minimun value in a column matrix with its index

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

Answer 1

[Dyuman Joshi]

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]

thank you so much !

Answer 2

[Voss]

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

[kurdistan mohsin]

thanks alot!

[Voss]

You're welcome!