Question

Here I have two matrix, one indicating cost and the other determines when to take into comparison.

cost =      [0.2 0.0 0.3; 0.4 0 0; 0.5 0 0];
available = [1   1   0  ; 1   0 0; 0   0 0];
available = logical(available);

I want to get the index of the min available element in the cost matrix, which in this case would compare 0.2, 0.0 and 0.4 and return the index of 0.0, which is (1, 2) or 4 in the cost matrix.

I tried

mul = cost .* available;     % Zero if not available, but I can't know if it is zero because cost is zero
mul(~mul) = nan;             % Set zero to be NaN
[minVal, minId] = min(mul)

This will help to get the min non-zero cost but if there exists zero elements which are available, it would be wrong.

So is there a better way to do so?

Answer 1

Here are two possible solutions. Both essentially involve converting all non-available costs to Inf.

%#Set up an example
Cost =      [0.2 0 0.3; 0.4 0 0; 0.5 0 0];
Available = [1   1   0; 1   0 0; 0   0 0];

%#Transform non-available costs to Inf
Cost(Available == 0) = Inf;

%#Obtain indices using find
[r, c] = find(Cost == min(min(Cost)))

%#Obtain linear indices and convert using ind2sub
[~, I1] = min(Cost(:));
[r2, c2] = ind2sub(size(Cost), I1);

Both solutions will only return the first minimum value in the instance that there is not a unique minimum. Also, the method will fail in the perverse case that all the available costs are Inf (but I guess you’ve got bigger problems if all your costs are infinite…).

I’ve done a few speed tests, and the second method is definitely faster, no matter what the dimensions of Cost, so should be strictly preferred. Also, if you only want linear indices and not subscript indices then you can of course drop the call to ind2sub. However, this doesn’t give you huge savings in efficiency, so if there is a preference for subscript indices then you should use them.