MATLAB Translating from Mathematica to Matlab

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The discussion focuses on translating a Mathematica code snippet into Matlab, specifically for generating a random process using the Karhunen-Loeve expansion. Participants suggest breaking down the code into simpler expressions for easier translation and highlight potential syntax issues, such as missing brackets. The correct Matlab equivalent involves using element-wise operations and matrix multiplication, with suggestions for initializing random variables more efficiently. The final recommended approach emphasizes avoiding for loops in Matlab, advocating for vectorized operations to improve performance. The conversation concludes with a clear direction on how to structure the Matlab code for accurate results.
Frank Einstein
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Hello everyone.

I was wondering if someone could tell me how to write the next line of Mathematica code in Matlab.
Code:
x = Sum [ Sqrt [ vals[ [ j ] ] ] * vecs [ [ j, All ] ] * z [ [ j ] ] , {j,1,10}
vals is a eigenvalues vector, vecs is a eigenvector matrix and z is a vector of random variables of length 10

Thanks for your answer.
 
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Why not decompose it into simpler expressions and then convert them yourself?

Also it looks like you might be missing a square bracket or something.

I spaced it out for readability in your post.
 
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jedishrfu said:
Why not decompose it into simpler expressions and then convert them yourself?

Also it looks like you might be missing a square bracket or something.

I spaced it out for readability in your post.

Indeed, I forgot to close the right square bracket.

Believe me. i have tried to do it myself. That code is from a colleague and represents the generation of a random process with a Karhunen Loeve expansion.

My attempts have not given propper results, this is my code:

for i=1:long
z(i)=normrnd(0,1);
f(i)=0;
end for i=1:long%times
for j=1:long%number of terms in the descomposition
f(i)=f(i)+sqrt(eigenvalues2(j))*z(j)*eigenvectors(i,j);
end

end

However, since this hasn't worked, I have tried to do it manually:

% f5(1)=eigenvectors(1,5)*sqrt(eigenvalues(5))*z(5)+eigenvectors(1,4)*sqrt(eigenvalues(4))*z(4)+eigenvectors(1,3)*sqrt(eigenvalues(3))*z(3)+eigenvectors(1,2)*sqrt(eigenvalues(2))*z(2)+eigenvectors(1,1)*sqrt(eigenvalues(1))*z(1);
% f5(2)=eigenvectors(2,5)*sqrt(eigenvalues(5))*z(5)+eigenvectors(2,4)*sqrt(eigenvalues(4))*z(4)+eigenvectors(2,3)*sqrt(eigenvalues(3))*z(3)+eigenvectors(2,2)*sqrt(eigenvalues(2))*z(2)+eigenvectors(2,1)*sqrt(eigenvalues(1))*z(1);
% f5(3)=eigenvectors(3,5)*sqrt(eigenvalues(5))*z(5)+eigenvectors(3,4)*sqrt(eigenvalues(4))*z(4)+eigenvectors(3,3)*sqrt(eigenvalues(3))*z(3)+eigenvectors(3,2)*sqrt(eigenvalues(2))*z(2)+eigenvectors(3,1)*sqrt(eigenvalues(1))*z(1);
% f5(4)=eigenvectors(4,5)*sqrt(eigenvalues(5))*z(5)+eigenvectors(4,4)*sqrt(eigenvalues(4))*z(4)+eigenvectors(4,3)*sqrt(eigenvalues(3))*z(3)+eigenvectors(4,2)*sqrt(eigenvalues(2))*z(2)+eigenvectors(4,1)*sqrt(eigenvalues(1))*z(1);
% f5(5)=eigenvectors(5,5)*sqrt(eigenvalues(5))*z(5)+eigenvectors(5,4)*sqrt(eigenvalues(4))*z(4)+eigenvectors(5,3)*sqrt(eigenvalues(3))*z(3)+eigenvectors(5,2)*sqrt(eigenvalues(2))*z(2)+eigenvectors(5,1)*sqrt(eigenvalues(1))*z(1);

However, if I calculate the correlation matrix for some realizations of this, the covariance matrix doesn't look like the kernel at all.

So I don't know what I should try next.
 
My take is:

Matlab:
for j = 1:10
   x=vals(j)*vecs(j)*z(j)
   sum=sum+sqrt(x)
end

I don’t know what the vecs[[j,all]] that’s something you’ll have to get from Mathematica. it looks to be related to matrix multiply against the z vector?
 
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jedishrfu said:
My take is:

Matlab:
for j = 1:10
   x=vals(j)*vecs(j)*z(j)
   sum=sum+sqrt(x)
end

I don’t know what the vecs[[j,all]] that’s something you’ll have to get from Mathematica. it looks to be related to matrix multiply against the z vector?

Thank you very much for your answer. I will keep working on it.
 
vecs[[j, all]] should translate to vecs(j,:) or vecs(:,j). the semi-colon tells MATLAB to select all of the elements in that dimension (j,:) being everything in the jth row, and (:,j) being everything in the jth column.

So something like
sum = [];
for j = 1:10
sum = sum + sqrt(vals(j) .* vecs(:,j) .* z(j)) ;
end

the .* means to multiply element-wise ( so 2 .* [1 1 1] becomes [2 2 2])
vecs is a matrix, and so taking one row or column leaves it as an array, and so you need array operations like .*, ./ etc.
 
Hepth said:
and so taking one row or column leaves it as an array, and so you need array operations like .*, ./ etc.
In this particular case, you do not need them. The only thing with elements in your expression is vecs(:,j), vals(j) and z(j) are both numbers. 2*[1 1 1] also evaluates to [2 2 2].

Frank Einstein said:
My attempts have not given propper results, this is my code:

for i=1:long
z(i)=normrnd(0,1);
f(i)=0;
endfor i=1:long%times
for j=1:long%number of terms in the descomposition
f(i)=f(i)+sqrt(eigenvalues2(j))*z(j)*eigenvectors(i,j);
end

end
In general, you should try to avoid for loops in matlab. It was written to take care of matrix operations.

What OP wants to do, given that vals is a N length row vector, z is a N length row vector, and vecs is a NxN matrix is likely on the form:

Code:
f = sqrt(vals.*z)*sqrt(vecs);
The first sqrt(vals.*z) is an N array containing sqrt(vals(j)*z(j)) in element j. The second sqrt contains the square root elements of vecs, which should also go into the sum. The matrix multiplication performs the sum. You may need to add primes on vals and z to get them to row vector format and you may need a prime on vecs if it is the other index you want to contract with.
 
Frank Einstein said:
z(i)=normrnd(0,1);
Same thing with initialisations. There is no point in calling your random number generator multiple times. The first few lines in the randn function help read:
Code:
randn Normally distributed pseudorandom numbers.
    R = randn(N) returns an N-by-N matrix containing pseudorandom values drawn
    from the standard normal distribution.  randn(M,N) or randn([M,N]) returns
    an M-by-N matrix.
so
Code:
z = randn(1,N);
will do perfectly fine to get a row vector with N numbers distributed according to ##\mathcal N(0,1)##.

Frank Einstein said:
f(i)=0;
Likewise, this should not be in a for loop. It should be
Code:
f = zeros(1,N);
 

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