Find the First element that satisfies a condition

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Tyler
Tyler 2016-3-12
评论： Jay Edelman 2023-8-11
I have an nxn matrix and would like to find the first element in each row that satisfies a given condition, and gives me NaN if there are none that meet the condition. The find command will give me all elements that satisfy it, but I only care about the first one in each row.
For instance, if my matrix is A = [7 3 9;5 8 8;0,2,6] and my condition is A<5, I would like to use a function that gives me [3 NaN 0] or at least tells me where in the matrix I can find those values and use it to index A.
Thanks!

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采纳的回答

Stephen23
Stephen23 2016-3-12
编辑：Stephen23 2016-3-12
No loops or find is required, just some logical indexing does the trick:
>> A = [7 3 9;5 8 8;0,2,6];
>> B = A.'; % important: arrange the data columnwise
>> X = B<5;
>> Y = cumsum(X,1)==1 & X;
>> Z = any(Y,1);
>> out(Z) = B(Y);
>> out(~Z) = NaN
out =
3 NaN 0
  2 个评论
Tyler
Tyler 2016-3-12
Thanks! Is there a way to output only the index of each row?
Jay Edelman
Jay Edelman 2023-8-11
A = [7 3 9;5 8 8;0,2,6];
X = A < 5;
[r,c] = find(cumsum(X>0,2)==1 & X);
idxs = NaN(height(X),1);
idxs(r) = c;

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更多回答（3 个）

Andrei Bobrov
Andrei Bobrov 2016-3-12
编辑：Andrei Bobrov 2016-3-14
t = A < 5;
x = cumsum(t,2) == 1 & t;
out = sum(x.*A,2);
out(~any(t,2)) = nan
idx = x*(1:size(A,2))'
  4 个评论
显示 2更早的评论
Olubunmi Anthony Jeyifous
Olubunmi Anthony Jeyifous 2018-11-11
Hi I have a similar problem. I have a long column vector and i would like to know the first row number the condition is met.

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Bill Tubbs
Bill Tubbs 2021-4-12
If you only want to find the locations in A where the condition is true and then index those values you can use find:
>> elements_idx = find(A<5)
elements_idx =
3
4
6
>> A(elements_idx)
ans =
0
3
2
Note: the indices in elements_idx are not (row, col) indices as you might expect for a matrix but linear indices starting in the top-left element and incrementing down through each column from left to right.
MHN
MHN 2016-3-12
编辑：MHN 2016-3-12
If you dont mind a little coding:
A = [7 3 9;5 8 8;0,2,6];
FirstCon = zeros(size(A,1),1);
i=1;
B = A'; % I am going to use the power of Matlab for processing a whole column, so I have to reverse the original matrix, so the rows become columns.
for I = B
temp = I(I<5);
if isempty(temp)
FirstCon(i,1) = NaN;
else
FirstCon(i,1) = temp(1);
end
i=i+1;
end

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