gives delX, delY of attraction caused by the goal point
dGoal = sqrt((gx-vx)^2 + (gy-vy)^2); % distance bw goal and current position
thetaG = atan2((gy-vy),(gx-vx)); % angle between goal and current position
% delXG = 0; delYG = 0;
if dGoal<goalR
delXG = 0; delYG = 0;
elseif ((goalS + goalR) >= dGoal) && (dGoal >= goalR)
delXG = alpha*(dGoal - goalR)*cos(thetaG);
delYG = alpha*(dGoal - goalR)*sin(thetaG);
else
delXG = alpha*goalS*cos(thetaG);
delYG = alpha*goalS*sin(thetaG);
end
end
Here is ObsDelta.m:
% This function gives delX, delY of repulsion caused by the obstacle
inf = 10;
dObs = sqrt((ox-vx)^2 + (oy-vy)^2); % distance bw goal and current position
thetaO = atan2((oy-vy),(ox-vx)); % angle between goal and current position
% delXO = 0; delYO = 0;
if dObs<obsRad
delXO = -(sign(cos(thetaO)))*inf;
delYO = -(sign(sin(thetaO)))*inf;
elseif (dObs < (obsS + obsRad)) && (dObs>=obsRad)
delXO = -beta*(obsS + obsRad - dObs)*cos(thetaO);
delYO = -beta*(obsS + obsRad - dObs)*sin(thetaO);
else
delXO = 0;
delYO = 0;
end
end
This is where I am getting the error
*>> circles
error: 'x' undefined near line 112, column 112
error: called from
circles at line 112 column 1*
This is Circles.m that this error comes from. From what I understand, this would not have this error in Matlab, but I cannot afford Matlab. Therefore, I have no way of knowing. In any case, I am trying to figure out how to solve this error.
function [ h ] = circles(x,y,r,varargin)
% h = circles(x,y,r,varargin) plots circles of radius r at points x and y.
% x, y, and r can be scalars or N-D arrays.
%
% Chad Greene, March 2014. Updated August 2014.
% University of Texas Institute for Geophysics.
%
%% Syntax
% circles(x,y,r)
% circles(...,'points',numberOfPoints)
% circles(...,'rotation',degreesRotation)
% circles(...,'ColorProperty',ColorValue)
% circles(...,'LineProperty',LineValue)
% h = circles(...)
%
%% Description
%
% circles(x,y,r) plots circle(s) of radius or radii r centered at points given by
% x and y. Inputs x, y, and r may be any combination of scalar,
% vector, or 2D matrix, but dimensions of all nonscalar inputs must agree.
%
% circles(...,'points',numberOfPoints) allows specification of how many points to use
% for the outline of each circle. Default value is 1000, but this may be
% increased to increase plotting resolution. Or you may specify a small
% number (e.g. 4 to plot a square, 5 to plot a pentagon, etc.).
%
% circles(...,'rotation',degreesRotation) rotates the shape by a given
% degreesRotation, which can be a scalar or a matrix. This is useless for
% circles, but may be desired for polygons with a discernible number of corner points.
%
% circles(...,'ColorProperty',ColorValue) allows declaration of
% 'facecolor' or 'facealpha'
% as name-value pairs. Try declaring any fill property as name-value pairs.
%
% circles(...,'LineProperty',LineValue) allows declaration of 'edgecolor',
% 'linewidth', etc.
%
% h = circles(...) returns the handle(s) h of the plotted object(s).
