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

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.

[madhan ravi]

provide your datas to test

[Al in St. Louis]

Here you go. This works, but the real data has thousands of elements. The principle is the same.

obj.posIx=[1, 5, 4, 3, 2];
t=[9 13 1 1 1];
mPosIx = obj.posIx (t == 13 | t == 9);
mIxPos = zeros (1, max (mPosIx));
for idx = 1:length(mPosIx)
    mIxPos (mPosIx (idx)) = idx;
end

[madhan ravi]

what do you mean by this?

      mIxPos (mPosIx (idx)) = idx;

[Al in St. Louis]

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

Answer 1

[Bruno Luong]

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

[Al in St. Louis]

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);

Answer 2

[Al in St. Louis]

This works:

mIxPos = zeros (1, max (mPosIx));
[~,col,v]=find(mPosIx)
mIxPos (v) = col;