Update counts according to a label - sas

Question

suppose to have the following:
data have;
input ID :$20. Label :$20. Hours :$20. Days :$20.;
cards;
0001 w 3144 3
0001 w 23 54
0001 p 12 1
0002 m 456 34
0002 w 2 1
0002 s 231 45
0002 w 98 23
0003 w 12 6
0003 w 98 76
;
Is there a way to, for each ID, sum the Hours, so get the total and then split it by the days but only when the label is == w? If the label is not w put a missing.
Desired output:
data have;
input ID :$20. Label :$20. Hours :$20. Days :$20.;
cards;
0001 w 167.3158 3
0001 w 3011.684 54
0001 p . 1
0002 m . 34
0002 w 32.79167 1
0002 s . 45
0002 w 754.2084 23
0003 w 8.048778 6
0003 w 101.9512 76
;
In other words: for 0001 in the desired output example I added: 3144+23+12 = 3179, the 54+3=57 that are the days where the label is "w" then I divided 3179 by 57 and multiplied the result for 3 and 54 but not for 1 respectively.
Thank you in advance

Answer 1

Same idea with #Stu Sztukowski, but use DOW-Loop skill:
data want;
do until(last.id);
set have;
by id notsorted;
sum_of_hours=sum(sum_of_hours,input(hours,best.));
sum_of_days_w=sum(sum_of_days_w,(label='w')*input(days,best.));
end;
do until(last.id);
set have;
by id notsorted;
if label='w' then hours=cats(sum_of_hours*(input(days,best.)/sum_of_days_w));
else hours='';
output;
end;
run;

Answer 2

The calculation you need to do looks like this in code form:
if(id = 'w') then hours = sum_of_hours_w/sum_of_days * days
else hours = .
All we need to do is get the sum of hours and days where label = 'w', then merge it back with our original table by id. The table to do this calculation would look like this:
id
label
hours
days
sum_of_hours
sum_of_days_w
0001
w
3144
3
3179
57
0001
w
23
54
3179
57
You can accomplish this all in a single SQL step.
proc sql;
create table want as
select t1.id
, t1.label
, CASE(t1.label)
when('w') then t2.sum_hours/t2.sum_days_w * t1.days
else .
END as hours
, t1.days
from have as t1
/* Get the sum of all hours and days where label = 'w' */
LEFT JOIN
(select id
, sum( (label = 'w')*days ) as sum_days_w
, sum(hours) as sum_hours
from have
group by id
) as t2
ON t1.id = t2.id
;
quit;