Pass function name as argument in mapply? - r

I would like to pass a function name as an argument in mapply:
f2 <- function(a, b) a + b^2
f <- function(a, b, func) func(a, b)
f(1, 3, f2) ## returns 10
mapply(f2, 1:2, 3) ## returns [1] 10 11
mapply(function(a, b) f(a, b, f2), 1:2, 3) ## returns [1] 10 11
mapply(f, 1:2, 3, f2) ## fails
The final mapply call generates the error
Error in dots[[3L]][[1L]] : object of type 'closure' is not subsettable
Is there any way to do this?

mapply assumes you want to iterate over all the vectors you pass after the first function. But you want to use the same value of f2 for every iteration. You can do that useing the MoreArgs= parameter
mapply(f, 1:2, 3, MoreArgs=list(func=f2))
You don't have the same problem with the 3 because R will perform vector recycling to expand 3 to c(3,3) to match the same length as c(1,2). Functions in R don't have the same implicit recycling behaviors. But if you want the value to always stay the same, it's better to put it in the MoreArgs parameter

1) Wrap the function in a list:
mapply(f, 1:2, 3, list(f2))
## [1] 10 11
2) Typically functions that have function arguments use match.fun so that one can pass either the function or a character string containing its name. For example, mapply itself does that so the above line of code could equally be written as: mapply("f", 1:2, 3, list(f2)) . If f were written that way then we could simply specify the name of f2 as a character string, namely "f2" .
f <- function(a, b, func) {
func <- match.fun(func)
func(a, b)
}
mapply(f, 1:2, 3, "f2")
## [1] 10 11

Related

apply a list of functions to a single argument in R

Hi I am trying to apply a list of functions to a single argument in R. For example,
flist <- list(F,G,H) #F,G,H are function objects
and say I want as a result a list or vector
(F(x),G(x),H(x)) where x is a scalar number.
Do you know how i can achieve that?
The most efficient way (it seems) to achieve this would be using a single lapply (instead of 3 different functions), such as
flist <- list(mean, unique, max) # Example functions list
MyScalar <- 1 # Some scalar
lapply(flist, function(f) f(MyScalar))
# [[1]]
# [1] 1
#
# [[2]]
# [1] 1
#
# [[3]]
# [1] 1
Though, if all the functions give the same size/class result, you could improve it even more using vapply
vapply(flist, function(x) x(MyScalar), FUN.VALUE = double(1))
## [1] 1 1 1
f <- function(x) x^1
g <- function(x) x^2
h <- function(x) x^3
l <- list(f, g, h)
sapply(l, do.call, list(2))
## [1] 2 4 8
do.call allows for function delegation with variable-length argument lists.
For example, c(1, 2, 3) can be called like so: do.call(c, list(1, 2, 3)).
(s|l)apply just iterates through a list and applies the specified function to each item. So the first iteration through l will be: do.call(l[[1]], list(2)), which is equivalent to l[[1]](2), which is equivalent to f(2).

Why can't I assign to multiple variables using mapply/assign? [duplicate]

