Effacer les filtres
Effacer les filtres

Find the index range of ones/trues?

19 vues (au cours des 30 derniers jours)
Yi-xiao Liu
Yi-xiao Liu le 14 Avr 2021
Commenté : Walter Roberson le 15 Avr 2021
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)?

Réponse acceptée

Walter Roberson
Walter Roberson le 14 Avr 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 commentaires
Yi-xiao Liu
Yi-xiao Liu le 14 Avr 2021
Brilliant, I didn't know strfind operates on numerical arrays
Walter Roberson
Walter Roberson le 15 Avr 2021
It isn't documented, but it is a long-standing trick that is very convenient.

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Plus de réponses (1)

SungJun Cho
SungJun Cho le 14 Avr 2021
Modifié(e) : SungJun Cho le 14 Avr 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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