Matrix padding with logical index

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

Answer 1

[Cedric]

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

[Piment]

thank you very much Cedric!

Answer 2

[Roger Stafford]

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));