I want to assign multiple variables in a single line in R. Is it possible to do something like this?
values # initialize some vector of values
(a, b) = values[c(2,4)] # assign a and b to values at 2 and 4 indices of 'values'
Typically I want to assign about 5-6 variables in a single line, instead of having multiple lines. Is there an alternative?
I put together an R package zeallot to tackle this very problem. zeallot includes an operator (%<-%) for unpacking, multiple, and destructuring assignment. The LHS of the assignment expression is built using calls to c(). The RHS of the assignment expression may be any expression which returns or is a vector, list, nested list, data frame, character string, date object, or custom objects (assuming there is a destructure implementation).
Here is the initial question reworked using zeallot (latest version, 0.0.5).
library(zeallot)
values <- c(1, 2, 3, 4) # initialize a vector of values
c(a, b) %<-% values[c(2, 4)] # assign `a` and `b`
a
#[1] 2
b
#[1] 4
For more examples and information one can check out the package vignette.
There is a great answer on the Struggling Through Problems Blog
This is taken from there, with very minor modifications.
USING THE FOLLOWING THREE FUNCTIONS
(Plus one for allowing for lists of different sizes)
# Generic form
'%=%' = function(l, r, ...) UseMethod('%=%')
# Binary Operator
'%=%.lbunch' = function(l, r, ...) {
Envir = as.environment(-1)
if (length(r) > length(l))
warning("RHS has more args than LHS. Only first", length(l), "used.")
if (length(l) > length(r)) {
warning("LHS has more args than RHS. RHS will be repeated.")
r <- extendToMatch(r, l)
}
for (II in 1:length(l)) {
do.call('<-', list(l[[II]], r[[II]]), envir=Envir)
}
}
# Used if LHS is larger than RHS
extendToMatch <- function(source, destin) {
s <- length(source)
d <- length(destin)
# Assume that destin is a length when it is a single number and source is not
if(d==1 && s>1 && !is.null(as.numeric(destin)))
d <- destin
dif <- d - s
if (dif > 0) {
source <- rep(source, ceiling(d/s))[1:d]
}
return (source)
}
# Grouping the left hand side
g = function(...) {
List = as.list(substitute(list(...)))[-1L]
class(List) = 'lbunch'
return(List)
}
Then to execute:
Group the left hand side using the new function g()
The right hand side should be a vector or a list
Use the newly-created binary operator %=%
# Example Call; Note the use of g() AND `%=%`
# Right-hand side can be a list or vector
g(a, b, c) %=% list("hello", 123, list("apples, oranges"))
g(d, e, f) %=% 101:103
# Results:
> a
[1] "hello"
> b
[1] 123
> c
[[1]]
[1] "apples, oranges"
> d
[1] 101
> e
[1] 102
> f
[1] 103
Example using lists of different sizes:
Longer Left Hand Side
g(x, y, z) %=% list("first", "second")
# Warning message:
# In `%=%.lbunch`(g(x, y, z), list("first", "second")) :
# LHS has more args than RHS. RHS will be repeated.
> x
[1] "first"
> y
[1] "second"
> z
[1] "first"
Longer Right Hand Side
g(j, k) %=% list("first", "second", "third")
# Warning message:
# In `%=%.lbunch`(g(j, k), list("first", "second", "third")) :
# RHS has more args than LHS. Only first2used.
> j
[1] "first"
> k
[1] "second"
Consider using functionality included in base R.
For instance, create a 1 row dataframe (say V) and initialize your variables in it. Now you can assign to multiple variables at once V[,c("a", "b")] <- values[c(2, 4)], call each one by name (V$a), or use many of them at the same time (values[c(5, 6)] <- V[,c("a", "b")]).
If you get lazy and don't want to go around calling variables from the dataframe, you could attach(V) (though I personally don't ever do it).
# Initialize values
values <- 1:100
# V for variables
V <- data.frame(a=NA, b=NA, c=NA, d=NA, e=NA)
# Assign elements from a vector
V[, c("a", "b", "e")] = values[c(2,4, 8)]
# Also other class
V[, "d"] <- "R"
# Use your variables
V$a
V$b
V$c # OOps, NA
V$d
V$e
here is my idea. Probably the syntax is quite simple:
`%tin%` <- function(x, y) {
mapply(assign, as.character(substitute(x)[-1]), y,
MoreArgs = list(envir = parent.frame()))
invisible()
}
c(a, b) %tin% c(1, 2)
gives like this:
> a
Error: object 'a' not found
> b
Error: object 'b' not found
> c(a, b) %tin% c(1, 2)
> a
[1] 1
> b
[1] 2
this is not well tested though.
A potentially dangerous (in as much as using assign is risky) option would be to Vectorize assign:
