I'm using octave's ezmesh to plot a linear regression defined as follows:
f = #(x,y) 1 * theta(1) + x * theta(2) + y * theta(3) + x * y * theta(4)
For some fixed vector theta:
octave:275> theta
theta =
9.4350e+00
1.7410e-04
3.3702e-02
1.6498e-07
I'm using a domain of [0 120000 0 1400], and can evaluate:
octave:276> f(0, 0)
ans = 9.4350
octave:277> f(120000, 1400)
ans = 105.23
However, if I run:
octave:278> ezmesh(f, [0 120000 0 1400])
The resulting mesh has a z value of around 570 for (0, 0) and just under 640 for (120000, 1400). I'm baffled. What could be causing this?
EDIT: Even if I simplify f to the following, similar behavior occurs:
octave:308> f = #(x, y) (x * y)
Why is ezmesh not handling multiplication as expected (by me), so that the function evaluates as I expect, and the values change when the function is used inside of ezmesh?
ezmesh invokes the function handle on a matrix of values (to benefit from vectorization performance). Use .* for multiplication.
Related
I am testing MATLAB capabilities in solving equations for a project that I intend to do, so I gave it a test run with something simple, but the results that it gives me are incorrect. I tried to solve two non-linear equations with two unknowns, one of the solutions is correct the other is not.
syms theta d x y
eq1 = d * cos(theta) == x;
eq2 = d * sin(theta) == y;
sol = solve(eq1, eq2, theta, d)
sol.theta
sol.d
The solutions for d are correct, but for theta I get:
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*atan((x + (x^2 + y^2)^(1/2))/y)
And the correct answer for theta is simply atan(y/x)
Then when I evaluate these solutions with x = 1, y = 0, I get:
eval(sol.d)
eval(sol.theta)
d = 1, -1
theta = NaN, -3.1416
Solutions for d are correct, but theta in that scenario should be 0.
What am I doing wrong?
EDIT: solving it by hand it looks like this: Divide the y equation by the x equation
y/x = (d * sin(theta)) / (d * cos(theta))
y/x = sin(theta)/cos(theta)
y/x = tan(theta)
theta = atan(y/x)
Even if matlab solves it in some other way and gets a different expression, it should still yield the same final result when I use numbers and it PARTIALLY does.
For x = 1 and y = 0, theta should be 0, => this doesnt work, it gives NaN (explanation bellow)
for x = 1 and y = 1, theta should be 45 degrees => this works
for x = 0 and y = 1 theta should be 90 degrees => this works
And I just checked it again with the 45 and 90 degree values for x and y and it works, but for x = 1 and y = 0 it still gives NaN as one of the answers and that is because it gets a 0/0 from the way it is expressing it
-2*atan((x - (x^2 + y^2)^(1/2))/y)
-2*(1 - (1^2 + 0^2))^(1/2)/0
-2*(1 - 1)^(1/2)/0
0/0
but if its in the form of atan(y/x) the result is
theta = atan(0/1)
theta = atan(0)
theta = 0
Did you mean to solve this:
syms a b theta d real
eq1 = a==d * cos(theta) ;
eq2 = b==d * sin(theta) ;
[sol] = solve([eq1 eq2],[d theta] ,'IgnoreAnalyticConstraints', true,'Real',true,'ReturnConditions',true);
When solving the equations with symbolic x and y, the solver will find a solution with a certain condition, which can be obtained using the argument 'ReturnCondition':
syms x y theta d real
eq1 = d*cos(theta) == x;
eq2 = d*sin(theta) == y;
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
This gives the following result for sol
>> sol.d
(x^2 + y^2)^(1/2)
-(x^2 + y^2)^(1/2)
>> sol.theta
2*pi*k - 2*atan((x - (x^2 + y^2)^(1/2))/y)
2*pi*k - 2*atan((x + (x^2 + y^2)^(1/2))/y)
>> sol.parameters
k
>> sol.conditions
y ~= 0 & in(k, 'integer')
y ~= 0 & in(k, 'integer')
As you can see, y = 0 does not fulfill this general solution given by the solver, resulting in your problem for y = 0. You can find solutions for y = 0 by either making y numeric instead of symbolic, or by adding an assumption:
syms x y theta d real
assume(y==0)
sol = solve([eq1; eq2],[d theta],'ReturnConditions',true);
I guess its easier to just set y=0 numeric, for this one condition, since there are already 4 possible solutions and conditions for the three lines above.
