Find the index range of ones/trues?

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For example, given a logical vector:
[0,0,1,0,0,1,1,0,1,1,1,0,0...]
I need a two-column matrix:
[3,3;
6,7;
9,11;...]
that register the start and end index of each "1 blocks". Is there any convenient way to do that (ideally vectorized)?

Accepted Answer

Walter Roberson
Walter Roberson on 14 Apr 2021
test_vector = [0,0,1,0,0,1,1,0,1,1,1,0,1,1,1,1];
starts = strfind([0 test_vector], [0 1]);
stops = strfind([test_vector 0], [1 0]);
out = [starts(:), stops(:)]
out = 4×2
3 3 6 7 9 11 13 16
  2 Comments
Yi-xiao Liu
Yi-xiao Liu on 14 Apr 2021
Brilliant, I didn't know strfind operates on numerical arrays
Walter Roberson
Walter Roberson on 15 Apr 2021
It isn't documented, but it is a long-standing trick that is very convenient.

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More Answers (1)

SungJun Cho
SungJun Cho on 14 Apr 2021
Edited: SungJun Cho on 14 Apr 2021
I made a brief function that gives you a two-column matrix of the start and end indices when a logical vector is given as an input. I used the iteration method here, but I am pretty sure there is a recursion method to solve this problem. Hope this helps.
test_vector = [0,0,1,0,0,1,1,0,1,1,1,0,1,1,1,1];
output = get_true(test_vector);
function [output] = get_true(test_vector)
test_array = find(test_vector == 1);
test_idx = zeros(length(test_array)-1,2);
for ii = 1:length(test_array)
if ii == length(test_array)
break;
end
test_idx(ii,1) = test_array(ii);
test_idx(ii,2) = test_array(ii+1);
end
output = zeros(length(test_idx),2);
for idx = 1:length(test_idx)
idx_start = test_idx(idx,1);
idx_end = test_idx(idx,2);
if idx_start+1 < idx_end
output(idx,1) = idx_start;
output(idx,2) = idx_end;
end
end
output = sort(output(output~=0));
output = vertcat(test_array(1), output, test_array(end));
output = reshape(output,2,length(output)/2)';
end
Best,
SungJun

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