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# How to remove items from two arrays that index each other without for loop

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Al in St. Louis on 23 Oct 2018
Commented: Al in St. Louis on 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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Al in St. Louis on 23 Oct 2018
It's a line of code that assigns the index of the for loop, idx, to the correct position. Here's the output of running the code on my test data. I get two vectors with the correct values in the correct positions. I just want to get rid of the for loop.
>> mIxPos
mIxPos =
1 0 0 0 2
>> mPosIx
mPosIx =
1 5

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### Accepted Answer

Bruno Luong on 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
##### 1 CommentShowHide None
Al in St. Louis on 23 Oct 2018
That's much better than calling find and saving two of its three outputs. My function has been reduced to:
t = [obj.network.node(obj.posIx).type];
mPosIx = obj.posIx (t == 13 | t == 9);
mIxPos (mPosIx) = 1:numel (mPosIx);

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### More Answers (1)

Al in St. Louis on 23 Oct 2018
This works:
mIxPos = zeros (1, max (mPosIx));
[~,col,v]=find(mPosIx)
mIxPos (v) = col;
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