The most efficient way to locate the first 1's in each row of a matrix

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Ulrik William Nash
Ulrik William Nash 2019 年 10 月 19 日
コメント済み: Jos (10584) 2019 年 10 月 21 日
I have a potentially large r by c matrix, m, consisting of 1's and 0's. What is the most efficient way create a r by 1 vector, v, containing the index of the first 1 in each row of m.
For example,
m = [0 1 1;1 0 0; 0 0 1]
should result in
v = [2;1;3]

回答 (2 件)

Star Strider
Star Strider 2019 年 10 月 19 日
The max function will return the indices of the first instances of the maximum value it encounters, in the desired dimension. Since your matrix only has (0,1), it is appropriate here.
This:
m = [0 1 1;1 0 0; 0 0 1];
[~,v] = max(m,[],2)
produces:
v =
2
1
3
  1 件のコメント
Jos (10584)
Jos (10584) 2019 年 10 月 21 日
nice! One can shorten it to:
[~, v] = max(m')
... but what if there is no 1 in a row

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Jos (10584)
Jos (10584) 2019 年 10 月 19 日
m = [0 1 1 1 ;1 0 1 0; 0 0 1 1 ; 0 0 0 1]
% a simple solution
[r, c] = find(m') ;
clear v1
v1(r,1) = c
% another solution
[v2, ~] = find(cumsum(cumsum(m,2),2)==1)

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