查找最小正值及其在矩阵各列中的位置

Question

I need to find the minimum positive values in each column and its position inside the column of a certain matrix. So if I have:

A = [1 4
     2 3
     3 6]

I need to obtain the values 1 and 3, and the positions 1 and 2. Doing this inside a for loop I obtain correctly the minimum values and its position, but it also catches the negative values:

for bit = 1:2
    [y(bit),x(bit)] = min(A(:,bit));
end

And if I use:

[y(bit),x(bit)] = min(A(A(:,bit)>0));

I don't receive the expected result. What I'm doing wrong? Thanks.

Answer 1

This can be easily achieved using inf and min ...

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New method using inf and no looping

Take some random example:

% Generated using A = randi([-100, 100], 10, 3)
A = [ 31    41   -12
     -93   -94   -24
      70   -45    53
      87   -91    59
      36   -81   -63
      52    65    -2
      49    39   -11
     -22   -37    29
      31    90    42
     -66   -94    51];

Set all negative values to positive infinity, which will ensure they are never the minimum value in the column.

A(A<=0) = inf; 
% if you want to preserve A, use A2=A; A2(A<=0)=inf;

Now you can just use the min function as expected.

[mins, idx] = min(A);
% mins = 31, 39, 29: as expected
% idx  = 1, 7, 8: the indices of the above values in each column as expected.

By default, min will get the column-wise minimum as you want.
To specify this explicitly, use min(A,[],1) , see the documentation for more details.

Note that you could achieve the same result by using NaN instead of inf .

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Your method

In response to why you were getting an unexpected result, it's because you weren't selecting the column of A in your loop, the second attempt should be corrected to

[y(bit),x(bit)] = min(A(A(:,bit)>0, bit));

However , this will still give an unexpected result! The minimums will be correct, but their indices will be lower than expected. This is because the indices will only count the positive values in each column, so you will get the nth positive number rather than the nth number. The easiest "workaround" is to abandon this method and use the quicker one above which doesn't require looping.