Converting 4D matrix to 2D with multiple for-loop

Hello, I'm trying to convert 4D matrix to 2D matrix instead of for-loop.

Here is my code.

X = randi(10,100,10,100);
A = ones*(10,100);
SUM = zeros(10,100);

for k=1:10
    for l=1:100
        for m=1:10
            for n=1:100
                SUM(m,n) = SUM(m,n) + A(k,l) * X(k,l,m,n);
            end
        end
    end
end

I want to know how to change 4D to 2D for simpler calculation without for-loops. Thank you.

Answer 1

[Andrei Bobrov]

X = randi(100,10,100,10,100);
A = randi(20,10,100);
[k,l,m,n] = size(X);

z = bsxfun(@times,X,A);
out = reshape(sum(reshape(z,k*l,1,m,n),1),m,n);

or

out = squeeze(sum(reshape(z,k*l,1,m,n)));

[Stephen23]

+1 very nice

[Andrei Bobrov]

Thank you Stephen!

[James Choi]

Fantastic code!! Thank you so much.

This is much simpler than my code and also its performance is absolutely better than for loop.

You really help me improve entire calculation efficiency.

I think someone who want to optimize such as fmincon, they should use bsxfun and reshape.

Thank you again.

Answer 2

[John BG]

1.- your first command

randi(10,100,10,100)

does not generate a 4D matrix, but 3D the first input field of randi is the range [1:10] within the output values fall within. 4D would be

X=randi(10,10,100,10,100)

2.- reshaping matrices is fairly simple with command reshape

[s1,s2,s3,s4]=size(X)

Total=s1*s2*s3*s4

to a square matrix

X2=reshape(X,[Total^.5,Total^.5])

or to

X3=reshape(X,[10,Total/10])

3.- However, your 4 for loops do more than 'reshaping'. You take a slice of X, multiply it by A, and then cumulatively sum it, don't you?

check if the following 2 for loops are ok:

X=randi(10,10,100,10,100)
A = ones(10,100)

 [s1,s2,s3,s4]=size(X)

SUM = zeros(s3,s4)
        for m=1:s3
            for n=1:s4
                X2=X(:,:,m,n)
                X3=A.*X2
                SUM = SUM+ X3
            end
        end

If this answer helps in any way to solve your question please click on the thumbs-up vote link above, thanks in advance

John

[e-mail]

[James Choi]

Thank you for your kind explanation. It really help me understand how matlab code works.

[John BG]

Just compared the result from the really compact squeeze answer, and the 2 for loops answer, and they do not seem to be the same,

which one is correct?

John

[e-mail]

[Andrei Bobrov]

Hi John! Simple example:

X = randi(20,4,5,3,2);
A = randi(20,4,5);

SUM = zeros(3,2);
for k=1:4
    for l=1:5
        for m=1:3
            for n=1:2
                SUM(m,n) = SUM(m,n) + A(k,l) * X(k,l,m,n);
            end
        end
    end
end

[k,l,m,n] = size(X);
z = bsxfun(@times,X,A);
out = reshape(sum(reshape(z,k*l,1,m,n),1),m,n);

[John BG]

Thanks Andrei