# Assign a Sub Array to Array Knowing the Number of Dimensions at Run Time

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Royi Avital am 10 Feb. 2022
Kommentiert: Royi Avital am 11 Feb. 2022
Assume we have tA and tB with the same number of dimesions. We also have all(size(tB) <= size(tA)) == true.
The task is to embed tB in tA. Something like: tA(1:size(tB, 1), 1:size(tB, 2), ..., 1:size(tB, n)) = tB. Yet since we know the number of dimensions only at runtime, it can't be written explicitly like that.
The question, is there an efficient way to do so without eval or explicitly computer the cartesian product and use sub2ind()?
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Royi Avital am 10 Feb. 2022
Bearbeitet: Royi Avital am 11 Feb. 2022
OK, It turns out it can be done using Cell Arrays:
vSizeB = size(tB);
numDims = length(vSizeB); %<! Equals to ndims(tB)
cIdx = cell(numDims, 1);
for ii = 1:numDims
cIdx{ii} = 1:vSizeB(ii);
end
tA(cIDx{:}) = tB;
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Stephen23 am 11 Feb. 2022
Bearbeitet: Stephen23 am 11 Feb. 2022
"Your code won't work as you need the 1:vSizeB(ii) vector and not only a single number"
Lets try it:
tA = nan(5,4,3);
tB = randi(9,4,3,2);
tmp = arrayfun(@(n)1:n,size(tB),'uni',0);
tA(tmp{:}) = tB
tA =
tA(:,:,1) = 4 9 5 NaN 2 1 5 NaN 2 2 2 NaN 7 5 4 NaN NaN NaN NaN NaN tA(:,:,2) = 6 1 8 NaN 9 4 5 NaN 5 4 4 NaN 4 3 3 NaN NaN NaN NaN NaN tA(:,:,3) = NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
Royi Avital am 11 Feb. 2022
@Stephen, Well you edited the comment. It was, at first, only num2cell().

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Rik am 10 Feb. 2022
There is probably a better way, but you can fill a cell array with the index vectors (simple loop with ndims), and then use this:
tA(inds{:}) = tB;
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