Find column number for every row in matrix

Hello,

I'm trying to extract the column number of the first positive value in a matrix for every row, without using any loops.

For instance

-1 4 1

1 -1 -1

-5 4 -1

I want the output column vector to be:

2

1

2

I've attempted using find(), but I cannot restrict it to when there are multiple positive values in the same row.

Answer 1

[Stephen23]

M = [-1,4,1;1,-1,-1;-5,4,-1]

M = 3×3

-1 4 1 1 -1 -1 -5 4 -1

Method one: logical array, CUMSUM, and FIND:

X = M>=0;
Y = X&cumsum(X,2)==1;
[C,~] = find(Y.')

C = 3×1

2 1 2

Method two: NUM2CELL, ARRAYFUN, and FIND:

C = cellfun(@(v)find(v>=0,1,'first'),num2cell(M,2))

C = 3×1

2 1 2

Method three: CUMPROD and SUM:

C = 1+sum(cumprod(M<0,2),2)

C = 3×1

2 1 2

Method four: CUMPROD, ONES, and MTIMES:

C = 1+cumprod(M<0,2)*ones(size(M,2),1)

C = 3×1

2 1 2

Method five: CUMSUM and MAX:

[~,C] = max(cumsum(M>=0,2)==1,[],2)

C = 3×1

2 1 2

Method six: CUMMAX, SIGN, MIN, and SUM:

C = 1-sum(min(0,sign(cummax(M,2))),2)

C = 3×1

2 1 2

Method seven: FIND and ACCUMARRAY:

[R,C] = find(M>=0);
C = accumarray(R,C,[],@min)

C = 3×1

2 1 2

Method eight: is left as an exercise for the reader.

Answer 2

[the cyclist]

Here is one way:

M = [-1     4     1;
      1    -1    -1;
     -5     4    -1];

[r,c] = find(M>0);

[~,j] = unique(r);

out = c(j)

out = 3×1

2 1 2

Answer 3

[David Hill]

a=[ -1     4     1
     1    -1    -1
    -5     4    -1];
A=(a>0)';
[b,d]=find(A);
[~,e]=unique(d);
b(e)

ans = 3×1

2 1 2

Answer 4

[Les Beckham]

A = [  -1     4     1
        1    -1    -1
       -5     4    -1 ];
[row, col] = find(A > 0)

row = 4×1

2 1 3 1

col = 4×1

1 2 2 3

[~,idx,~] = unique(row, 'first')

idx = 3×1

2 1 3

col(idx)

ans = 3×1

2 1 2

Answer 5

[John D'Errico]

Simple. One line.

A = randi([-5,5],[10,7])

A = 10×7

0 -2 -5 -1 -3 1 0 2 1 -5 5 5 5 -2 3 0 4 -5 -2 5 -2 -3 5 -5 1 3 5 0 5 4 -4 -3 4 -5 -1 4 1 -4 -1 -1 -3 -5 2 4 -1 0 -2 5 1 0 1 3 0 3 5 -5 3 4 -1 1 -1 1 2 -2 5 -4 -1 1 -3 -1

[~,col] = max(A > 0,[],2)

col = 10×1

6 1 1 2 1 1 1 2 1 2

Why does it work? A > 0 returns an array that is true (1) only when a number is positive.

Max returns the location of the first such element (among equal maxes) that it finds. And since the A>0 matrix is a boolean one, it is composed of only 0 and 1 elements.

Will this fail if there are no positive elements at all in a row? Well, yes. In that case, it will return 1 for those rows.

But that is true of almost any of the schemes suggested by others. If you want a scheme that will return perhaps a NaN for those rows where there were no positive elements, you could patch it by one more test. I'll regenerate a new matrix with only a few columns, but many negative elements, to insure at least one row will have all non-positive elements.

A = randi([-10,5],[10,3])

A = 10×3

-4 2 -8 -1 -3 -4 -6 5 -2 -5 -10 4 4 -1 2 -4 4 0 -2 -4 -4 -9 -9 -1 -7 3 2 3 3 -7

[~,col] = max(A > 0,[],2);
col(~any(A>0,2)) = NaN

col = 10×1

2 NaN 2 3 1 2 NaN NaN 2 1

As you can see, in this final result, rows {2,7,8} had no positive elements at all, so a NaN was generated for those rows.

Or, perhaps you don't want to see a NaN generated. What we might do here is to append an extra column, that is ALWAYS positive. This is a common trick one can employ.

[~,col] = max([A,ones(size(A,1),1)] > 0,[],2)

col = 10×1

2 4 2 3 1 2 4 4 2 1

So for a 3 column array, it returns the number 4 where there were NO positive elements.

In the end, it is important that you decide what to do when NO elements are positive in a row, as you should expect to see a bug in your code at some point.

[Les Beckham]

Except that it doesn't quite work. Row 3 has no positive elements, but this approach returns a 1 for that row.