minimun value in a column matrix with its index
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hi, I write the bellow code to find the minimum value in a column vector but it should not consider zero as a minimum , but how to find the exact indexing of the minimum value?
in the example the row index of minimum value exept zero should be 2 , why it give me 1?
can anybody correct it for me so the row_idx = 2 !
A=[0; 5 ;6 ;9 ;55]
[min_value,row_idx]=min(A(A~=0))
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Accepted Answer
Dyuman Joshi
on 10 Jun 2022
Edited: Dyuman Joshi
on 10 Jun 2022
A=[0; 5 ;6 ;9 ;55]
A(A~=0) %when you do this, your vector changes
Method 1
minvalue=min(A(A~=0))
rowidx=find(A==minvalue)
Method 2
A(A==0)=nan
[min_value, row_idx]=min(A)
More Answers (1)
Voss
on 10 Jun 2022
A(A~=0) is a 4-by-1 vector:
A=[0; 5 ;6 ;9 ;55];
A(A~=0)
When you do min on that you find that 5 is the minimum, which occurs at index 1, as you can see right there.
To get the correct index in A (not in A(A~=0)) of the non-zero minimum, you can store the indices of the non-zero elements of A:
nz_idx = find(A~=0)
then use min:
[min_value,temp_idx] = min(A(nz_idx))
and then get the index in A:
row_idx = nz_idx(temp_idx)
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