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bknitch
Quartz | Level 8

I have a data set that's been transposed already. I'm trying to find a way to replace the 1's with the numeric value it corresponds to in the variable field. I'm trying to un-transpose the variable and make the table long and skinny rather than wide. I'm also trying to replace the variable names with one name as 'COND_C'.

 

There are roughly 90 variables that start at CC1 and end in CC189. I tried it with an if statement but it did not capture it correctly. 

 

ID CC1 CC2 CC6 CC8 CC9 CC100 CC108
123 1 0 0 0 0 1 0
124 0 1 0 0 0 1 0
125 1 0 0 1 0 0 0

 

This is what I would like the output to look like. 

ID COND_C
123 1
123 100
124 2
124 100
125 1
125 8
1 ACCEPTED SOLUTION

Accepted Solutions
r_behata
Barite | Level 11
data have;
input ID$ CC1	CC2	CC6	CC8	CC9	CC100	CC108;
cards;
123 1 0 0 0 0 1 0
124 0 1 0 0 0 1 0
125 1 0 0 1 0 0 0
;
run;

data want;
set have;
array C_[*]  CC:;

do i = 1 to dim(C_);
	if C_[i] > 0 then 
		COND_C=input(compress(vname(C_[i]),'CC'),8.);
	else continue;
	output;
end;

drop CC:;
run;

View solution in original post

3 REPLIES 3
Reeza
Super User

Do a wide to long proc transpose and filter out all the 0's.

proc transpose data=have out=want(where=(col1 ne 0) rename = _name_ = cond_c);
by ID;
var CC1-CC189;
run;
r_behata
Barite | Level 11
data have;
input ID$ CC1	CC2	CC6	CC8	CC9	CC100	CC108;
cards;
123 1 0 0 0 0 1 0
124 0 1 0 0 0 1 0
125 1 0 0 1 0 0 0
;
run;

data want;
set have;
array C_[*]  CC:;

do i = 1 to dim(C_);
	if C_[i] > 0 then 
		COND_C=input(compress(vname(C_[i]),'CC'),8.);
	else continue;
	output;
end;

drop CC:;
run;
bknitch
Quartz | Level 8
This was perfect, thank you for this!

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