## Data step

Frequent Contributor
Posts: 141

# Data step

```data final;
set newl6;
d1=0.95*dv;
if cmt=11 then dv=d1;
run;

data final1;
set final;
d2=0.05*dv;
if cmt=23 then dv=d2;
run;```

I ran the attached code on this data set but the data set final1 code did not populate cmt=23.  Can someone tell me why since it worked on CMT=11?   Is there a better way to populate the desired CMTS 11 and 23 with the d1 and d2 values?

subj           time         dv             cmt        amt               evid        mdv         wt               d1              d2

 1 1 0 1 30000 1 1 23 0 0 1 1 0 2 30000 1 1 23 0 0 1 1 0 3 0 2 1 23 0 0 1 1 0 4 0 2 1 23 0 0 1 1 0 5 0 2 1 23 0 0 1 1 0 6 0 2 1 23 0 0 1 1 0 7 0 2 1 23 0 0 1 1 0 8 0 2 1 23 0 0 1 1 0 9 0 2 1 23 0 0 1 1 0 10 0 2 1 23 0 0 1 1 0.4864 11 0 2 0 23 0.4864 0.02432 1 1 0 12 0 2 1 23 0 0 1 1 0 13 30000 1 1 23 0 0 1 1 0 14 30000 1 1 23 0 0 1 1 0 15 0 2 1 23 0 0 1 1 0 16 0 2 1 23 0 0 1 1 0 17 0 2 1 23 0 0 1 1 0 18 0 2 1 23 0 0 1 1 0 19 0 2 1 23 0 0 1 1 0 20 0 2 1 23 0 0 1 1 0 21 0 2 1 23 0 0 1 1 0 22 0 2 1 23 0 0 1 1 0 23 0 2 1 23 0 0 1 1 0 24 0 2 1 23 0 0
Super User
Posts: 2,041

## Re: Data step

[ Edited ]

combine the 2 datsteps

```data final;
set newl6;
d1=0.95*dv;
if cmt=11 then dv=d1;
run;

data final1;
set final;
d2=0.05*dv;
if cmt=23 then dv=d2;
run;```

to

```data final;
set newl6;
d1=0.95*dv;d2=.05*dv;
if cmt=11 then dv=d1;else if cmt=23 then dv=d2;
run;

```
Frequent Contributor
Posts: 141

## Re: Data step

I changed the code to include the value for d2 but it didn't work. I still
got a dv=0 for cmt=23 though the value for d2 was in the d2 column.

data final;

set newl6;

d1=0.95*dv;

d2=0.05*dv;

if cmt=11 then dv=d1;

else if cmt=23 then dv=d2;

Super User
Posts: 13,876

## Re: Data step

Since your variables do not align with the "data" it is very hard to read.

You should show the values before any calculation. Since your are overwriting the value of dv we can't see what the actual starting. But if the original value of dv was 0 where cmt=23 then d2 = 0* 0.05 = 0, so dv=d2 would = 0. A further clue that dv was 0 is that d1 =0 for cmt=23 also. So the final value of dv =0 is not a failed assignment.

If you expected a different value then show a starting value for dv not = 0 and the result. Some example code. Also see how to show data as a data step so we can test something.

```data work.newl6;
input subj time dv cmt amt evid mdv wt;
datalines;
1 1 0.4864 11 0 2 0 23
1 1 .5 23 0 2 1 23

data work.final;
set work.newl6;
d1=0.95*dv;
if cmt=11 then dv=d1;
run;

data work.final1;
set work.final;
d2=0.05*dv;
if cmt=23 then dv=d2;
run;```
Frequent Contributor
Posts: 141

## Re: Data step

``````dATA NORMA;
SET NORM2A;
IF TIME=1 AND Y GT 11 THEN DELETE;
IF TIME=2 AND Y GT 28 THEN DELETE;
IF TIME=3 AND Y GT 27 THEN DELETE;
IF TIME=4 AND Y GT 26 THEN DELETE;
IF TIME=5 AND Y GT 15 THEN DELETE;
IF TIME=6 AND Y GT 15 THEN DELETE;
IF TIME=8 AND Y GT 15 THEN DELETE;
IF TIME=24 AND Y GT 15 THEN DELETE;
rename id=wsubj;
RUN;

proc print data=norma; run;

proc sort data=norma;
by wsubj time ;run;

data newL2;
do time=0 to 24 by 1;
DO CMT=1 TO 24;
do wsubj= 1 to 14;
OUTPUT;
END;
end;
end;
RUN;

proc sort data=newl2;
by wsubj time ;run;

data newl3;
retain wsubj time cmt;
set newl2;
if time=7 then delete;
if time=9 then delete;
if time=11 then delete;
if time=13 then delete;
if time=14 then delete;
if time=15 then delete;
if time=16 then delete;
if time=17 then delete;
if time=18 then delete;
if time=19 then delete;
if time=20 then delete;
if time=21 then delete;
if time=22 then delete;
if time=23 then delete;
keep wsubj time cmt;
run;

proc sort data=newl3;
by wsubj time ;run;

Data newl4;
merge  norma newl3;
by wsubj time ;
y1=0.95*y;
y2=0.05*y;
run;

``````

I was able to get the desired result by restructuring the SAS code to that which I posted above.  The desired data was obtained by merging norma and newl3 and taking the desired fractions of y.

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