i am trying to create a function in which i want to remove one of the variables passed to it.
now R works in annoying ways in that it copies the object instead of giving a reference.
(technically the copying only happens if you make a change... but meh...)
a=function(b){
rm(b)
# rm(b)
}
test=123
a(test) # will remove b, not test
# you can verify that by adding the 2nd rm(b)
i tried
a=function(b){
rm(match.call()[[2]])
}
but that gives the error:
Error in rm(match.call()[[3]]) :
... must contain names or character strings
Try this:
Foo <- function(x){
Sx <- deparse(substitute(x))
rm(list=Sx,envir=sys.frame(-1))
}
##
Z <- 123
ls()
##
[1] "Foo" "Z"
##
Foo(x=Z)
ls()
[1] "Foo"
similar to nrussell's answer, here's the line from cgwtools::askrm which does an arbitrary function call on the selected object:
call(fn, as.name(thenam)), envir = parent.frame(1))
(and, yes, I'm plugging my own toolkit here :-) )
Related
Within a loop, I try to remove a list of data frames simply with
rm(a,b,c,d)
However, in case an data-frame (e.g. b) does not exist in the global environment, I get a warning
In rm(a,b,c,d,...:
object 'b' not found
How can I suppress this warning to not pop-up?
Use suppressWarnings
suppressWarnings(rm(a,b,c,d))
#RonakShah has the correct answer here, though it is possible to avoid generating a warning at all by defining a function that checks for the existence of variables before trying to remove them:
rm2 <- function(...)
{
names <- sapply(as.list(match.call()[-1]), deparse);
rm(list = names[sapply(names, exists)], envir = parent.frame())
}
So you can do:
x <- 1; y <- 2;
ls()
#> [1] "rm2" "x" "y"
rm2(x, y, z) # Note no warning generated since no attempt made to remove z
ls()
[1] "rm2"
I have a workspace with lots of objects and I would like to remove all but one. Ideally I would like to avoid having to type rm(obj.1, obj.2... obj.n). Is it possible to indicate remove all objects but these ones?
Here is a simple construct that will do it, by using setdiff:
rm(list=setdiff(ls(), "x"))
And a full example. Run this at your own risk - it will remove all variables except x:
x <- 1
y <- 2
z <- 3
ls()
[1] "x" "y" "z"
rm(list=setdiff(ls(), "x"))
ls()
[1] "x"
Using the keep function from the gdata package is quite convenient.
> ls()
[1] "a" "b" "c"
library(gdata)
> keep(a) #shows you which variables will be removed
[1] "b" "c"
> keep(a, sure = TRUE) # setting sure to TRUE removes variables b and c
> ls()
[1] "a"
I think another option is to open workspace in RStudio and then change list to grid at the top right of the environment(image below). Then tick the objects you want to clear and finally click on clear.
I just spent several hours hunting for the answer to a similar but slightly different question - I needed to be able to delete all objects in R (including functions) except a handful of vectors.
One way to do this:
rm(list=ls()[! ls() %in% c("a","c")])
Where the vectors that I want to keep are named 'a' and 'c'.
Hope this helps anyone searching for the same solution!
To keep all objects whose names match a pattern, you could use grep, like so:
to.remove <- ls()
to.remove <- c(to.remove[!grepl("^obj", to.remove)], "to.remove")
rm(list=to.remove)
Replace v with the name of the object you want to keep
rm(list=(ls()[ls()!="v"]))
hat-tip: http://r.789695.n4.nabble.com/Removing-objects-and-clearing-memory-tp3445763p3445865.html
To keep a list of objects, one can use:
rm(list=setdiff(ls(), c("df1", "df2")))
This takes advantage of ls()'s pattern option, in the case you have a lot of objects with the same pattern that you don't want to keep:
> foo1 <- "junk"; foo2 <- "rubbish"; foo3 <- "trash"; x <- "gold"
> ls()
[1] "foo1" "foo2" "foo3" "x"
> # Let's check first what we want to remove
> ls(pattern = "foo")
[1] "foo1" "foo2" "foo3"
> rm(list = ls(pattern = "foo"))
> ls()
[1] "x"
require(gdata)
keep(object_1,...,object_n,sure=TRUE)
ls()
From within a function, rm all objects in .GlobalEnv except the function
initialize <- function(country.name) {
if (length(setdiff(ls(pos = .GlobalEnv), "initialize")) > 0) {
rm(list=setdiff(ls(pos = .GlobalEnv), "initialize"), pos = .GlobalEnv)
}
}
assuming you want to remove every object except df from environment:
rm(list = ls(pattern="[^df]"))
let's think in different way, what if we wanna remove a group?
try this,
rm(list=ls()[grep("xxx",ls())])
I personally don't like too many tables, variables on my screen, yet I can't avoid using them. So I name the temporary things starting with "xxx", so I can remove them after it is no longer used.
