DROP doesn't seem to work in PROC TRANSPOSE

[kbauphealth20]

I transposed the following data and included DROP = _NAME_ to drop the variable that was formerly used to describe the labels of a set of values:

Original list of variables: SSN, VisitDt, Measure, Value

PROC TRANSPOSE
       DATA = WORK.V1
       OUT = WORK.V2 (DROP = _NAME_
                                                  RENAME = (COL1 = HtIn
                                                                      COL2 = WtKg
                                                                      COL3 = SBP
                                                                      COL4 = DBP)
                                                  );
       VAR Value;
       BY SSN VisitDt;

RUN;

The new list of variables is:  SSN, VisitDt, _LABEL_, HtIn, WtKg, SBP, DBP

Why is _LABEL_ in the output? Shouldn't it have dropped? Any ideas on how to fix this? Thx!

[Reeza]

Why would you assume that _label_ would be dropped if _name_ was dropped? It's not mentioned in the DROP statement so if it's the output without any other DROP/RENAME I'd expect that you'd have to explicitly list it in the DROP statement as well.

FYI - personally I'd consider it dangerous to rename variables in that fashion, I'd highly recommend the usage of an ID and/or IDLABEL statement in your PROC TRANSPOSE code.

[kbauphealth20]

That's good to know. I was taught that it improves efficiency to rename in the set statement. Is the concern incorrectly renaming a variable and making it permanent?

[unison · Accepted solution]

If no other variables in output start with "_", then drop all _NAME_ _LABEL_ by specifying DROP= _:

Example:

proc transpose data=sashelp.class out=want(drop=_:);
by name;
id sex;
var height;
run;