Reconstruction of a "valid" expression within a function

You are passing just a string to subset().  At the very least you need 
to parse it (but still this does not work easily with subset() -- see 
below).  But are you sure you need to do this?  subset() for dataframes 
already accepts subset expressions involving the columns of the 
dataframe, e.g.:

 > df <- data.frame(x=1:10,y=rep(1:5,2))
 > subset(df, y==2)
   x y
2 2 2
7 7 2
 >

However, it's tricky to get subset() to work with an expression for its 
subset argument.  This is because of the way it evaluates its subset 
expression (look at the code for subset.data.frame()).

 > subset(df, parse(text="df$y==2"))
Error in subset.data.frame(df, parse(text = "df$y==2")) :
         'subset' must evaluate to logical
 > subset(df, parse(text="y==2"))
Error in subset.data.frame(df, parse(text = "y==2")) :
         'subset' must evaluate to logical
 >

It's a little tricky in general passing R language expressions around, 
because many functions that work with expressions work with the 
unevaluated form of the actual argument, rather than with an R language 
expression as the value of a variable.  E.g.:

 > with(df, y==2)
  [1] FALSE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE
 > cond <- parse(text="y==2")
 > cond
expression(y == 2)
 > with(df, cond)
expression(y == 2)

One way to make these types of functions work with R language 
expressions as the value of a variable is to use do.call():

 > do.call("with", list(df, cond))
  [1] FALSE  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE FALSE FALSE
 >

So, returning to subset(), you can give it an expression that is stored 
in the value of a variable like this:

 > do.call("subset", list(df, cond))
   x y
2 2 2
7 7 2
 >

However, if you're a beginner at R, I suspect that you'll get much 
further if you avoid such meta-language constructs and just find a way 
to make subset() work for you without trying to paste together R 
language expressions.

Hope this helps,

-- Tony Plate

Hello all,

I have some trouble in reconstructing a valid expression within a
function,
here is my question.

I am building a function :

SUB<-function(DF,subset=TRUE) {
#where DF is a data frame, with Var1, Var2, Fact1, Fact2, Fact3
#and subset would be an expression, eg. Fact3 == 1 

#in a first time I want to build a subset from DF
#I managed to, with an expression like eg. DF$Fact3,
# but I would like to skip the DF$ for convenience
# so I tried something like this :

tabsub<-deparse(substitute(subset))
dDF<-deparse(substitute(DF))

if (tabsub[1]!="TRUE") {
subset<-paste(dDF,"$",tabsub,sep="")}

#At this point, I have a string that seems to be the expression that I
want
sDF<-subset(DF, subset)
}

#But I have an error message :

Error in r & !is.na(r) : operations are possible only for numeric or

logical types


I can not understand why is that, even after I've tried to convert
properly the string into an expression.
I've been all the day trying to sort that problem ...
Maybe this attempt is ackward and I have not understood what is really
behind an expression. 
But if anyone could give me a tip concerning this problem or point me to
relevant references, I would really appreciate.

Thanks
Pascal Boisson
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