# remove all objects but selected
rm(list = ls()[which("key_function" != ls())])
How about this?
# Removes all objects except the specified & the function itself.
rme <- function(except=NULL){
except = ifelse(is.character(except), except, deparse(substitute(except)))
rm(list=setdiff(ls(envir=.GlobalEnv), c(except,"rme")), envir=.GlobalEnv)
}
The following will remove all the objects from your console
rm(list = ls())
Using a basic function such as this:
myname<-function(z){
nm <-deparse(substitute(z))
print(nm)
}
I'd like the name of the item to be printed (or returned) when iterating through a list e.g.
for (csv in list(acsv, bcsv, ccsv)){
myname(csv)
}
should print:
acsv
bcsv
ccsv
(and not csv).
It should be noted that acsv, bcsv, and ccsvs are all dataframes read in from csvs i.e.
acsv = read.csv("a.csv")
bcsv = read.csv("b.csv")
ccsv = read.csv("c.csv")
Edit:
I ended up using a bit of a compromise. The primary goal of this was not to simply print the frame name - that was the question, because it is a prerequisite for doing other things.
I needed to run the same functions on four identically formatted files. I then used this syntax:
for(i in 1:length(csvs)){
cat(names(csvs[i]), "\n")
print(nrow(csvs[[i]]))
print(nrow(csvs[[i]][1]))
}
Then the indexing of nested lists was utilized e.g.
print(nrow(csvs[[i]]))
which shows the row count for each of the dataframes.
print(nrow(csvs[[i]][1]))
Then provides a table for the first column of each dataframe.
I include this because it was the motivator for the question. I needed to be able to label the data for each dataframe being examined.
The list you have constructed doesn't "remember" the expressions it was constructed of anymore. But you can use a custom constructor:
named.list <- function(...) {
l <- list(...)
exprs <- lapply(substitute(list(...))[-1], deparse)
names(l) <- exprs
l
}
And so:
> named.list(1+2,sin(5),sqrt(3))
$`1 + 2`
[1] 3
$`sin(5)`
[1] -0.9589243
$`sqrt(3)`
[1] 1.732051
Use this list as parameter to names, as Thomas suggested:
> names(mylist(1+2,sin(5),sqrt(3)))
[1] "1 + 2" "sin(5)" "sqrt(3)"
To understand what's happening here, let's analyze the following:
> as.list(substitute(list(1+2,sqrt(5))))
[[1]]
list
[[2]]
1 + 2
[[3]]
sqrt(5)
The [-1] indexing leaves out the first element, and all remaining elements are passed to deparse, which works because of...
> lapply(as.list(substitute(list(1+2,sqrt(5))))[-1], class)
[[1]]
[1] "call"
[[2]]
[1] "call"
Note that you cannot "refactor" the call list(...) inside substitute() to use simply l. Do you see why?
I am also wondering if such a function is already available in one of the countless R packages around. I have found this post by William Dunlap effectively suggesting the same approach.
I don't know what your data look like, so here's something made up:
csvs <- list(acsv=data.frame(x=1), bcsv=data.frame(x=2), ccsv=data.frame(x=3))
for(i in 1:length(csvs))
cat(names(csvs[i]), "\n")
I'm working on an R package that has a number of functions that follow a non-R-standard practice of modifying in place the object passed in as an argument. This normally works OK, but fails when the object to be modified is on a list.
An function to give an example of the form of the assignments:
myFun<-function(x){
xn <- deparse(substitute(x))
ev <- parent.frame()
# would do real stuff here ..
# instead set simple value to modify local copy
x[[1]]<-"b"
# assign in parent frame
if (exists(xn, envir = ev))
on.exit(assign(xn, x, pos = ev))
# return invisibly
invisible(x)
}
This works:
> myObj <-list("a")
> myFun(myObj)
> myObj
[[1]]
[1] "b"
But it does not work if the object is a member of a list:
> myObj <-list("a")
> myList<-list(myObj,myObj)
> myFun(myList[[1]])
> myList
[[1]]
[[1]][[1]]
[1] "a"
[[2]]
[[2]][[1]]
[1] "a"
After reading answers to other questions here, I see the docs for assign clearly state:
assign does not dispatch assignment methods, so it cannot be used to set elements of vectors, names, attributes, etc.
