Find the index range of ones/trues?

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 个评论
显示无隐藏无

Yi-xiao Liu 2021-4-14

Brilliant, I didn't know strfind operates on numerical arrays

Walter Roberson 2021-4-15

It isn't documented, but it is a long-standing trick that is very convenient.

请先登录，再进行评论。

Answer 2

SungJun Cho 2021-4-14

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/801531-find-the-index-range-of-ones-trues#answer_674786

编辑：SungJun Cho 2021-4-14

在 MATLAB Online 中打开

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