Hi,
all the non-zero elements are in column 1 - not in column 6:
m=rand(4,960);
m = [m; zeros(20,960)];
m=permute(reshape(m, 4,6,960), [1 2 3]); % need this 3d matrix for some operation
m1=squeeze(m(:,1,:));
Sometimes for debugging it is helpful to have a look to the results that your code produces - See here for m after performing the permute and reshape:
val(:,:,1) =
0.5108 0 0 0 0 0
0.5300 0 0 0 0 0
0.1545 0 0 0 0 0
0.2341 0 0 0 0 0
val(:,:,2) =
0.6191 0 0 0 0 0
0.3090 0 0 0 0 0
0.2135 0 0 0 0 0
0.2826 0 0 0 0 0
val(:,:,3) =
0.5438 0 0 0 0 0
0.7191 0 0 0 0 0
0.0393 0 0 0 0 0
0.6480 0 0 0 0 0
You can continue this up to 960 - but the result keeps the same. So now you know the reason for your problem, but what do you want to achieve with permute? Note that:
>> A = [1 2 3; 4 5 6; 7 8 9]
A =
1 2 3
4 5 6
7 8 9
>> B = permute(A, [1 2 3])
B =
1 2 3
4 5 6
7 8 9
In other words - your permute does not do anything - if you leave the permute command away the resulting matrix m keeps the same:
m1 = permute(reshape(m, 4,6,960), [1 2 3]);
m2 = reshape(m, 4,6,960);
sum(sum(sum(m1 == m2))) == 4*6*960
ans =
logical
1
This seems to be the original problem you have - your permute call does not change your matrix the way that you expect it to do.
Best regards
Stephan