assignVec <- Vectorize("assign",c("x","value"))
#.GlobalEnv is probably not what one wants in general; see below.
assignVec(c('a','b'),c(0,4),envir = .GlobalEnv)
a b
0 4
> b
[1] 4
> a
[1] 0
Or I suppose you could vectorize it yourself manually with your own function using mapply that maybe uses a sensible default for the envir argument. For instance, Vectorize will return a function with the same environment properties of assign, which in this case is namespace:base, or you could just set envir = parent.env(environment(assignVec)).
As others explained, there doesn't seem to be anything built in. ...but you could design a vassign function as follows:
vassign <- function(..., values, envir=parent.frame()) {
vars <- as.character(substitute(...()))
values <- rep(values, length.out=length(vars))
for(i in seq_along(vars)) {
assign(vars[[i]], values[[i]], envir)
}
}
# Then test it
vals <- 11:14
vassign(aa,bb,cc,dd, values=vals)
cc # 13
One thing to consider though is how to handle the cases where you e.g. specify 3 variables and 5 values or the other way around. Here I simply repeat (or truncate) the values to be of the same length as the variables. Maybe a warning would be prudent. But it allows the following:
vassign(aa,bb,cc,dd, values=0)
cc # 0
list2env(setNames(as.list(rep(2,5)), letters[1:5]), .GlobalEnv)
Served my purpose, i.e., assigning five 2s into first five letters.
Had a similar problem recently and here was my try using purrr::walk2
purrr::walk2(letters,1:26,assign,envir =parent.frame())
https://stat.ethz.ch/R-manual/R-devel/library/base/html/list2env.html:
list2env(
list(
a=1,
b=2:4,
c=rpois(10,10),
d=gl(3,4,LETTERS[9:11])
),
envir=.GlobalEnv
)
If your only requirement is to have a single line of code, then how about:
> a<-values[2]; b<-values[4]
I'm afraid that elegent solution you are looking for (like c(a, b) = c(2, 4)) unfortunatelly does not exist. But don't give up, I'm not sure! The nearest solution I can think of is this one:
attach(data.frame(a = 2, b = 4))
or if you are bothered with warnings, switch them off:
attach(data.frame(a = 2, b = 4), warn = F)
But I suppose you're not satisfied with this solution, I wouldn't be either...
R> values = c(1,2,3,4)
R> a <- values[2]; b <- values[3]; c <- values[4]
R> a
[1] 2
R> b
[1] 3
R> c
[1] 4
Another version with recursion:
let <- function(..., env = parent.frame()) {
f <- function(x, ..., i = 1) {
if(is.null(substitute(...))){
if(length(x) == 1)
x <- rep(x, i - 1);
stopifnot(length(x) == i - 1)
return(x);
}
val <- f(..., i = i + 1);
assign(deparse(substitute(x)), val[[i]], env = env);
return(val)
}
f(...)
}
example:
> let(a, b, 4:10)
[1] 4 5 6 7 8 9 10
> a
[1] 4
> b
[1] 5
> let(c, d, e, f, c(4, 3, 2, 1))
[1] 4 3 2 1
> c
[1] 4
> f
[1] 1
My version:
let <- function(x, value) {
mapply(
assign,
as.character(substitute(x)[-1]),
value,
MoreArgs = list(envir = parent.frame()))
invisible()
}
example:
> let(c(x, y), 1:2 + 3)
> x
[1] 4
> y
[1]
Combining some of the answers given here + a little bit of salt, how about this solution:
assignVec <- Vectorize("assign", c("x", "value"))
`%<<-%` <- function(x, value) invisible(assignVec(x, value, envir = .GlobalEnv))
c("a", "b") %<<-% c(2, 4)
a
## [1] 2
b
## [1] 4
I used this to add the R section here: http://rosettacode.org/wiki/Sort_three_variables#R
Caveat: It only works for assigning global variables (like <<-). If there is a better, more general solution, pls. tell me in the comments.
For a named list, use
list2env(mylist, environment())
For instance:
mylist <- list(foo = 1, bar = 2)
list2env(mylist, environment())
will add foo = 1, bar = 2 to the current environement, and override any object with those names. This is equivalent to
mylist <- list(foo = 1, bar = 2)
foo <- mylist$foo
bar <- mylist$bar
This works in a function, too:
f <- function(mylist) {
list2env(mylist, environment())
foo * bar
}
mylist <- list(foo = 1, bar = 2)
f(mylist)
However, it is good practice to name the elements you want to include in the current environment, lest you override another object... and so write preferrably
list2env(mylist[c("foo", "bar")], environment())
Finally, if you want different names for the new imported objects, write:
list2env(`names<-`(mylist[c"foo", "bar"]), c("foo2", "bar2")), environment())
which is equivalent to
foo2 <- mylist$foo
bar2 <- mylist$bar

