Number of distinct values across variables

[Viktor Emonds]

Hello,

I have what feels like a very elementary problem, yet so far I have failed to find a proper solution to it. For various family members, a closed question on ancestry was asked and I simply would like to construct a variable that counts the number of distinct values across those variables for each respondent, disregarding missings. For example, if the variable is missing for 2 family members and has the value 4 in the other cases, the new variable should be 1, if the row of values is 3 3 4 3 5 . . , it should be 3 etc.

Does anyone know an easy way to do this? Thanks!

Best,

Viktor

[Carole J. Wilson]

See rownvals in help egenmore

rownvals(numvarlist) [ , missing ] returns the number of distinct values in each observation for a set
of numeric variables numvarlist. Thus if the values in one observation for five numeric variables
are 1, 1, 2, 2, 3 the function returns 3 for that observation. Missing values, i.e. any of . .a
... .z, are ignored unless the missing option is specified. (Stata 9 required.)

[Nick Cox]

Carole is naturally right. There is a write-up of this problem within http://www.stata-journal.com/sjpdf.h...iclenum=pr0046 Section 7.

Fond though I am of that function, it's worth stressing another way to do it too. reshape, work out the number of distinct values within groups and reshape back.

clear 
input y1 y2 y3 
1    1  1
1    2  1
1    3  1
2    1  1
2    2  1
2    3  1
3    1  1
3    2  1
3    3  1
end 
gen id = _n 
reshape long y, i(id) 
egen tag = tag(y id) 
egen nvals = total(tag), by(id)
drop tag 
reshape wide y, i(id)   
list, sep(0)  

     +---------------------------+
     | id   y1   y2   y3   nvals |
     |---------------------------|
  1. |  1    1    1    1       1 |
  2. |  2    1    2    1       2 |
  3. |  3    1    3    1       2 |
  4. |  4    2    1    1       2 |
  5. |  5    2    2    1       2 |
  6. |  6    2    3    1       3 |
  7. |  7    3    1    1       2 |
  8. |  8    3    2    1       3 |
  9. |  9    3    3    1       2 |
     +---------------------------+

[Viktor Emonds]

Thank you very much for the quick and useful answers, much appreciated!