Replace the value in the columns of a matrix

In matrix A = [-1 2 3; 4 -5 6; 7 8 -9] order for k1 = 1: size (A, 1) A (k1, find (A (k1,:)> 0, 1, 'first')) = 1 makes replacement of the first positive value in each row 1. If it changes the expression so: for k1 = 1: size (A, 1) A (k1, find (A (k1,:)> 0, 1, 'last')) = 1 will replaced the last positive value in each row. How to do so, to replace the last positive value of 1, but in each column?

[the cyclist]

Is this a homework problem? It seems a bit hard to believe that you could write the code for doing rows, but not understand how to do it for columns.

[Mario Marinov]

Not quite so, but thanks!

Answer 1

[Kirby Fears]

You can switch the loop to iterate over the columns instead of the rows by changing

 k1=1:size(A,1)

to

 k1=1:size(A,2)

Then switch the indexing of A similarly, changing

 A(k1, find (A (k1,:)> 0, 1, 'last'))

to

 A(find(A(:,k1)> 0, 1, 'last'),k1)

[Mario Marinov]

Thank you so much! Work and understand why!