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
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;
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;
Are you ready for the spotlight? We're accepting content ideas for SAS Innovate 2025 to be held May 6-9 in Orlando, FL. The call is open until September 25. Read more here about why you should contribute and what is in it for you!
Learn how use the CAT functions in SAS to join values from multiple variables into a single value.
Find more tutorials on the SAS Users YouTube channel.