How to find first '1' in every row

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Kasmayanti Sakaria
Kasmayanti Sakaria 2017-11-7
编辑： Jan 2019-5-23
Hi,
I have a matrix,
A = [ 0 0 0 0 0 0 1 1 1 1 0 0; 0 0 0 0 0 1 1 1 1 0 0 0; 0 0 0 0 0 0 1 1 1 1 0 0]
My question is, how to find the first '1' in each row. I want the output will show like this:
B = [7; 6; 7]
Meaning that, for the first row, the number 1 found on column number 7, second row found in column number 6 and so on.
Thank you in advance for any answered

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采纳的回答

Matt J
Matt J 2017-11-7
[~,B]=max(A,[],2);
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显示 2更早的评论
ME
ME 2019-5-20
perfect solution! Thanks!

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更多回答（3 个）

Jos (10584)
Jos (10584) 2019-5-20
Not better than using max (for this type of input), but just to show you an alternative:
A = [ 0 0 0 0 0 0 1 1 1 1 0 0; 0 0 0 0 0 1 1 1 1 0 0 0; 0 0 0 0 0 0 1 1 1 1 0 0]
[c, ~] = find(cumsum(A, 2).' == 1)
  2 个评论
ME
ME 2019-5-21
another good solution, but for larger matrices slower
tic; [~, B] = max(A,[],2); toc;
tic; [c, ~] = find(cumsum(A, 2).' == 1); toc;
size(A)
Elapsed time is 0.014103 seconds.
Elapsed time is 0.153354 seconds.
ans =
1000 12501
Jos (10584)
Jos (10584) 2019-5-21
of course it is slower, as it requires three internal loops (cumsum, ==, & find) rather than one :-D

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Walter Roberson
Walter Roberson 2019-5-23
A = [ 0 0 0 0 0 0 1 1 1 1 0 0; 0 0 0 0 0 1 1 1 1 0 0 0; 0 0 0 0 0 0 1 1 1 1 0 0];
c = sum(cumprod(~A,2),2)+1;
I do sometimes use this techique to do a vectorized "find" if the first or last element along a dimension. However it does force you to think carefully about what value is "reasonable" for the case where there is no match. In this form of the code, a row with no 1 would give you a result which is 1 greater than the number of columns.
Jan
Jan 2019-5-23
编辑：Jan 2019-5-23
Speed test:
A = randi([0,1], 1000, 12501);
tic; % MAX
[~, B] = max(A,[],2);
toc;
>> 0.007 sec
tic; % CUMSUM
[c, ~] = find(cumsum(A, 2).' == 1);
toc;
>> 0.160 sec
tic; % FOR over rows
n = size(A,1);
d = zeros(1, n);
for k = 1:n
d(k) = find(A(k, :), 1);
end
toc
>> 0.193 sec
tic; % FOR over columns
At = A.';
n = size(At, 2);
e = zeros(1, n);
for k = 1:n
e(k) = find(At(:, k), 1);
end
toc
>> 0.119 sec
tic; % FOR, Accumulate
n = size(A, 2);
d = zeros(size(A,1), 1);
for k = 1:n
m = A(:, k);
d(m & ~d) = k;
if all(d)
break;
end
end
toc
>> 0.009 sec % Worst case: 0.220 sec!!!

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