[R] Turning strings into expressions

Peter Dalgaard p.dalgaard at biostat.ku.dk
Thu Nov 25 18:53:16 CET 2004


"Dimitris Rizopoulos" <dimitris.rizopoulos at med.kuleuven.ac.be> writes:

> Hi Alexander,
> 
> you could try:
> 
> my.string <- "A==1 & B==2"
> (my.frame <- data.frame(A=sample(1:2, 20, TRUE), B=sample(1:2, 20,
> TRUE)))
> subset(my.frame, eval(parse(text=my.string)))

Hmm, considering the nonstandard evaluation that is going on inside
subset(), I think I'd rather try

 my.frame <- data.frame(A=sample(1:2, 20, TRUE), B=sample(1:2, 20,TRUE))
 my.string <- "A==1 & B==2"
 l <- as.list(parse(text=my.string))
 names(l)<-"sub"
 eval(substitute(subset(my.frame, sub), l))
   A B
18 1 2


(or perhaps l <- list(sub=parse(text=my.string)[[1]]) is less
cryptic).

Point being that this way you'll literally evaluate

> substitute(subset(my.frame, sub), l)
subset(my.frame, A == 1 & B == 2)


-- 
   O__  ---- Peter Dalgaard             Blegdamsvej 3  
  c/ /'_ --- Dept. of Biostatistics     2200 Cph. N   
 (*) \(*) -- University of Copenhagen   Denmark      Ph: (+45) 35327918
~~~~~~~~~~ - (p.dalgaard at biostat.ku.dk)             FAX: (+45) 35327907




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