Find the First element that satisfies a condition

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!

Answer 1

[Stephen23]

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

[Tyler]

Thanks! Is there a way to output only the index of each row?

[Jay Edelman]

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;

Answer 2

[Andrei Bobrov]

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))'

[Star Strider]

Andrei, you never cease to amaze me with the efficiency of you code.

My only criticism is that you need to be a bit more careful when copying it to paste to your Answer:

A = [7 3 9;5 8 8;0,2,6];
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))'

Add the last parenthesis and the transpose operator and it works.

[Andrei Bobrov]

Thank you Star for your comment.

[Star Strider]

My pleasure.

That is brilliant code!

[Olubunmi Anthony Jeyifous]

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.

Answer 3

[Bill Tubbs]

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.

Answer 4

[MHN]

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