[R] Setting elements in data frame

[Peter Dalgaard BSA]

Kaspar Pflugshaupt <pflugshaupt at cirsium.ethz.ch> writes:

> On Thursday 22 February 2001 14:17, Peter Dalgaard BSA wrote:
> 
> 
> > This seems to work, save for a (sort of spurious) warning that 6000 is
> > not a valid replacement value for the factor column:
> >
> > x[,]<-lapply(x,function(z){z[z==6]<-6000;z})
> >
> > (Why the comma? Search me...)
> >
> 
> Works also without the comma:
> 
> x <- data.frame(lapply(x,function(x) {x[x==6]<-6000;x}))
> 
> but still gives the warning.

Yes. I was trying to avoid the data.frame step by assigning the list
coming out of lapply back to the original frame, thus keeping all
attributes and class info.

You can get rid of the warning with 

x[2:3]<-lapply(x[2:3],function(z){z[z==6]<-6000;z})

or generically

s <-sapply(x,is.numeric)
x[s] <- lapply(x[s],function(z){z[z==6]<-6000;z})

Hmmmm... How about

recode<-function(x,value,replacement){
	x[if (is.na(value)) is.na(x) else x==value] <- replacement
	x
}
s <-sapply(x,is.numeric)
x[s] <- lapply(x[s], recode, 6, 6000)
x[s] <- lapply(x[s], recode, NA, 1000)

-- 
   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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