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

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Al in St. Louis 2018년 10월 23일
댓글: Al in St. Louis 2018년 10월 23일
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 2018년 10월 23일
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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### 채택된 답변

Bruno Luong 2018년 10월 23일
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
##### 댓글 수: 1표시숨기기 없음
Al in St. Louis 2018년 10월 23일
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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### 추가 답변(1개)

Al in St. Louis 2018년 10월 23일
This works:
mIxPos = zeros (1, max (mPosIx));
[~,col,v]=find(mPosIx)
mIxPos (v) = col;

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R2018a

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