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Sathish_jammy
Lapis Lazuli | Level 10

I have a dataset to calculate by the given formulae. Kindly suggest a code to resolve it.

data Have;
input ID Val Type$;
cards;
1 10.5 normal
2 12.4 abnormal
3 16.6 normal
4 10.8 normal
5 13.9 abnormal
6 17.1 normal
7 14.4 abnormal
8 11.2 abnormal
run;

Formula: count (abnormal)/sum(Val) * 1000 : i.e: 4/51.9*1000

1 ACCEPTED SOLUTION

Accepted Solutions
Tom
Super User Tom
Super User

What if there are no abnormal types?

proc sql;
 select sum(type='abnormal') as N
      , sum(case when (type='abnormal') then val else . end) as D
      , calculated N/calculated D *1000 as want
   from have 
 ;
quit;
       N         D      want
----------------------------
       4      51.9  77.07129

View solution in original post

8 REPLIES 8
FreelanceReinh
Jade | Level 19

Hello @Sathish_jammy,

 

With PROC SQL you can do it in one step:

proc sql;
select divide(1000, m)
from (select mean(val) as m from have where type='abnormal');
quit;

 

Tom
Super User Tom
Super User

What if there are no abnormal types?

proc sql;
 select sum(type='abnormal') as N
      , sum(case when (type='abnormal') then val else . end) as D
      , calculated N/calculated D *1000 as want
   from have 
 ;
quit;
       N         D      want
----------------------------
       4      51.9  77.07129
mkeintz
PROC Star

@Tom wrote:

What if there are no abnormal types?

And what if there is a missing VAL for an abnormal type?

--------------------------
The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set

Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for
Allow PROC SORT to output multiple datasets

--------------------------
mkeintz
PROC Star

What you are asking for is the reciprocal of the mean of VAL for the subgroup type='abnormal'.

 

If your task expands to generating such a calculation for many subgroups/many variables, or if you might simultaneously want this for a hierarchy of subgroups, you might consider using PROC SUMMARY followed by a data step.   Let's say you want this not only for "abnormal" but also for "normal" and for the entire population, and maybe you have a second variable, NEWVAL:

 

data Have;
input ID Val Type$;
newval=val + 3*uniform(12531366);
cards;
1 10.5 normal
2 12.4 abnormal
3 16.6 normal
4 10.8 normal
5 13.9 abnormal
6 17.1 normal
7 14.4 abnormal
8 11.2 abnormal
run;

proc summary data=have ;
  class type;
  var  val newval;
  output out=need (where=(_stat_='MEAN'));
run;

data want;
  set need (drop=_:);
  val_result=1000*(1/val);
  newval_result=1000*(1/newval);
run;

Run it and take a look at the final and intermediate result datasets.

 

 

--------------------------
The hash OUTPUT method will overwrite a SAS data set, but not append. That can be costly. Consider voting for Add a HASH object method which would append a hash object to an existing SAS data set

Would enabling PROC SORT to simultaneously output multiple datasets be useful? Then vote for
Allow PROC SORT to output multiple datasets

--------------------------
Ksharp
Super User
data Have;
input ID Val Type$;
cards;
1 10.5 normal
2 12.4 abnormal
3 16.6 normal
4 10.8 normal
5 13.9 abnormal
6 17.1 normal
7 14.4 abnormal
8 11.2 abnormal
;
run;


proc sql;
 select count(val) as N
      , sum(val) as D
      , calculated N/calculated D *1000 as want
   from have 
   	where type='abnormal'
 ;
quit;
PaigeMiller
Diamond | Level 26

@Ksharp wrote:
data Have;
input ID Val Type$;
cards;
1 10.5 normal
2 12.4 abnormal
3 16.6 normal
4 10.8 normal
5 13.9 abnormal
6 17.1 normal
7 14.4 abnormal
8 11.2 abnormal
;
run;


proc sql;
 select count(*) as N
      , sum(val) as D
      , calculated N/calculated D *1000 as want
   from have 
   	where type='abnormal'
 ;
quit;

For other data sets, which may have missing values of variable VAL, the above code gives the wrong answer, while code using PROC MEANS or PROC SUMMARY still gives the right answer when missing values are present.

--
Paige Miller
Ksharp
Super User

Ha Paige, I just edited my code to fix the problem you are talking about .

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