# looking for convenient way to extract matrix from 1*n*n

2 views (last 30 days)
Yu Li on 4 Jun 2023
Commented: Yu Li on 4 Jun 2023
Hi:
I have a 3*3*3 matrix, and I want to extract the data after its summation.
for example:
A=zeros(5,5,5)
B=sum(A,1);
now B is a 1*5*5 matrix, I want to assign it to a 5*5 matrix C but could not find a way, so looking for help here.
I tried
C=B(1,:,:)
but it does not work.
Thanks!

Ayush on 4 Jun 2023
Hi YuLi,
You can use squeeze function here:
A=zeros(5,5,5);
size(A) ;%5x5x5
B=sum(A,1);
size(B) %1x5x5
C=squeeze(B);
size(C) %5x5
squeeze function removes all singleton dimensions. You can read more about them here.
It might interest you : Adding extra dimension
Now as, here we are squeezing the dimension, we can also add another dimension to already existing matrix.
A=randi([-10,10],4,4);
size(A) %4x4
C(:,:,2)=C;
size(C) %4x4x2
Hope it helps.
Yu Li on 4 Jun 2023
thank you.

David Goodmanson on 4 Jun 2023
Edited: David Goodmanson on 4 Jun 2023
Hi YL,
A=zeros(5,5,5)
B=squeeze(sum(A,1));
Squeeze removes all singleton dimensions.
Yu Li on 4 Jun 2023
David, thank you too.

Walter Roberson on 4 Jun 2023
A = zeros(5,5,5)
A =
A(:,:,1) = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A(:,:,2) = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A(:,:,3) = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A(:,:,4) = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A(:,:,5) = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
B = permute(sum(A,1), [2 3 1]);
size(B)
ans = 1×2
5 5
squeeze() is a convenience function that ends up invoking permute(). squeeze() will not always give you the result you might expect:
C = squeeze(ones(1,1,5)), size(C)
C = 5×1
1 1 1 1 1
ans = 1×2
5 1
Notice that squeeze() removed all of the singular dimensions, not just the first singular dimension.
Yu Li on 4 Jun 2023
thank you Walter, Ayush's answer come first so I accepted his. but your answer extend my understanding, thank you.