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lapply to list of variables

4 messages · Ana, Uwe Ligges, Dennis Murphy +1 more

Ana
# 
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!
Uwe Ligges
#
On 08.11.2011 17:59, Ana wrote:
This is a list of length one character vectors rather than a "list of 
variables".
To make it work, redefine:

func <-function(x){
  x <- get(x)
  is.na(x[x<=10]) <- TRUE
  x
}
This is not R syntax.
Then your function will take an infinite amount of time - or you will 
get amazing reputation in computer sciences.

Uwe Ligges
Dennis Murphy
# 
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:
1 day later
Kenn Konstabel
# 
Hi hi,

It is much easier to deal with lists than a large number of separate
objects. So the first answer to your question
.. 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 ...
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: