Pre-locating an array

Susan (view profile)

on 10 Apr 2019
Latest activity Commented on by Susan

on 10 Apr 2019

James Tursa (view profile)

Hi MATLAB guys,
I got stuck at some point and really appreciate your help.
Do you know how can I pre-locate the value of X in the following code? X is a matrix and for each i and j its dimentions will change. For example for given i and j, X has a dimention of Ri * Tj.
N = [1 2 3 4 5 6 7 8 9 10];
M = [11 22 33];
X === zeros %%%%pre-locating here
for k = 1 : 10
for m = 1 : N(k)
for i = 1 : 3;
for j = 1: M(i)
X(:,:, j,i,m,k) = sth;
end
end
end
end

Rik

Rik (view profile)

on 10 Apr 2019
Do you know the size of sth in advance?
Susan

Susan (view profile)

on 10 Apr 2019
Thanks for your reply. It depends on the i and j and for different i and j the size of this matrix is different. And, yes, I know the sizes in advance.

James Tursa (view profile)

on 10 Apr 2019
Edited by James Tursa

James Tursa (view profile)

on 10 Apr 2019

Numeric arrays must be rectangular. You cannot have different dimensions for different slices. To get that behavior you would need to use something like a cell array to store your different sized variables. E.g.,
X{j,i,m,k} = sth;
But, even though the contents of the cell array elements do not all have to be the same size, the cell array itself must be rectangular. You could still use it in your above scheme as long as it was OK to have a bunch of empty cells in the spots you were not using.

Susan

Susan (view profile)

on 10 Apr 2019
Thanks for your reply. I would like to use cell. As you said, since the dimensions of matrix X is changing, the best way to store it is cell array. I just don't know how to do that. Any idea? Thanks
Rik

Rik (view profile)

on 10 Apr 2019
You can use the code James suggested, and you can do it like this:
N = [1 2 3 4 5 6 7 8 9 10];
M = [11 22 33];
X=cell(max(M(:)),numel(M),max(N(:)),numel(N));
for k = 1 : numel(N)
for m = 1 : N(k)
for i = 1 : numel(M)
for j = 1: M(i)
X{j,i,m,k} = sth;
end
end
end
end
Susan

Susan (view profile)

on 10 Apr 2019
Thank you so much! I appreciate your help.