Look for minimum value in array

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Hello,
I have a function like this:
i=1;
Min=ones(1,200);
while i>=200
min_y = min(y(:,i));
pos_y = find(y(:,i)==min_y);
phi0s = x(pos_y ,i);
Min(1,i) = phi0s;
i=i+1;
end
Its job is to find a min value of y and as a results give a corresponding x value.
x and y are arrays with a dimension of n x 200 - let say that I stored results of 200 experiment in those two arrays and now I need to find smallest y value in each column and corresponding x.
Can I improve it somehow or even better run it without a loop? I will be grateful for any tips!

Accepted Answer

Rik
Rik on 2 Sep 2020
Loops aren't slow. You generally only need to remove them if there are native functions that accept array inputs, which happens to be the case here:
%generate random data
n=4;
x=rand(n,200);
y=rand(n,200);
%your corrected code
i=1;
Min=ones(1,200);
while i<=200%why aren't you using a for-loop here?
min_y = min(y(:,i));
pos_y = find(y(:,i)==min_y);
phi0s = x(pos_y ,i);
Min(1,i) = phi0s;
i=i+1;
end
%vectorized
[~,idx]=min(y,[],1);
ind=sub2ind(size(x),idx,1:size(y,2));
Min2=x(ind);
isequal(Min2,Min)
  4 Comments
Image Analyst
Image Analyst on 2 Sep 2020
And the 1 can be omitted
Min(i) = phi0s;
And you should probably use a for loop like he suggested, and use "col" or "column" for the column index rather than i which is the imaginary variable sqrt(-1). Or use the vectorized approach. It's more compact and cryptic than a for loop though. So use whatever you're comfortable with. For loops will be fast, particularly if there are only a microscopic 200 iterations.
Mateusz Brzezinski
Mateusz Brzezinski on 2 Sep 2020
Ok, thanks I will do that

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More Answers (1)

David Hill
David Hill on 2 Sep 2020
[Y,idx]=min(y);
X=x(length(y)*(0:length(idx)-1)+idx);

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