column number in which the minimum value for each row is located

[Maria Navarro]

Hi all,

I would like to obtain a matrix c that stores instead of the minimum value for each row, the column number in which the minimum value for each row is located.
For instance in the following matrix a

100 200 500 300
20 10 50 30
3 5 2 1

the command b=rowmin(a) will give that the minimum values for each row are

100
10
1

How could I obtain a matrix c that gives me the column number in which the minimum value for each row is located?

1
2
4

Thank you very much in advance for your help

Best wishes,
Maria

[William Lisowski]

Welcome to Statalist!

A question: what you you expect to be done if the minimum is not unique in a given row?

[Christophe Kolodziejczyk]

See help mata minindex() for a function which finds the indices of the minina and how the function handles ties. The function is only for vectors so you will have to loop over columns.

[Hua Peng (StataCorp)]

Also look at -help mf_st_matrix- since it looks like you are dealing with Stata matrix instead of Mata matrix, you need use -st_matrix()- to get the Stata matrix into Mata, then use -minindex()- as Chistophe suggested to get the result.

[Maria Navarro]

Thank you very much for your help!

minindex is giving exactly the vector that I need (see output attached) but I would like to store the printed results (vector j': column number/numbers in which the minimum value is located) in a matrix where each row will be the result obtained in each iteration. I will really appreciate your help.

for (i=1;i<=rows(a);i++) {
    minindex(a[i,],1,j,w)
    j'
    }

[Christophe Kolodziejczyk]

This solution only works if you have no ties

b = J(rows(a),1,.)
for (i=1;i<=rows(a);i++) {
    b[i,1] = minindex(a[i,],1,j,w)

    }

if you do have ties, one solution is to use a vector of pointers

b = J(rows(a),1,NULL)
for (i=1;i<=rows(a);i++) {
    b[i,1] = &minindex(a[i,],1,j,w)

    }

you refer to the result of the ith row by using the dereferencing operator

i=1
*b[1,i]

or you can just create a matrix with enough columns to contain the maximum number of ties (which you have to determine in advance)