# Matrix padding with logical index

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Piment on 6 Jul 2014
Answered: Roger Stafford on 6 Jul 2014
I have a matrix(6 by 3) of different row lengths:
0.9969 0.9724 0.3951
0.3590 0.5865 0.3983
0.6252 0.7780 0.7513
NaN 0.7277 0.5224
NaN NaN 0.4904
NaN NaN 0.0887
and a logical index matrix that has same or longer row length(7 by 3) like:
1 0 1
1 1 1
0 1 0
0 0 1
0 1 1
0 1 1
1 0 1
how can I get the following results(7 by 3) without loop:
0.9969 0 0.3951
0.3590 0.9724 0.3983
0 0.5865 0
0 0 0.7513
0 0.7780 0.5224
0 0.7277 0.4904
0.6252 0 0.0887
Thanks very much in advance

Cedric Wannaz on 6 Jul 2014
Edited: Cedric Wannaz on 6 Jul 2014
Assuming that the first array is A, the logical array is B, and you want to build C:
C = zeros( size( B )) ;
C(find( B )) = A(~isnan( A )) ;
EDIT : if, for any reason, you needed the row index in A of elements of C, you could get them as follows
>> cumsum( B ) .* B
ans =
1 0 1
2 1 2
0 2 0
0 0 3
0 3 4
0 4 5
3 0 6

#### 1 Comment

Piment on 6 Jul 2014
thank you very much Cedric!

### More Answers (1)

Roger Stafford on 6 Jul 2014
In case it is of interest to you, the following code does not depend on placing NaNs in the first array. It uses the 1's in the second array to determine where to place elements from the first array as they are taken out in sequential order in each column. The only requirement is that for each column there be enough rows in the first array to match the number of 1's in that column of the second array. I call the first array x, the second one y, and the result z.
[r1,c] = find(y ~= 0);
f = find([diff(c)~=0])+1;
r2 = ones(size(c,1),1);
r2(f) = r2(f)-diff([1;f]);
r2 = cumsum(r2);
z = zeros(size(y));
z(r1+size(z,1)*(c-1)) = x(r2+size(x,1)*(c-1));