Finding last non-zero rows for individual columns of a matrix without loops

I have the following code:

A =[1     2     3;
     1     0     1;
     0     5     1;
     1     0     2;
     2     3     2];
 
 [row,~] = find(A(1:4,[1,2]),1,'last');

What I was hoping to achieve through the above code is to find the last non-zero rows for individual columns. So my expected answer was

 row = [4 3] % since for the first column, the first non-zero element is on 4th row and similarly on 3rd row for the second column

However, with the above code, the answer I am getting is

 row = 3

I have done it with a for loop, but with the size of data I have, it is computationally expensive to do loops. I was hoping to get it done without loops.

Thank you in advance for your kind contribution.

Answer 1

[per isakson]

Try this

[r,c] = find( A==0 );
[~,ixa] = unique( c, 'last' );
>> r(ixa)
ans =
     3
     4

[J AI]

This seems to work just fine. Thanks a lot for your contribution.

[J AI]

I apologize for a slight mistake in my question, because the answer I was looking for was [4,3] instead of [3,4] (I have modified my question) since I am looking for non-zero elements, and not zero elements. However, your approach is still correct with a slight modification like this one:

[r,c] = find( A(1:4,1:2)~=0 )
[~,ixa] = unique( c, 'last' )
r(ixa)

Thank you once again!