How do I integrate something like (x + y) (or an expression with any number of variables) with respect to just one of the variables, let's say x from 0 to 1, and get a function of the other variable (1/2 + y in this case) back?
If you mean symbolically, then use:
syms x y
f=int(x+y,x,0,1)
which gives
f =
y + 1/2
then get f(y=4) from subs(f,4) which gives
ans =
9/2
If you have more than two variables, use:
syms x y z
f=int(x+y+z,x,0,1)
which gives
f =
y + z + 1/2
then for f(y=4,z=5) use subs(f,{y,z},[4,5]) gives
ans =
19/2
I want to write my own 2 Dimensional DFT function with reduced loops.
What I try to implement is Discrete Fourier Transform:
Using the separability property of transform (actually exponential function), we can write this as multiplication of two 1 dimensional DFT. Then, we can calculate the exponential terms for rows (the matrix wM below) and columns (the matrix wN below) of transform. Then, for summation process we can multiply them as "F = wM * original_matrix * wN"
Here is the code I wrote:
f = imread('cameraman.tif');
[M, N, ~] = size(f);
wM = zeros(M, M);
wN = zeros(N, N);
for u = 0 : (M - 1)
for x = 0 : (M - 1)
wM(u+1, x+1) = exp(-2 * pi * 1i / M * x * u);
end
end
for v = 0 : (N - 1)
for y = 0 : (N - 1)
wN(y+1, v+1) = exp(-2 * pi * 1i / N * y * v);
end
end
F = wM * im2double(f) * wN;
The first thing is I dont want to use 2 loops which are MxM and NxN times running. If I used a huge matrix (or image), that would be a problem. Is there any chance to make this code faster (for example eliminating the loops)?
The second thing is displaying the Fourier Transform result. I use the codes below to display the transform:
% // "log" method
fl = log(1 + abs(F));
fm = max(fl(:));
imshow(im2uint8(fl / fm))
and
% // "abs" method
fa = abs(F);
fm = max(fa(:));
imshow(fa / fm)
When I use the "abs" method, I see only black figure, nothing else. What is wrong with "abs" method you think?
And the last thing is when I compare the transform result of my own function with MATLAB' s fft2() function', mine displays darker figure than MATLAB' s result. What am I missing here? Implementation misktake?
The transform result of my own function:
The transform result of MATLAB fft2() function:
I am happy you solved your problem but unfortunately you answer is not completely right. Indeed it does the job, but as I commented, im2double will normalize everything to 1, therefore showing the scaled result you have. What you want (if you are looking for performance) is not doing im2doubleand then multiply by 255, but directly casting to double().
You can eliminate loops by using meshgrid.
For example:
M = 1024;
tic
[ mX, mY ] = meshgrid( 0 : M - 1, 0 : M - 1 );
wM1 = exp( -2 * pi * 1i / M .* mX .* mY );
toc
tic
for u = 0 : (M - 1)
for x = 0 : (M - 1)
wM2( u + 1, x + 1 ) = exp( -2 * pi * 1i / M * x * u );
end
end
toc
all( wM1( : ) == wM2( : ) )
The timing on my system was:
Elapsed time is 0.130923 seconds.
Elapsed time is 0.493163 seconds.
I'm trying build a matlab function that will evaluate a function and vector that are sent in as parameters. I'm having a hard time trying to figure out how to send in the function so that it can be evaluated in the matlab function. I figured out how to do it without the function but I'm a little lost trying to evaluate it within a matlab function. Below is my code...
