How do I find the index of the first element equal to 0 for each row of a matrix in a fast way

Hi,

I have a NxL matrix, how can I obtain a vector containing the index of the first element equal to 0 of each row of the matrix in in a fast way? Each row always has an element equal to 0 so don't worry about possible errors of that kind. I would like to use the command 'find' in such a way that with one single call it gives me back the entire vector idx which, at the moment, I obtain in the following way:

(M is the given matrix)

idx=zeros(N,1)
for k=1:N
    idx(k)=find(M(k,:)==0,1);
end

Answer 1

[Walter Roberson]

You cannot do it with a single find()

A vectorized way to do it is

M = ones(5,10); M(randperm(50,10)) = 0;
M

M = 5×10

1 1 1 1 0 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 1 1 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1

idx = sum(cumprod(logical(M),2),2)+1;

idx

idx = 5×1

5 3 2 11 1

[Walter Roberson]

Sometimes when people have these kinds of questions, their setup does not have negative values or complex values. In that case, you can get the results in a single call using min() with two outputs.

Answer 2

[Torsten]

[~,idx] = max(M==0,[],2)

or

idx = arrayfun(@(k)find(M(k,:)==0,1),1:size(M,1))

[Walter Roberson]

Interesting use of max() !