I was trying to plot data.
Firstly, I have load the data from a file
data = load('ex1data1.txt'); % read comma separated data
X = data(:, 1); y = data(:, 2);
m = length(y); % number of training examples
Then i have called the function plotData
function figure=plotData(x, y)
figure; % open a new figure window
if(is_vector(x) && is_vector(y))
figure=plot(x,y,'rx',MarkerSize,10);
xlabel('Profit in $10,000s');
ylabel('Population of city in 10,000s');
endif
endfunction
But Iam getting an error. which says:
x is undefined
Thank you in advance.
The problem is in the following statement:
X = data(:, 1); y = data(:, 2);
you have defined X variable but when you call
plotData(x, y)
you are using lowercase X
I think if change the statement: plotData(X, y) will solve your problem
Related
So I am trying to solve this problem from coursera regarding onevsAll classification. It comes under the ml course under Andrew NG. I am not able to find mistake in my code and it keeps showing this error
error: fmincg: operator +: nonconformant arguments (op1 is 401x1, op2 is 4x1)
error: called from
fmincg at line 87 column 5
oneVsAll at line 60 column 13
ex3 at line 77 column 13
I have tried reading the fmincg code but I cannot understand the problem.
function [all_theta] = oneVsAll(X, y, num_labels, lambda)
% Some useful variables
m = size(X, 1);
n = size(X, 2);
% You need to return the following variables correctly
all_theta = zeros(num_labels, n + 1);
% Add ones to the X data matrix
X = [ones(m, 1) X];
for c = 1:num_labels
initial_theta = zeros(n+1,1) ;
options = optimset('GradObj', 'on', 'MaxIter', 50);
[theta] = fmincg (#(t)(lrCostFunction(t, X, (y == c), lambda)),initial_theta, options);
all_theta(c, :) = theta;
end;
Program is showing error mentioned above regarding the fmincg file.
I am trying to make a function in Octave where you give octave a function f(x,y) as a string, a change in X, a change in Y, a starting point, and the size of a matrix, the function will create a matrix populated with the values of f(x,y) at each point in the matrix.
This is for an application that displays a 3d graph, using the matrix to map each value to a block
# funcStr: The function whose Z values are being calculated
# dx: the change in x that each block in the x direction represents
# dy: the change in y that each block in the y direction represents
# startPt: the point (in an array of x, y) that center block represents
# res: the side length (in blocks) of the plane
pkg load symbolic
syms x y
function[zValues] = calculateZValues(funcStr, dx, dy, startPt, res)
zValues = zeros(res);
eqn = #(x, y) inline(funcStr);
startX = startPt{1};
startY = startPt{2};
for yOffset = 1:res
for xOffset = 1:res
xCoord = startX + dx * xOffset;
yCoord = startY + dy * yOffset;
zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
endfor
endfor
endfunction
The error I am getting is:
>> calculateZValues("x*y", 1, 1, {0,0}, 10)
parse error near line 20 of file /home/rahul/Documents/3dGraph/graph/calculateZValues.m
anonymous function bodies must be single expressions
>>> zValues(res * yOffset + xOffset) = double(subs(eqn, #(x, y), {xCoord, yCoord}));
I have no idea what the issue is. I have replaced the #(x,y) part with {x,y} in the line referenced by the error but it says nothing or it raises an error about the function subs not being declared. I have also tried moving the pkg and syms lines above the function header
I have a problen in scilab
How can I plot functions containing if and < like
function y = alpha(t)
if (t < 227.8) then
y = 0.75;
elseif (t < 300) then
y = 2.8 - 0.009 .* t;
else
y = 0.1;
end
endfunction
and
function [r]=minus_alpha(t)
r = 1 - alpha(t)
endfunction
When I use
x = linspace(0,300)
plot(x, alpha(x))
I got the error message
WARNING: Transposing row vector X to get compatible dimensions
plot2d: falsche Größe für Eingangsargument: inkompatible Größen.
Error 999 : in plot2d called by plot
Sorry for german mix. Thank you.
You can avoid explicit loop and be more efficient using the followin code
function y = alpha(t)
y=0.1*ones(t);
y(t<227.8)=0.75;
i=t>=227.8&t<300;
y(i)=2.8 - 0.009 .* t(i);
endfunction
It is really sad to see a great majority of Scilab community is not aware of vectorized operations. You can change your function to:
function y = alpha(t)
y = 0.1;
if t < 227.8 then
y = 0.75;
elseif t < 300 then
y = 2.8 - 0.009 * t;
end
y = 1 - y;
endfunction
and then use feval to broadcast the function over the sequence:
x = linspace(0, 300);
plot2d(x, feval(x, alpha));
which results:
rule of thumb if you are using for loop you need to revise your code and if someone is offering you a code where there is unnecessary for loop you shouldn't probably use it.