Since there is an existing codebase using these functions, we cannot abandon the modify-in-place syntax. Does anyone have suggestions for workarounds or alternative approaches for modifying objects which are members of a list in a parent frame?
UPDATE:
I've considered trying to roll my own assignment function, something like:
assignToListInEnv<-function(name,env,value){
# assume name is something like "myList[[1]]"
#check for brackets
index<-regexpr('[[',name,fixed=TRUE)[1]
if(index>0){
lname<-substr(name,0,index-1)
#check that it exists
if (exists(lname,where=env)){
target<-get(lname,pos=env)
# make sure it is a list
if (is.list(target)){
eval(parse(text=paste('target',substr(name,index,999),'<-value',sep='')))
assign(lname, target, pos = env)
} else {
stop('object ',lname,' is not a list in environment ',env)
}
} else {
stop('unable to locate object ',lname,' in frame ',env)
}
}
}
But it seems horrible brittle, would need to handle many more cases ($ and [ as well as [[) and would probably still fail for [[x]] because x would be evaluated in the wrong frame...
Since it was in the first search results to my query, here's my solution :
You can use paste() with "<<-" to create an expression which will assign the value to your list element when evaluated.
assignToListInEnv<-function(name, value, env = parent.frame()){
cl <- as.list(match.call())
lang <- str2lang(paste(cl["name"], "<<-", cl["value"]))
eval(lang, envir = env)
}
EDIT : revisiting this answer because it got a vote up
I'm not sure why I used <<- instead of <-. If using the 'env' argument, <<-with assign to the parent.frame of that env.
So if you always want it to be the first parent.frame it can just be :
assignToListInParentFrame<-function(name, value){
cl <- as.list(match.call())
paste(cl["name"], "<<-", cl["value"]) |>
str2lang() |>
eval()
}
and if you want to precise in which env to modify the list :
assignToListInEnv<-function(name, value, env){
cl <- as.list(match.call())
paste(cl["name"], "<-", cl["value"]) |>
str2lang() |>
eval(envir = env)
}
I have loaded in a R console different type of objects.
I can remove them all using
rm(list=ls())
or remove only the functions (but not the variables) using
rm(list=lsf.str())
My question is:
is there a way to remove all variables except the functions
Here's a one-liner that removes all objects except for functions:
rm(list = setdiff(ls(), lsf.str()))
It uses setdiff to find the subset of objects in the global environment (as returned by ls()) that don't have mode function (as returned by lsf.str())
The posted setdiff answer is nice. I just thought I'd post this related function I wrote a while back. Its usefulness is up to the reader :-).
lstype<-function(type='closure'){
inlist<-ls(.GlobalEnv)
if (type=='function') type <-'closure'
typelist<-sapply(sapply(inlist,get),typeof)
return(names(typelist[typelist==type]))
}
You can use the following command to clear out ALL variables. Be careful because it you cannot get your variables back.
rm(list=ls(all=TRUE))
Here's a pretty convenient function I picked up somewhere and adjusted a little. Might be nice to keep in the directory.
list.objects <- function(env = .GlobalEnv)
{
if(!is.environment(env)){
env <- deparse(substitute(env))
stop(sprintf('"%s" must be an environment', env))
}
obj.type <- function(x) class(get(x, envir = env))
foo <- sapply(ls(envir = env), obj.type)
object.name <- names(foo)
names(foo) <- seq(length(foo))
dd <- data.frame(CLASS = foo, OBJECT = object.name,
stringsAsFactors = FALSE)
dd[order(dd$CLASS),]
}
> x <- 1:5
> d <- data.frame(x)
> list.objects()
# CLASS OBJECT
# 1 data.frame d
# 2 function list.objects
# 3 integer x
> list.objects(env = x)
# Error in list.objects(env = x) : "x" must be an environment
I wrote this to remove all objects apart from functions from the current environment (Programming language used is R with IDE R-Studio):
remove_list=c() # create a vector
for(i in 1:NROW(ls())){ # repeat over all objects in environment
if(class(get(ls()[i]))!="function"){ # if object is *not* a function
remove_list=c(remove_list,ls()[i]) # ..add to vector remove_list
}
}
rm(list=remove_list) # remove all objects named in remove_list
Notes-
The argument "list" in rm(list=) must be a character vector.
The name of an object in position i of the current environment is returned from ls()[i] and the object itself from get(ls()[i]). Therefore the class of an object is returned from class(get(ls()[i]))