%
%
%% EXAMPLES:
%
% Example 1:
% circles(5,10,3)
%
% % Example 2:
% x = 2:7;
% y = [5,15,12,25,3,18];
% r = [3 4 5 5 7 3];
% figure
% circles(x,y,r)
%
% % Example 3:
% figure
% circles(1:10,5,2)
%
% % Example 4:
% figure
% circles(5,15,1:5,'facecolor','none')
%
% % Example 5:
% figure
% circles(5,10,3,'facecolor','green')
%
% % Example 6:
% figure
% h = circles(5,10,3,'edgecolor',[.5 .2 .9])
%
% % Example 7:
% lat = repmat((10:-1:1)',1,10);
% lon = repmat(1:10,10,1);
% r = .4;
% figure
% h1 = circles(lon,lat,r,'linewidth',4,'edgecolor','m','facecolor',[.6 .4 .8]);
% hold on;
% h2 = circles(1:.5:10,((1:.5:10).^2)/10,.12,'edgecolor','k','facecolor','none');
% axis equal
%
% % Example 8: Circles have corners
% This script approximates circles with 1000 points. If all those points
% are too complex for your Pentium-II, you can reduce the number of points
% used to make each circle. If 1000 points is not high enough resolution,
% you can increase the number of points. Or if you'd like to draw
% triangles or squares, or pentagons, you can significantly reduce the
% number of points. Let's try drawing a stop sign:
%
% figure
% h = circles(1,1,10,'points',8,'color','red');
% axis equal
% % and we see that our stop sign needs to be rotated a little bit, so we'll
% % delete the one we drew and try again:
% delete(h)
% h = circles(1,1,10,'points',8,'color','red','rot',45/2);
% text(1,1,'STOP','fontname','helvetica CY',...
% 'horizontalalignment','center','fontsize',140,...
% 'color','w','fontweight','bold')
%
% figure
% circles([1 3 5],2,1,'points',4,'rot',[0 45 35])
%
%
% TIPS:
% 1. Include the name-value pair 'facecolor','none' to draw outlines
% (non-filled) circles.
%
% 2. Follow the circles command with axis equal to fix distorted circles.
%
% See also: fill, patch, and scatter.
%% Check inputs:
assert(isnumeric(x),'Input x must be numeric.')
assert(isnumeric(y),'Input y must be numeric.')
assert(isnumeric(r),'Input r must be numeric.')
if ~isscalar(x) && ~isscalar(y)
assert(numel(x)==numel(y),'If neither x nor y is a scalar, their dimensions must match.')
end
if ~isscalar(x) && ~isscalar(r)
assert(numel(x)==numel(r),'If neither x nor r is a scalar, their dimensions must match.')
end
if ~isscalar(r) && ~isscalar(y)
assert(numel(r)==numel(y),'If neither y nor r is a scalar, their dimensions must match.')
end
%% Parse inputs:
% Define number of points per circle:
tmp = strcmpi(varargin,'points')|strcmpi(varargin,'NOP')|strcmpi(varargin,'corners')|...
strncmpi(varargin,'vert',4);
if any(tmp)
NOP = varargin{find(tmp)+1};
tmp(find(tmp)+1)=1;
varargin = varargin(~tmp);
else
NOP = 1000; % 1000 points on periphery by default
end
% Define rotation
tmp = strncmpi(varargin,'rot',3);
if any(tmp)
rotation = varargin{find(tmp)+1};
assert(isnumeric(rotation)==1,'Rotation must be numeric.')
rotation = rotation*pi/180; % converts to radians
tmp(find(tmp)+1)=1;
varargin = varargin(~tmp);
else
rotation = 0; % no rotation by default.
end
% Be forgiving if the user enters "color" instead of "facecolor"
tmp = strcmpi(varargin,'color');
if any(tmp)
varargin{tmp} = 'facecolor';
end
%% Begin operations:
% Make inputs column vectors:
x = x(:);
y = y(:);
r = r(:);
rotation = rotation(:);
% Determine how many circles to plot:
numcircles = max([length(x) length(y) length(r) length(rotation)]);
% Create redundant arrays to make the plotting loop easy:
if length(x)<numcircles
x(1:numcircles) = x;
end
if length(y)<numcircles
y(1:numcircles) = y;
end
if length(r)<numcircles
r(1:numcircles) = r;
end
if length(rotation)<numcircles
rotation(1:numcircles) = rotation;
end
% Define an independent variable for drawing circle(s):
t = 2*pi/NOP*(1:NOP);
% Query original hold state:
holdState = ishold;
hold on;
% Preallocate object handle:
h = NaN(size(x));
% Plot circles singly:
for n = 1:numcircles
h(n) = fill(x(n)+r(n).*cos(t+rotation(n)), y(n)+r(n).*sin(t+rotation(n)),'',varargin{:});
end
% Return to original hold state:
if ~holdState
hold off
end
% Delete object handles if not requested by user:
if nargout==0
clear h
end
end
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
%GRADIENTDESCENT Performs gradient descent to learn theta
% theta = GRADIENTDESCENT(X, y, theta, alpha, num_iters) updates theta by
% taking num_iters gradient steps with learning rate alpha
% Initialize some useful values
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
% ====================== YOUR CODE HERE ======================
% Instructions: Perform a single gradient step on the parameter vector
% theta.