How to use apply() with a function that takes more than one input?

I have a 10 x 5 data frame and a function that receives 2 inputs a and b.
a is a vector and b is an integer.
The function fun calculates the mean of the vector a and multiplies it by b and returns the result. In the following code, I try to apply() this function to every column of x but it does not seem to work. Help, please!
x = data.frame(rnorm(10), rnorm(10), rnorm(10), rnorm(10), rnorm(10))
fun = function(a, b)
{
c = mean(a) * b
return(c)
}
apply(x, 2, fun(x,2))
If you want to pass more than one parameter to an "apply"-ied function where one of them is the column vector and the other one is a constant, you can do it in two ways;
apply(x, 2, fun, b=2)
OR:
apply(x, 2, function(x) {fun(x, 2)} )
The possibly seen-as-odd behavior of R is that the expression fun(x,2) is not a function whereas function(x) {fun(x, 2)} is.
apply(x, 2, fun, b=2)
#------------------
rnorm.10. rnorm.10..1 rnorm.10..2 rnorm.10..3 rnorm.10..4
-0.06806881 0.32749640 -0.14400234 -0.41493410 -0.02669955
Here the problem is simple since you have constant value for b. However, if you have two or more than two inputs, you can use these as lists and then use Map function. For your example:
set.seed(1)
mydata<-data.frame(rnorm(10), rnorm(10), rnorm(10), rnorm(10), rnorm(10))
a<-as.list(names(mydata))
b<-as.list(rep(2,5)) # you can keep b<-2 it doesn't change the results since b is constant
myout<-Map(function(x,y) y*mean(mydata[,x]),a,b)
>myout
[[1]]
[1] 0.2644056
[[2]]
[1] 0.4976899
[[3]]
[1] -0.2673465
[[4]]
[1] 0.2414604
[[5]]
[1] 0.2682734

Replace the argument with a list in R [duplicate]

This question already has an answer here:
Can you pass a vector to a vararg?: Vector to sprintf
(1 answer)
Closed 9 years ago.
Some times there is a function which has an arbitrary number of arguments. For example, the bootstrapPage function of package shiny. If I have a data.frame and I want to create one widget for one row, then I have not figured out a pretty way to pass the number of arguments according to the row number of the data.frame. So far, I generate the script and use the trick of eval(parse(text="..."))
In fact, the structure of arguments passed to function in R (key and value) is similar to a list, so I am wondering if there is a way to pass the argument as a list in R.
More specifically, if I have a function f and a list argv, is there a way to pass the objects in argv to f according to the matching of the name of argv and the name of the arguments of f, and the position in argv and the position in the arguments of f?
For example, let
f <- function(a, b) a + b
argv <- list(a=1, b=2)
How should I pass the argv to f which is equivalent to f(a=argv$a, b=argv$b)?
Or if we have:
f <- function(a, b, ...) { # some codes }
argv <- list(a = 1, b = 2, 3, 4, 5)
How should I pass the argv to f which is equivalent to f(a=argv$a, b=argv$b, argv[[3]], argv[[4]], argv[[5]])?
Thanks!
You are looking for do.call.
do.call(f, argv)
Here are some examples
> args <- list(n = 10, mean = 3, sd = .5)
> do.call(rnorm, args)
[1] 3.589416 3.393031 2.928506 2.503925 3.316584 2.581787 2.686507 3.178877
[9] 3.083885 2.821506
> do.call(rnorm, list(10, 3, .5))
[1] 3.964526 2.838760 2.436684 3.068581 1.842332 3.739046 4.050525 3.097042
[9] 3.665041 3.535947
> f <- function(a, b) a + b
> argv <- list(a=1, b=2)
> do.call(f, argv)
[1] 3
> f <- function(a, b, ...){print(a);print(b);print(sum(...))}
> argv <- list(a=1,b=2, 3, 4, 5)
> do.call(f, argv)
[1] 1
[1] 2
[1] 12