All the proposed answers are overcomplicated considering that the function alpha in the original demand is piecewise-affine. In Scilab in can be coded that way:
x = linspace(0,400,1000);
plot(x,linear_interpn(x,[227.8 300],[0.75 0.1]))
i.e. you just have to know the nodes coordinates (here abscissae) and value of the function at nodes. The function linear_interpn does also multilinear interpolation, it is worth knowing it guys...
If you check the output of your alpha(x), you will see that it is just a scalar (not a vector). I guess you wanted something like this, so it's necessary to iterate through t to compute each value of y based on the value of t:
clc;
clear;
function y = alpha(t)
for i=1:size(t,"*")
if t(i) < 227.8 then
y(i) = 0.75;
elseif t(i) < 300 then
y(i) = 2.8 - 0.009 * t(i);
else
y(i) = 0.1;
end
end
endfunction
x = linspace(0,300);
plot2d(x,alpha(x));
If you find the answer useful, please do not forget to accept it, so others will see that your problem is solved.
Before your answers (thank you) my workaround was a combination of indicator functions composed with floor and exp( -t^2):
function y = alpha(t)
y = floor(exp(-(t .* (t-T1)) / (T1*T1))) * 0.75
+ floor(exp(-((t-T2) .* (t- T1) / (2000)))) .* (2.8-0.009 .* t)
+ floor(exp(-((t-T2) .* (t-1000) / (200000))))*0.1
endfunction
I am trying to solve a system of two ODEs using Octave, and in particular the function lsode.
The code is the following:
function xdot = f (x,t)
a1=0.00875;
a2=0.075;
b1=7.5;
b2=2.5;
d1=0.0001;
d2=0.0001;
g=4*10^(-8);
K1=5000;
K2=2500;
n=2;
m=2;
xdot = zeros(2,1);
xdot(1) = a1+b1*x(1)^n/(K1^n+x(1)^n)-g*x(1)*x(2)-d1*x(1);
xdot(2) = a2+b2*x(1)^m/(K2^m+x(1)^m)-d2*x(2);
endfunction
t = linspace(0, 5000, 200)';
x0 = [1000; 1000];
x = lsode ("f", x0, t);
set term dumb;
plot(t,x);
I am getting continuously the same error, that "x" is not defined, and I do not know why. The error is the following:
warning: function name 'f' does not agree with function file name '/home /Simulation 1/sim.m'
error: 'x' undefined near line 17 column 17
error: called from
sim at line 17 column 9
It would we great that any of you could help me with this code.
You have two errors. One, you are not saving your source code with the proper name. Two, variable "x" is a vector, and nothing in your script indicates that. You should add a line "x = zeros(1,2);" right after "xdot = zeros(2,1);".
Try the following code:
function ODEs
t = linspace(0, 5000, 200);
x0 = [1000; 1000];
x = lsode (#f, x0, t);
fprintf('t = %e \t\t x = %e\n',t,x);
endfunction
function xdot = f(x,t)
a1=0.00875;
a2=0.075;
b1=7.5;
b2=2.5;
d1=0.0001;
d2=0.0001;
g=4*10^(-8);
K1=5000;
K2=2500;
n=2;
m=2;
xdot = zeros(2,1);
x = zeros(1,2);
xdot(1) = a1+b1*x(1)^n/(K1^n+x(1)^n)-g*x(1)*x(2)-d1*x(1);
xdot(2) = a2+b2*x(1)^m/(K2^m+x(1)^m)-d2*x(2);
endfunction
Save it as ODEs.m and execute it. It does not plot anything, but gives you an output with the results for the t range you supplied.
I have followed the tutorial on http://www.mit.edu/people/abbe/matlab/ode.html and prepared a function as follows:
function dxy = diffxy(xy)
%
%split xy into variables in our equations
%
x = xy(1);
xdot = xy(2);
y = xy(3);
%
% define the derivatives of these variables from equations
%
xdot = xdot;
ydot = 3*x + 2*y + 5;
xdoubledot = 3 - ydot + 2*xdot;
%
%return the derivatives in dxy in the right order
%
dxy = [xdot; xdoubledot; ydot]
end
When I call it using
[T, XY] = ode45('diffxy',0,10,[0 1 0])
I get an error
??? Error using ==> diffxy
Too many input arguments.
I also tried
XY= ode45(#diffxy,[0 10],[0;1;0])
Anybody have any idea?
haven't read the whole tutorial but aren't you supposed to define your function as
function dxy = diffxy(t, xy)
where t is time vector