Finding column index of the first instance of 1

I have a logical array like this:

I want to extrat the colum index of the cells where the first instance of 1 is detected. Like this:

For example, you see here that the first two rows show 4, because that is where 1 is first detected.

Answer 1

[Walter Roberson]

C = sum(cumprod(~X, 2),1) + 1;

C will be one more than the number of columns for any row that has no 1.

[Pelajar UM]

This works, with a little adjustment:

C = sum(cumprod(~X, 2),2) + 1;

Now imagine X was computed inside a for loop. In the example above, k=1:5000

Is there a way to stop the loop before 5000 when all the rows have at least a 1?

Basically, I have a difficult time bringing C inside the loop and getting reasonable results....

[Torsten]

idx = find(sum(A,2)==0)
if idx == []
  break
end

But I don't know if it's more time-consuming to check the above condition for each k or to make MATLAB build the matrix A completely and to check the condition after the loop has finished.

[Pelajar UM]

Thanks @Torsten. The second line gives a warning:

Unexpected use of '[' in a scalar context.

And it appears to break the loop already in k=1.

Instead we can use

if isempty(idx) == 1
    break
end

But checking idx in every loop takes longer.

Method 1: 22 seconds

Method 2: 528 seconds

[Walter Roberson]

isempty(idx)

Answer 2

[Jonas]

use the find() function together with a loop over each row

[Pelajar UM]

Like this?

Doesn't work, because it doesn't find the first instance. It finds all the indices that meet this condition.

for p=1:n %n is the length of the logical array X
    G(p,:)=find (X(p,:));
end