Assign multiple new variables on LHS in a single line

I want to assign multiple variables in a single line in R. Is it possible to do something like this?
values # initialize some vector of values
(a, b) = values[c(2,4)] # assign a and b to values at 2 and 4 indices of 'values'
Typically I want to assign about 5-6 variables in a single line, instead of having multiple lines. Is there an alternative?
I put together an R package zeallot to tackle this very problem. zeallot includes an operator (%<-%) for unpacking, multiple, and destructuring assignment. The LHS of the assignment expression is built using calls to c(). The RHS of the assignment expression may be any expression which returns or is a vector, list, nested list, data frame, character string, date object, or custom objects (assuming there is a destructure implementation).
Here is the initial question reworked using zeallot (latest version, 0.0.5).
library(zeallot)
values <- c(1, 2, 3, 4) # initialize a vector of values
c(a, b) %<-% values[c(2, 4)] # assign `a` and `b`
a
#[1] 2
b
#[1] 4
For more examples and information one can check out the package vignette.
There is a great answer on the Struggling Through Problems Blog
This is taken from there, with very minor modifications.
USING THE FOLLOWING THREE FUNCTIONS
(Plus one for allowing for lists of different sizes)
# Generic form
'%=%' = function(l, r, ...) UseMethod('%=%')
# Binary Operator
'%=%.lbunch' = function(l, r, ...) {
Envir = as.environment(-1)
if (length(r) > length(l))
warning("RHS has more args than LHS. Only first", length(l), "used.")
if (length(l) > length(r)) {
warning("LHS has more args than RHS. RHS will be repeated.")
r <- extendToMatch(r, l)
}
for (II in 1:length(l)) {
do.call('<-', list(l[[II]], r[[II]]), envir=Envir)
}
}
# Used if LHS is larger than RHS
extendToMatch <- function(source, destin) {
s <- length(source)
d <- length(destin)
# Assume that destin is a length when it is a single number and source is not
if(d==1 && s>1 && !is.null(as.numeric(destin)))
d <- destin
dif <- d - s
if (dif > 0) {
source <- rep(source, ceiling(d/s))[1:d]
}
return (source)
}
# Grouping the left hand side
g = function(...) {
List = as.list(substitute(list(...)))[-1L]
class(List) = 'lbunch'
return(List)
}
Then to execute:
Group the left hand side using the new function g()
The right hand side should be a vector or a list
Use the newly-created binary operator %=%
# Example Call; Note the use of g() AND `%=%`
# Right-hand side can be a list or vector
g(a, b, c) %=% list("hello", 123, list("apples, oranges"))
g(d, e, f) %=% 101:103
# Results:
> a
[1] "hello"
> b
[1] 123
> c
[[1]]
[1] "apples, oranges"
> d
[1] 101
> e
[1] 102
> f
[1] 103
Example using lists of different sizes:
Longer Left Hand Side
g(x, y, z) %=% list("first", "second")
# Warning message:
# In `%=%.lbunch`(g(x, y, z), list("first", "second")) :
# LHS has more args than RHS. RHS will be repeated.
> x
[1] "first"
> y
[1] "second"
> z
[1] "first"
Longer Right Hand Side
g(j, k) %=% list("first", "second", "third")
# Warning message:
# In `%=%.lbunch`(g(j, k), list("first", "second", "third")) :
# RHS has more args than LHS. Only first2used.
> j
[1] "first"
> k
[1] "second"
Consider using functionality included in base R.
For instance, create a 1 row dataframe (say V) and initialize your variables in it. Now you can assign to multiple variables at once V[,c("a", "b")] <- values[c(2, 4)], call each one by name (V$a), or use many of them at the same time (values[c(5, 6)] <- V[,c("a", "b")]).
If you get lazy and don't want to go around calling variables from the dataframe, you could attach(V) (though I personally don't ever do it).
# Initialize values
values <- 1:100
# V for variables
V <- data.frame(a=NA, b=NA, c=NA, d=NA, e=NA)
# Assign elements from a vector
V[, c("a", "b", "e")] = values[c(2,4, 8)]
# Also other class
V[, "d"] <- "R"
# Use your variables
V$a
V$b
V$c # OOps, NA
V$d
V$e
here is my idea. Probably the syntax is quite simple:
`%tin%` <- function(x, y) {
mapply(assign, as.character(substitute(x)[-1]), y,
MoreArgs = list(envir = parent.frame()))
invisible()
}
c(a, b) %tin% c(1, 2)
gives like this:
> a
Error: object 'a' not found
> b
Error: object 'b' not found
> c(a, b) %tin% c(1, 2)
> a
[1] 1
> b
[1] 2
this is not well tested though.
A potentially dangerous (in as much as using assign is risky) option would be to Vectorize assign:
assignVec <- Vectorize("assign",c("x","value"))
#.GlobalEnv is probably not what one wants in general; see below.
assignVec(c('a','b'),c(0,4),envir = .GlobalEnv)
a b
0 4
> b
[1] 4
> a
[1] 0
Or I suppose you could vectorize it yourself manually with your own function using mapply that maybe uses a sensible default for the envir argument. For instance, Vectorize will return a function with the same environment properties of assign, which in this case is namespace:base, or you could just set envir = parent.env(environment(assignVec)).
As others explained, there doesn't seem to be anything built in. ...but you could design a vassign function as follows:
vassign <- function(..., values, envir=parent.frame()) {
vars <- as.character(substitute(...()))
values <- rep(values, length.out=length(vars))
for(i in seq_along(vars)) {
assign(vars[[i]], values[[i]], envir)
}
}
# Then test it
vals <- 11:14
vassign(aa,bb,cc,dd, values=vals)
cc # 13
One thing to consider though is how to handle the cases where you e.g. specify 3 variables and 5 values or the other way around. Here I simply repeat (or truncate) the values to be of the same length as the variables. Maybe a warning would be prudent. But it allows the following:
vassign(aa,bb,cc,dd, values=0)
cc # 0
list2env(setNames(as.list(rep(2,5)), letters[1:5]), .GlobalEnv)
Served my purpose, i.e., assigning five 2s into first five letters.
Had a similar problem recently and here was my try using purrr::walk2
purrr::walk2(letters,1:26,assign,envir =parent.frame())
https://stat.ethz.ch/R-manual/R-devel/library/base/html/list2env.html:
list2env(
list(
a=1,
b=2:4,
c=rpois(10,10),
d=gl(3,4,LETTERS[9:11])
),
envir=.GlobalEnv
)
If your only requirement is to have a single line of code, then how about:
> a<-values[2]; b<-values[4]
I'm afraid that elegent solution you are looking for (like c(a, b) = c(2, 4)) unfortunatelly does not exist. But don't give up, I'm not sure! The nearest solution I can think of is this one:
attach(data.frame(a = 2, b = 4))
or if you are bothered with warnings, switch them off:
attach(data.frame(a = 2, b = 4), warn = F)
But I suppose you're not satisfied with this solution, I wouldn't be either...
R> values = c(1,2,3,4)
R> a <- values[2]; b <- values[3]; c <- values[4]
R> a
[1] 2
R> b
[1] 3
R> c
[1] 4
Another version with recursion:
let <- function(..., env = parent.frame()) {
f <- function(x, ..., i = 1) {
if(is.null(substitute(...))){
if(length(x) == 1)
x <- rep(x, i - 1);
stopifnot(length(x) == i - 1)
return(x);
}
val <- f(..., i = i + 1);
assign(deparse(substitute(x)), val[[i]], env = env);
return(val)
}
f(...)
}
example:
> let(a, b, 4:10)
[1] 4 5 6 7 8 9 10
> a
[1] 4
> b
[1] 5
> let(c, d, e, f, c(4, 3, 2, 1))
[1] 4 3 2 1
> c
[1] 4
> f
[1] 1
My version:
let <- function(x, value) {
mapply(
assign,
as.character(substitute(x)[-1]),
value,
MoreArgs = list(envir = parent.frame()))
invisible()
}
example:
> let(c(x, y), 1:2 + 3)
> x
[1] 4
> y
[1]
Combining some of the answers given here + a little bit of salt, how about this solution:
assignVec <- Vectorize("assign", c("x", "value"))
`%<<-%` <- function(x, value) invisible(assignVec(x, value, envir = .GlobalEnv))
c("a", "b") %<<-% c(2, 4)
a
## [1] 2
b
## [1] 4
I used this to add the R section here: http://rosettacode.org/wiki/Sort_three_variables#R
Caveat: It only works for assigning global variables (like <<-). If there is a better, more general solution, pls. tell me in the comments.
For a named list, use
list2env(mylist, environment())
For instance:
mylist <- list(foo = 1, bar = 2)
list2env(mylist, environment())
will add foo = 1, bar = 2 to the current environement, and override any object with those names. This is equivalent to
mylist <- list(foo = 1, bar = 2)
foo <- mylist$foo
bar <- mylist$bar
This works in a function, too:
f <- function(mylist) {
list2env(mylist, environment())
foo * bar
}
mylist <- list(foo = 1, bar = 2)
f(mylist)
However, it is good practice to name the elements you want to include in the current environment, lest you override another object... and so write preferrably
list2env(mylist[c("foo", "bar")], environment())
Finally, if you want different names for the new imported objects, write:
list2env(`names<-`(mylist[c"foo", "bar"]), c("foo2", "bar2")), environment())
which is equivalent to
foo2 <- mylist$foo
bar2 <- mylist$bar

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