%
% Hint: While debugging, it can be useful to print out the values
% of the cost function (computeCost) and gradient here.
%
%theta(iter)=theta(iter)-0.01*(1/m)*(((theta(1)+theta(2))*X-y)*X(iter,2))
theta=theta-(alpha*(1/m)*(X*theta-y)*X(iter,2);
% ============================================================
% Save the cost J in every iteration
J_history(iter) = computeCost(X, y, theta);
end
end
theta=theta-(alpha*(1/m)*(X*theta-y)*X(iter,2);
The parentheses are not balanced as far as I can tell?
You're missing a closing parenthesis ) are you not?
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
%GRADIENTDESCENT Performs gradient descent to learn theta
% theta = GRADIENTDESENT(X, y, theta, alpha, num_iters) updates theta by
% taking num_iters gradient steps with learning rate alpha
% Initialize some useful values
m = length(y); % number of training examples
J_history = zeros(num_iters, 1);
for iter = 1:num_iters
% ====================== YOUR CODE HERE ======================
% Instructions: Perform a single gradient step on the parameter vector
% theta.
%
% Hint: While debugging, it can be useful to print out the values
% of the cost function (computeCost) and gradient here.
%
hypothesis = x*theta;
theta_0 = theta(1) - alpha(1/m)*sum((hypothesis-y)*x);
theta_1 = theta(2) - alpha(1/m)*sum((hypothesis-y)*x);
theta(1) = theta_0;
theta(2) = theta_1;
% ============================================================
% Save the cost J in every iteration
J_history(iter) = computeCost(X, y, theta);
end
end
I keep getting this error
error: gradientDescent: subscript indices must be either positive integers less than 2^31 or logicals
on this line right in-between the first theta and =
theta_0 = theta(1) - alpha(1/m)*sum((hypothesis-y)*x);
I'm very new to octave so please go easy on me, and
thank you in advance.
This is from the coursera course on Machine Learning from Week 2
99% sure your error is on the line pointed out by topsig, where you have alpha(1/m)
it would help if you gave an example of input values to your function and what you hoped to see as an output, but I'm assuming from your comment
% taking num_iters gradient steps with learning rate alpha
that alpha is a constant, not a function. as such, you have the line alpha(1/m) without any operator in between. octave sees this as you indexing alpha with the value of 1/m.
i.e., if you had an array
x = [3 4 5]
x*(2) = [6 8 10] %% two times each element in the array
x(2) = [4] %% second element in the array
what you did doesn't seem to make sense, as 'm = length(y)' which will output a scalar, so
x = [3 4 5]; m = 3;
x*(1/m) = x*(1/3) = [1 1.3333 1.6666] %% element / 3
x(1/m) = ___error___ %% the 1/3 element in the array makes no sense
note that for certain errors it always indicates that the location of the error is at the assignment operator (the equal sign at the start of the line). if it points there, you usually have to look elsewhere in the line for the actual error. here, it was yelling at you for trying to apply a non-integer subscript (1/m)
Given a 2D circle with 2 angles in the range -PI -> PI around a coordinate, what is the value of the smallest angle between them?
Taking into account that the difference between PI and -PI is not 2 PI but zero.
An Example:
Imagine a circle, with 2 lines coming out from the center, there are 2 angles between those lines, the angle they make on the inside aka the smaller angle, and the angle they make on the outside, aka the bigger angle.
