Hi
Can someone help me with this?
How can I apply a function to a list of variables.
something like this
listvar=list("Monday","Tuesday","Wednesday")
func=function(x){x[which(x<=10)]=NA}
lapply(listvar, func)
were
Monday=[213,56,345,33,34,678,444]
Tuesday=[213,56,345,33,34,678,444]
...
in my case I have a neverending list of vectors.
Thanks!
On 08.11.2011 17:59, Ana wrote:> Hi > > Can someone help me with this? > > How can I apply a function to a list of variables. > > something like this > > > listvar=list("Monday","Tuesday","Wednesday")This is a list of length one character vectors rather than a "list of variables".> func=function(x){x[which(x<=10)]=NA}To make it work, redefine: func <-function(x){ x <- get(x) is.na(x[x<=10]) <- TRUE x }> lapply(listvar, func) > > were > Monday=[213,56,345,33,34,678,444] > Tuesday=[213,56,345,33,34,678,444]This is not R syntax.> ... > > in my case I have a neverending list of vectors.Then your function will take an infinite amount of time - or you will get amazing reputation in computer sciences. Uwe Ligges> > Thanks! > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
Hi:
Here's another way of doing this on the simplified version of your example:
L <- vector('list', 3) # initialize a list of three components
## populate it
for(i in seq_along(L)) L[[i]] <- rnorm(20, 10, 3)
## name the components
names(L) <- c('Monday', 'Tuesday', 'Wednesday')
## replace values <= 10 with NA
lapply(L, function(x) replace(x, x <= 10, NA)
If you have a large number of atomic objects, you could create a
vector of the object names, make a list out of them (e.g., L <-
list(objnames)) and then mimic the lapply() as above. replace() only
works on vectors, though - Uwe's solution is more general.
HTH,
Dennis
On Tue, Nov 8, 2011 at 8:59 AM, Ana <rrasterr at gmail.com>
wrote:> Hi
>
> Can someone help me with this?
>
> How can I apply a function to a list of variables.
>
> something like this
>
>
> listvar=list("Monday","Tuesday","Wednesday")
> func=function(x){x[which(x<=10)]=NA}
>
> lapply(listvar, func)
>
> were
> Monday=[213,56,345,33,34,678,444]
> Tuesday=[213,56,345,33,34,678,444]
> ...
>
> in my case I have a neverending list of vectors.
>
> Thanks!
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Hi hi, It is much easier to deal with lists than a large number of separate objects. So the first answer to your question> How can I apply a function to a list of variables... might be to convert your "list of variables" to a regular list. Instead of ... monday <- 1:3 tuesday <- 4:7 wednesday <- 8:100 ... you could have: mydays <- list(monday=1:3, tuesday=4:7, wednesday=8:100) ... and things will be a lot easier. If you happen to have a thousand of such variables in your workspace then you can convert them to a list using ... listvar=list("Monday","Tuesday","Wednesday") mylist <- lapply(listvar, get) ... but this is not always nice to you, for example, if you have variables that have the same names as some functions in base packages. Try this: a<-1 b<-2 c<-3 mean<-4 listvar <- c("a", "b", "c", "mean") lapply(listvar, get) - a safer way would be lapply(lv, get, envir=.GlobalEnv) # and for a named list, the best I can think of is: structure(lapply(lv, get, envir=.GlobalEnv), .Names=lv) And then ...> func=function(x){x[which(x<=10)]=NA} > lapply(listvar, func)do you want actually to change your variables with this? Or just to have a list with your original variables where any x[x<=10] is set to NA? In the first case you'll need to do nonstandard tricks that everyone will say you should avoid (but you can use e.g. `assign`, or macros - see `defmacro` in package gtools). In the second case you just need to take care that your function would return a sensible value. An assignment returns the value that was assigned, in this case, it is always NA but if that's not what you meant, you can try func <- function(x){ x[x<=10] <- NA; x} lapply(listvar, func) hth On Tue, Nov 8, 2011 at 6:59 PM, Ana <rrasterr at gmail.com> wrote:> Hi > > Can someone help me with this? > > How can I apply a function to a list of variables. > > something like this > > > listvar=list("Monday","Tuesday","Wednesday") > func=function(x){x[which(x<=10)]=NA} > > lapply(listvar, func) > > were > Monday=[213,56,345,33,34,678,444] > Tuesday=[213,56,345,33,34,678,444] > ... > > in my case I have a neverending list of vectors. > > Thanks! > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >