Re: [R] lapply to list of variables
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 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@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.
>
__
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Re: [R] lapply to list of variables
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 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@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.
>
__
R-help@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.
Re: [R] lapply to list of variables
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@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