Both angles when added up make a full circle. Given that each angle can fit within a certain range, what is the smaller angles value, taking into account the rollover
This gives a signed angle for any angles:
a = targetA - sourceA
a = (a + 180) % 360 - 180
Beware in many languages the modulo operation returns a value with the same sign as the dividend (like C, C++, C#, JavaScript, full list here). This requires a custom mod function like so:
mod = (a, n) -> a - floor(a/n) * n
Or so:
mod = (a, n) -> (a % n + n) % n
If angles are within [-180, 180] this also works:
a = targetA - sourceA
a += (a>180) ? -360 : (a<-180) ? 360 : 0
In a more verbose way:
a = targetA - sourceA
a -= 360 if a > 180
a += 360 if a < -180
x is the target angle. y is the source or starting angle:
atan2(sin(x-y), cos(x-y))
It returns the signed delta angle. Note that depending on your API the order of the parameters for the atan2() function might be different.
If your two angles are x and y, then one of the angles between them is abs(x - y). The other angle is (2 * PI) - abs(x - y). So the value of the smallest of the 2 angles is:
min((2 * PI) - abs(x - y), abs(x - y))
This gives you the absolute value of the angle, and it assumes the inputs are normalized (ie: within the range [0, 2π)).
If you want to preserve the sign (ie: direction) of the angle and also accept angles outside the range [0, 2π) you can generalize the above. Here's Python code for the generalized version:
PI = math.pi
TAU = 2*PI
def smallestSignedAngleBetween(x, y):
a = (x - y) % TAU
b = (y - x) % TAU
return -a if a < b else b
Note that the % operator does not behave the same in all languages, particularly when negative values are involved, so if porting some sign adjustments may be necessary.
An efficient code in C++ that works for any angle and in both: radians and degrees is:
inline double getAbsoluteDiff2Angles(const double x, const double y, const double c)
{
// c can be PI (for radians) or 180.0 (for degrees);
return c - fabs(fmod(fabs(x - y), 2*c) - c);
}
See it working here:
https://www.desmos.com/calculator/sbgxyfchjr
For signed angle:
return fmod(fabs(x - y) + c, 2*c) - c;
In some other programming languages where mod of negative numbers are positive, the inner abs can be eliminated.
I rise to the challenge of providing the signed answer:
def f(x,y):
import math
return min(y-x, y-x+2*math.pi, y-x-2*math.pi, key=abs)
For UnityEngine users, the easy way is just to use Mathf.DeltaAngle.
Arithmetical (as opposed to algorithmic) solution:
angle = Pi - abs(abs(a1 - a2) - Pi);
I absolutely love Peter B's answer above, but if you need a dead simple approach that produces the same results, here it is:
function absAngle(a) {
// this yields correct counter-clock-wise numbers, like 350deg for -370
return (360 + (a % 360)) % 360;
}
function angleDelta(a, b) {
// https://gamedev.stackexchange.com/a/4472
let delta = Math.abs(absAngle(a) - absAngle(b));
let sign = absAngle(a) > absAngle(b) || delta >= 180 ? -1 : 1;
return (180 - Math.abs(delta - 180)) * sign;
}
// sample output
for (let angle = -370; angle <= 370; angle+=20) {
let testAngle = 10;
console.log(testAngle, "->", angle, "=", angleDelta(testAngle, angle));
}
One thing to note is that I deliberately flipped the sign: counter-clockwise deltas are negative, and clockwise ones are positive
There is no need to compute trigonometric functions. The simple code in C language is:
#include <math.h>
#define PIV2 M_PI+M_PI
#define C360 360.0000000000000000000
double difangrad(double x, double y)
{
double arg;
arg = fmod(y-x, PIV2);
if (arg < 0 ) arg = arg + PIV2;
if (arg > M_PI) arg = arg - PIV2;
return (-arg);
}
double difangdeg(double x, double y)
{
double arg;
arg = fmod(y-x, C360);
if (arg < 0 ) arg = arg + C360;
if (arg > 180) arg = arg - C360;
return (-arg);
}
let dif = a - b , in radians
dif = difangrad(a,b);
let dif = a - b , in degrees
dif = difangdeg(a,b);
difangdeg(180.000000 , -180.000000) = 0.000000
difangdeg(-180.000000 , 180.000000) = -0.000000
difangdeg(359.000000 , 1.000000) = -2.000000
difangdeg(1.000000 , 359.000000) = 2.000000
No sin, no cos, no tan,.... only geometry!!!!
A simple method, which I use in C++ is:
double deltaOrientation = angle1 - angle2;
double delta = remainder(deltaOrientation, 2*M_PI);