I have an nx1 column vector, each element e_n an integer between 1 and p. I want to convert it to an nxp matrix with a 1 as the e_nth element of each row and the rest 0.
have = {6,9,1,...}
want = {0 0 0 0 0 1 0 0 0 ... ,
0 0 0 0 0 0 0 0 1 ... ,
1 0 0 0 0 0 0 0 0 ... ,
... }
What's the easiest way to do that?
The easiest way is to use the SUB2NDX function to convert (row,col) subscripts into indices. You didn't specify the dimensions of the final matrix, so in the following example I used a 10-column matrix. You can modify the dimension as you need to:
proc iml;
have = {6,9,1,2};
/* Somewhere you need to specify the columns of want?
Is it max(have)? Is is square? */
want = j(nrow(have),10,0);
rows = T(1:nrow(have));
idx = sub2ndx(dimension(want), rows||have);
want[idx] = 1;
print want;
The easiest way is to use the SUB2NDX function to convert (row,col) subscripts into indices. You didn't specify the dimensions of the final matrix, so in the following example I used a 10-column matrix. You can modify the dimension as you need to:
proc iml;
have = {6,9,1,2};
/* Somewhere you need to specify the columns of want?
Is it max(have)? Is is square? */
want = j(nrow(have),10,0);
rows = T(1:nrow(have));
idx = sub2ndx(dimension(want), rows||have);
want[idx] = 1;
print want;
Or how about the 'unique design' trick?
proc iml;
have = {6, 9, 1, 6, 2};
want = j( nrow(have), max(have), 0);
want[ , unique(have)] = design(have);
print want;
Or use this simple way to get index you need. proc iml; have = {6,9,1,2}; want = j(nrow(have),max(have),0); idx = t(have)+(0:nrow(have)-1)#max(have); want[idx] = 1; print want; quit;
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