This is what I'm trying to do...
x = [x1 x2]';
f = x(x1)^2 + 2 * (x2)^2
x = [5 10];
f = (5)^2 + 2 * (10)^2 % which I would like to return 225, not a column vector
This is what I have and what I have tried...
x = [5 10]';
% without using a matlab function
% k = 1
% f = x(k)^2 + 2 * x(k + 1)^2; % returns the correct answer of 225
f = x^2 + 2 * x^2 % complains about the scalar 2
f = x.^2 + 2 * x.^2 % returns a column vector [75; 300]
function [value] = evalFunction(f,x)
value = f(x);
I've tried...
f = #(x) x.^2 + 2 * (x+1).^2;
value = evalFunction(#f,x) %Error: "f" was previously used as a variable
So I tried...
f = #(x) x.^2 + 2 * (x+1).^2;
value = evalFunction(f,x) %value = [97;342]
I'm new to matlab so any help is appreciated. I've been doing some research and found some stuff here on stackoverflow but can't seem to get it to work. I've seen there are other ways to do this, but I will eventually be adding more code to the matlab evalFunction function so I'd like to do it this way. Thanks!
Anonymous functions and function handles plus array indexing. Taking x as a 2-element vector, define and use your function like:
f = #(x) x(1).^2 + 2 * x(2).^2;
value = evalFunction(f,x) % but you can just do f(x) if that is all you need
However, if evalFunction does nothing other than evaluate f at x, then you don't need it at all. Just do f(x).
Alternately,
f = #(x1,x2) x1.^2 + 2*x2.^2;
value = evalFunction(f,x1,x2); % here your function will call it by f(x1,x2)
You are probably coming at this from a C background - in Matlab, x+1 is the entire vector x with 1 added - not the element offset by 1.
The function you need is
f = #(x)x(1).^2 + 2 * (x(2)).^2;
or, to be a little more "matlab-like":
f = #(x) [1 2] * x(1:2)'.^2;
Which performs the element-wise square of the first two elements of x as a column vector, and then does the matrix multiplication with [1 2], resulting in
1 * x(1) .^2 + 2 * x(2) .^2;
Which seems to be what you were asking for.
caveat: did not have opportunity to test this...
I have to solve a system of ordinary differential equations of the form:
dx/ds = 1/x * [y* (g + s/y) - a*x*f(x^2,y^2)]
dy/ds = 1/x * [-y * (b + y) * f()] - y/s - c
where x, and y are the variables I need to find out, and s is the independent variable; the rest are constants. I've tried to solve this with ode45 with no success so far:
y = ode45(#yprime, s, [1 1]);
function dyds = yprime(s,y)
global g a v0 d
dyds_1 = 1./y(1) .*(y(2) .* (g + s ./ y(2)) - a .* y(1) .* sqrt(y(1).^2 + (v0 + y(2)).^2));
dyds_2 = - (y(2) .* (v0 + y(2)) .* sqrt(y(1).^2 + (v0 + y(2)).^2))./y(1) - y(2)./s - d;
dyds = [dyds_1; dyds_2];
return
where #yprime has the system of equations. I get the following error message:
YPRIME returns a vector of length 0, but the length of initial
conditions vector is 2. The vector returned by YPRIME and the initial
conditions vector must have the same number of elements.
Any ideas?
thanks
Certainly, you should have a look at your function yprime. Using some simple model that shares the number of differential state variables with your problem, have a look at this example.
function dyds = yprime(s, y)
dyds = zeros(2, 1);
dyds(1) = y(1) + y(2);
dyds(2) = 0.5 * y(1);
end
yprime must return a column vector that holds the values of the two right hand sides. The input argument s can be ignored because your model is time-independent. The example you show is somewhat difficult in that it is not of the form dy/dt = f(t, y). You will have to rearrange your equations as a first step. It will help to rename x into y(1) and y into y(2).
Also, are you sure that your global g a v0 d are not empty? If any one of those variables remains uninitialized, you will be multiplying state variables with an empty matrix, eventually resulting in an empty vector dyds being returned. This can be tested with
assert(~isempty(v0), 'v0 not initialized');
in yprime, or you could employ a debugging breakpoint.
the syntax for ODE solvers is [s y]=ode45(#yprime, [1 10], [2 2])
and you dont need to do elementwise operation in your case i.e. instead of .* just use *