How do I find the indices of nonzero elements for a matrix of logical vectors?

For example, I have the following matrix with logical vectors as rows:

A=[1 0 0 0; 0 0 1 0; 0 0 0 1]

A = 3×4

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

I would like to get a vector with the indices of the first nonzero element for each row, such that the answer for matrix A should be:

v=[1;3;4]

v = 3×1

1 3 4

I'd be grateful for any tip how to do this. Thank you

Answer 1

[Matt J]

A=[1 0 0 0; 0 0 1 0; 0 0 0 1];

[~,w]=max(A,[],2)

w = 3×1

1 3 4

[Distelfink]

Thank you! Yes, that works perfectly and is the general answer to finding the column indices of the first nonzero element per row - see also my related question, which also received that answer.

Answer 2

[Image Analyst]

Try find():

A=[1 0 0 0; 0 0 1 0; 0 0 0 1]

A = 3×4

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

[rows, columns] = find(A)

rows = 3×1

1 2 3

columns = 3×1

1 3 4

rows and columns are the (row, column) location of all the non-zero values in A.

[Distelfink]

I noticed that there is one slight issue with the solution: Suppose my matrix has one more row, such that it is

B=[1 0 0 0; 0 0 1 0; 0 0 0 1;1 0 0 0]

B = 4×4

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

Then

[~, columns]=find(B)

columns = 4×1

1 1 3 4

What I would like instead, however, is a vector w=[1;3;4;1]. Do you have an idea how to achieve this?

[Matt J]

B=[1 0 0 0; 0 0 1 0; 0 0 0 1;1 0 0 0];

[w,~]=find(B')

w = 4×1

1 3 4 1

[Torsten]

w = zeros(size(B,1),1);
for row = 1:size(B,1)
    w(row) = find(B(row,:),1,'first');
end
w

[Distelfink]

Thanks a lot, Torsten. Is there also a vectorized, i.e. non-loop solution?

[Torsten]

Use Matt's answer - it's more elegant.

But I wonder how you interprete the result if you have one or more rows of zeros.

[Distelfink]

Matt's solution works perfectly! I find it a little confusing why it works (why is the output of the find function for 'row' not sorted but for 'column' it is sorted?) - ...but I can live with that :) Thank you all!

[Matt J]

why is the output of the find function for 'row' not sorted but for 'column' it is sorted?

They are sorted lexicographically. Find() searches along columns first, then rows second.

[Image Analyst]

You should do it like I originally suggested, getting both the row and column for the 1's. If you don't, then it can be confusing. For example, using your protocol, what would you want for a matrix that had more than one 1 in a row and had rows with no 1's in them, like this:

B=[1 0 0 0; 0 0 0 0; 0 0 1 0; 1 0 1 1; 0 0 0 1; 1 0 0 0]

B = 6×4

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

If you do it like I originally suggested, there would be no problem or ambiguity.

[Distelfink]

Thanks, @Matt J. The lexicographic sorting by find() was the missing piece for me to understand why your solution works.

@Image Analyst: Thanks for your comment, too. I should have made clear in my question that in my case each row contains exactly one element equal to one, and that the matrix contains rows that are identical.

[Distelfink]

Just a quick question to the respondents, since I am new to the forum: What is the best practice for me to make clear in this thread that my question was resolved? As I mentioned, I originally asked the question in an imprecise way and thus the answer I was looking for (Matt's answer in the comments above) was only posted in a comment to the answer above. Should I "accept" Image Analyst's answer nevertheless? Should I edit my original question and hint towards the comments of the answer? Thank you

[Image Analyst]

I'd ask Matt to post his answer as an official Answer. If he does, then you can Accept his answer and award him reputation points. If you learned anything from any of the other Answers in a post you can optionally "Vote" for them to award that poster reputation points also. You can only Accept one answer but you can vote for as many as you want.