# minimun value in a column matrix with its index

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kurdistan mohsin on 10 Jun 2022
Commented: Voss on 10 Jun 2022
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]
A = 5×1
0 5 6 9 55
[min_value,row_idx]=min(A(A~=0))
min_value = 5
row_idx = 1

Dyuman Joshi on 10 Jun 2022
Edited: Dyuman Joshi on 10 Jun 2022
A=[0; 5 ;6 ;9 ;55]
A = 5×1
0 5 6 9 55
A(A~=0) %when you do this, your vector changes
ans = 4×1
5 6 9 55
Method 1
minvalue=min(A(A~=0))
minvalue = 5
rowidx=find(A==minvalue)
rowidx = 2
Method 2
A(A==0)=nan
A = 5×1
NaN 5 6 9 55
[min_value, row_idx]=min(A)
min_value = 5
row_idx = 2
kurdistan mohsin on 10 Jun 2022
thank you so much !

Voss on 10 Jun 2022
A(A~=0) is a 4-by-1 vector:
A=[0; 5 ;6 ;9 ;55];
A(A~=0)
ans = 4×1
5 6 9 55
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)
nz_idx = 4×1
2 3 4 5
then use min:
[min_value,temp_idx] = min(A(nz_idx))
min_value = 5
temp_idx = 1
and then get the index in A:
row_idx = nz_idx(temp_idx)
row_idx = 2
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Voss on 10 Jun 2022
You're welcome!