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Asked by Itzik Ben Shabat
on 7 Nov 2012

Hi, I am trying to creat a matrix of all possible coordinates given 2 coordinate vectors. my code is attached below. I know there is a better way to do this. can anyone help?

n=8 %can be any number Ycoord=[0.5:n-0.5]; Xcoord=[0.5:n-0.5]; for i=1:n for j=1:n handles.coordinates{i,j}={Xcoord(i),Ycoord(j)}; end end

thanks

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Answer by Jan Simon
on 7 Nov 2012

Edited by Jan Simon
on 7 Nov 2012

Accepted answer

No, there is no "better" way, as long as you want to store the data in a cell array. You could use `mat2cell`, but then the same loops are performed inside this function also. One small improvement is to omit the unnecessary square brackets around the vectors (MLint should mention this already):

Ycoord = 0.5:n-0.5; Xcoord = 0.5:n-0.5;

If you could store the values in a numerical array, simplifications are possible:

nX = length(Xcoord); nY = length(Ycoord); handles.coordinates = cat(3, repmat(Ycoord, nX, 1), repmat(Xcoord', 1, nY));

Then an pair of coordinates is restored by:

C = squeeze(handles.coordinates(i, j, :));

Answer by Daniel
on 7 Nov 2012

It depends what you mean by better. To me the best code is the the code that is easiest for me to maintain unless it is a time bottleneck. If the maintainable code makes my overall function take too long, then I look into optimizing.

For your code, I would preallocate handles.coordinates and probably save the data as doubles instead of a cell array. As for eliminating the loops, you can and you will see an improvement, but then the code might be less readable and harder to maintain for you. For n equal to 8 (or any number of that order of magnitude) efficiency doesn't really matter unless you are calling the code a lot.

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