# Alternative for nested for loop

14 views (last 30 days)
Fawad Farooq Ashraf on 20 Feb 2021
I need to implement this logic to make combinations of a matrix. This code is a bit cumbersome since I need to add more rows and columns at some point. Is there a simpler way to do this?
k = 0;
for x1 = 1:2
for x2 = 1:2
for x3 = 1:2
for y1 = 1:2
for y2 = 1:2
for y3 = 1:2
for z1 = 1:2
for z2 = 1:2
for z3 = 1:2
k = k+1;
JV{k} = [J1(1,x1),J2(1,x2),J3(1,x3);
J1(2,y1),J2(2,y2),J3(2,y3);
J1(3,z1),J2(3,z2),J3(3,z3)];
end
end
end
end
end
end
end
end
end
J1, J2 & J3 are 3x2 matrices.
Rik on 20 Feb 2021
You can probably use a function that generates all permutations to simplify this. Maybe even ndgrid could help you out.

Adam Danz on 20 Feb 2021
Edited: Adam Danz on 21 Feb 2021
> Is there a simpler way to do this?
As long as you're preallocating the JV cell array, your approach is very simple, fast, and readable. The alternative below does the same thing but is much less readable and it's slower (there may be faster ways than mine).
However, depending on how you're using the JV cell array, it may be more efficient to pack those values into a 3D array instead. So, my recommendations are
1. keep the nested loops; it's fast, readable, and simple.
2. definitely preallocate the JV variable whether you're using a cell array or 3D array.
3. Consider using a 3D array for JV.
% Demo values; J1 J2 and J3 must be the same size
J1 = [1:3;4:6]';
J2 = J1.*10;
J3 = J2.*10;
allJ = cat(3,J1, J2, J3); % 3x2x3
colidx = flipud(combvec(1:2,1:2,1:2,1:2,1:2,1:2,1:2,1:2,1:2)); % * see note
rowidx = repmat([1 1 1 2 2 2 3 3 3]', 1, size(colidx,2));
pageIdx = repmat([1 2 3 1 2 3 1 2 3]', 1, size(colidx,2));
ind = sub2ind(size(allJ), rowidx, colidx, pageIdx);
JVec = allJ(ind);
JArray = permute(reshape(JVec, 3,3,[]),[2,1,3]); % you could stop here and use a 3D array
JV = arrayfun(@(page){JArray(:,:,page)}, 1:size(JArray,3));
I ran your nested conditional version with the same inputs and confirmed that both methods produce the same results.
Note: combvec requires the deeplearning toolbox. See this answer for an alternative.
Adam Danz on 23 Feb 2021
Edited: Adam Danz on 23 Feb 2021
I disagree with the documentation that those statements are equivalent (btw @Fawad han I edited your comment to fix the broken URL).
1. Problems when cell array size decreases
Here's an example of what Rik said.
for i = [ 5 4 3];
c{1,i} = [];
disp(numel(c))
end
5 5 5
Compare that to
for i = [ 5 4 3];
c = cell(1,i);
disp(numel(c))
end
5 4 3
2. Problems when var already exists but isn't a cell array
c = 1:20;
c{1,5} = []
Unable to perform assignment because brace indexing is not supported for variables of this type.
But this works,
c = cell(1,5)
1×5 cell array
{0×0 double} {0×0 double} {0×0 double} {0×0 double} {0×0 double}
Fawad Farooq Ashraf on 23 Feb 2021
I see. I've been using it wrong then. Thank you for correcting me.

### Categories

Find more on Logical in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!