Creating a new Variable - Grouping columns by an interaction

[jonatan_velarde]

Good night SAS friends:

Continuing the last post, i want to ask for some help to solve this problem, the data set is as follows:

data have;
input LAMBING_ORDER    ANIMAL     SEX$    DAM   SIRE ANIMAL_BIRTH_DATE:mmddyy10.    month_ABD    day_ABD    year_ABD    ANIMAL_BIRTH_TYPE;
format animal_birth_date yymmdd10.;
cards;
1    6870    M  1   2    10/13/2009    10    13    2009    1
1    5555    F  3   4    10/14/2009    10    14    2009    1
1    5556    M  5   6    10/14/2009    10    14    2009    2
1    5557    M  5   6    10/14/2009    10    14    2009    2
1    5558    M  7   8    10/15/2009    10    15    2009    2
1    5559    F  9  10    10/15/2009    10    15    2009    2
1    5560    F  9  10    10/15/2009    10    15    2009    2
1    5561    F  11 10    10/15/2009    10    15    2009    2
1    5562    F  11 10    10/15/2009    10    15    2009    2
1    5563    M  12 10    10/16/2009    10    14    2009    1
1    5564    F  12 10    10/16/2009    10    14    2009    1
2    6971    F  12 10    10/27/2010    10    13    2010    3
2    6972    F  12 10    10/27/2010    10    13    2010    3
2    6973    M  12 10    10/27/2010    10    13    2010    3
;

Using the informarion of SEX and birth_type, its needed to create a new variable associated to each animal considering the interaction of birth type with lamb sex resulting this vaues:

1 - one male lamb (this means: birth_type = 1 and sex = M)

2 - one female lamb, (this means: birth_type = 1 and sex = F) 

3 - two male lambs, (this means: birth_type = 2 and sex = M, from the same DAM) 

4 - two females lambs, (this means: birth_type = 2 and sex = F, from the same DAM) 

5 - one male lamb and one female lamb (this means: birth_type = 2 and sex = M and F, from the same DAM) 

6 - more than two lambs, independent of sex (this means: birth_type = > 3 ) 

Thank you very much for your help.

Regards

[Ksharp · Accepted solution]

You didn't post the output you need yet ? Assuming I understand what you mean.

data have;
input LAMBING_ORDER    ANIMAL     SEX$    DAM   SIRE ANIMAL_BIRTH_DATE:mmddyy10.    month_ABD    day_ABD    year_ABD    ANIMAL_BIRTH_TYPE;
format animal_birth_date yymmdd10.;
cards;
1    6870    M  1   2    10/13/2009    10    13    2009    1
1    5555    F  3   4    10/14/2009    10    14    2009    1
1    5556    M  5   6    10/14/2009    10    14    2009    2
1    5557    M  5   6    10/14/2009    10    14    2009    2
1    5558    M  7   8    10/15/2009    10    15    2009    2
1    5559    F  9  10    10/15/2009    10    15    2009    2
1    5560    F  9  10    10/15/2009    10    15    2009    2
1    5561    F  11 10    10/15/2009    10    15    2009    2
1    5562    F  11 10    10/15/2009    10    15    2009    2
1    5563    M  12 10    10/16/2009    10    14    2009    1
1    5564    F  12 10    10/16/2009    10    14    2009    1
2    6971    F  12 10    10/27/2010    10    13    2010    3
2    6972    F  12 10    10/27/2010    10    13    2010    3
2    6973    M  12 10    10/27/2010    10    13    2010    3
;
run;
data want;
length flag $ 40;
flag='more than two lambs';
do until(last.DAM);
 set have;
 by ANIMAL_BIRTH_TYPE DAM notsorted;
 if sex = 'M' then has_M=1;
 if sex = 'F' then has_F=1;
end;

do until(last.DAM);
 set have;
 by ANIMAL_BIRTH_TYPE DAM notsorted;
 if ANIMAL_BIRTH_TYPE = 1 and sex = 'M' then flag='one male lamb';
 if ANIMAL_BIRTH_TYPE = 1 and sex = 'F' then flag='one female lamb';
 if ANIMAL_BIRTH_TYPE = 2 and has_M and not has_F then flag='two male lambs';
 if ANIMAL_BIRTH_TYPE = 2 and has_F and not has_M then flag='two females lambs';
 if ANIMAL_BIRTH_TYPE = 2 and has_M and has_F then flag='one male lamb and one female lamb';
 output;
end;
drop has_:;
run;