How to remove items from two arrays that index each other without for loop
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Al in St. Louis
le 23 Oct 2018
Commenté : Al in St. Louis
le 23 Oct 2018
This is hard to explain in words. Here's my code, and I'd like to know whether there's a way to vectorize the part with a for loop:
t = [obj.network.node(obj.posIx).type];
mPosIx = obj.posIx (t == 13 | t == 9);
mIxPos = zeros (1, max (mPosIx));
for idx = 1:length(mPosIx)
mIxPos (mPosIx (idx)) = idx;
end
This code does exactly what I need it to do. It removes all items that aren't type 9 or 13. The mPosIx array points to the correct position in the node array. The mIxPos array either contains a zero where there is no corresponding element in the position array (which does not appear in this code), or it contains the correct index in the position array. The for loop just feels clunky, but I couldn't figure out another way to do it.
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Bruno Luong
le 23 Oct 2018
You can vectorize LHS as with RHS, the for loop becomes one line
mIxPos(mPosIx) = 1:length(mPosIx);
That returns the inverse of permutation mPosIx
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