On Sep 15, 2010, at 5:45 PM, Mark Ebbert wrote:
Dear R gurus,
I regularly come across a situation where I would like to apply a
function to a subset of data in a dataframe, but I have not found an
R function to facilitate exactly what I need. More specifically, I'd
like my function to have a context of where the data it's analyzing
came from. Here is an example:
### BEGIN ###
func<-function(x){
m<-median(x$x)
if(m > 2 & m < x$y){
return(T)
}
return(F)
}
The semantic question is what are you trying to test when you say "m <
x$y" ? "m" is a scalar and x is a vector. By default only the first
element of x$y will be compared (not actually callable in that manner.)
tmp<-
data.frame(x=1:10,y=c(rep(34,3),rep(35,3),rep(34,4)),z=c(rep("a",
3),rep("b",3),rep("c",4)))
res<-aggregate(tmp,list(z),func)
I see Dennis has tried to move you forward to the plyr strategy, but
some of us are mired in the traditonal ways:
?split # returns a dataframe in segments defined by a factor
> func<-function(x){
+ m<-median(x["x"], na.rm=TRUE)
+ if(m > 2 && m < x["y"]){
+ return(T)
+ }
+ return(F)
+ }
>
> tmp<-
data.frame(x=1:10,y=c(rep(34,3),rep(35,3),rep(34,4)),z=c(rep("a",
3),rep("b",3),rep("c",4)))
> res<-lapply(split(tmp,list(tmp$z)), func)
> res
$a
[1] FALSE
$b
[1] TRUE
$c
[1] TRUE
### END ###
The values in the example are trivial, but the problem is that only
one column is passed to my function at a time, so I can't determine
how 'm' relates to 'x$y'. Any tips/guidance is appreciated.
--
David Winsemius, MD
West Hartford, CT
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