How to use R's S3-classes together with parameters? - r

I fear I get something really wrong. The basics are from here
and a basic (minimal) example is understood (I think) and working:
fun.default <- function(x) { # you could add further fun.class1 (works)...
print("default")
return(x[1] + x[2])
}
my_fun <- function(x) {
print("my_fun")
print(x)
res <- UseMethod("fun", x)
print(res)
print("END my_fun...")
return(res)
}
x <- c(1, 2)
my_fun(x)
However, if I want to add parameters, something goes really wrong. Form the link above:
Once UseMethod has found the correct method, it’s invoked in a special
way. Rather than creating a new evaluation environment, it uses the
environment of the current function call (the call to the generic), so
any assignments or evaluations that were made before the call to
UseMethod will be accessible to the method.
I tried all variants I could think of:
my_fun_wrong1 <- function(x, y) {
print("my_fun_wrong1")
print(x)
x <- x + y
print(x)
res <- UseMethod("fun", x)
print(res)
print("END my_fun_wrong1...")
return(res)
}
x <- c(1, 2)
# Throws: Error in fun.default(x, y = 2) : unused argument (y = 2)
my_fun_wrong1(x, y = 2)
my_fun_wrong2 <- function(x) {
print("my_fun_wrong2")
print(x)
x <- x + y
print(x)
res <- UseMethod("fun", x)
print(res)
print("END my_fun_wrong2...")
return(res)
}
x <- c(1, 2)
y = 2
# Does not throw an error, but does not give my expetced result "7":
my_fun_wrong2(x) # wrong result!?
rm(y)
my_fun_wrong3 <- function(x, ...) {
print("my_fun_wrong3")
print(x)
x <- x + y
print(x)
res <- UseMethod("fun", x)
print(res)
print("END my_fun_wrong3...")
return(res)
}
x <- c(1, 2)
# Throws: Error in my_fun_wrong3(x, y = 2) : object 'y' not found
my_fun_wrong3(x, y = 2)
Edit after answer G. Grothendieck: Using fun.default <- function(x, ...) I get
Runs after change, but I don't understand the result:
my_fun_wrong1(x, y = 2)
[1] "my_fun_wrong1"
[1] 1 2
[1] 3 4 # Ok
[1] "default"
[1] 3 # I excpect 7
As before - I don't understand the result:
my_fun_wrong2(x) # wrong result!?
[1] "my_fun_wrong2"
[1] 1 2
[1] 3 4 # Ok!
[1] "default"
[1] 3 # 3 + 4 = 7?
Still throws an error:
my_fun_wrong3(x, y = 2)
[1] "my_fun_wrong3"
[1] 1 2
Error in my_fun_wrong3(x, y = 2) : object 'y' not found
I think, this question is really useful!

fun.default needs ... so that the extra argument is matched.
fun.default <- function(x, ...) {
print("default")
return(x[1] + x[2])
}
x <- c(1, 2)
my_fun_wrong1(x, y = 2)
## [1] "my_fun_wrong1"
## [1] 1 2
## [1] 5 6
## [1] 3
Also, any statements after the call to UseMethod in the generic will not be evaluated as UseMethoddoes not return so it is pointless to put code after it in the generic.
Furthermore, you can't redefine the arguments to UseMethod. The arguments are passed on as they came in.
Suggest going over the help file ?UseMethod although admittedly it can be difficult to read.
Regarding the quote from ?UseMethod that was added to the question, this just means that the methods can access local variables defined in the function calling UseMethod. It does not mean that you can redefine arguments. Below ff.default refers to the a defined in ff.
a <- 0
ff <- function(x, ...) { a <- 1; UseMethod("ff") }
ff.default <- function(x, ...) a
ff(3)
## [1] 1

Related

Finding and writing environment

I am writing a R equivalent to Pythons 'pop' method. I know 99th percentile has one but I'd prefer my own (practice/understanding/consistency etc).
For reference, pop() takes an object and removes the first item from the object whilt also returning it. So
> l <- c(1,3,5)
> x <- pop(l)
> print(l)
> 3, 5
> print(x)
> 1
I am using assign() to replace the input object with one less the first value and returning said first value from the function.
My question is, how do I get the environment of the input object and use this environment within assign()?
I have tried using pryr::where() which returns 'R_GlobalEnv' but I can't use this value in assign(). Instead the only value I can get to work in assign() is 'globalenv()'.
Posted from mobile so let me know if something doesn't work.
You can implement this in base R, though it's not advised. R is a functional language and functions with side effects are not expected by end-users.
pop <- function(vec)
{
vec_name <- deparse(substitute(vec))
assign(vec_name, vec[-1], envir = parent.frame())
vec[1]
}
a <- c(2, 7, 9)
a
#> [1] 2 7 9
pop(a)
#> [1] 2
a
#> [1] 7 9
pop(a)
#> [1] 7
a
#> [1] 9
Created on 2020-08-15 by the reprex package (v0.3.0)
The following answer is based in this R-Help post, function pop with function getEnvOf from this SO post, both adapted to the question's problem.
getEnvOf <- function(what, which=rev(sys.parents())) {
what <- as.character(substitute(what))
for (frame in which)
if (exists(what, frame=frame, inherits=FALSE))
return(sys.frame(frame))
return(NULL)
}
pop <- function(x){
y <- as.character(substitute(x))
e <- getEnvOf(y)
if(length(x) > 0) {
val <- x[[length(x)]]
assign(y, x[-length(x)], envir = parent.env(e))
val
} else {
msg <- paste(sQuote(y), "length is not > 0")
warning(msg)
NULL
}
}
y <- c(1,3,5)
pop(y)
This also works with lists.
z <- list(1, 2, 5)
pop(z)
w <- list(1, c(2, 4, 6), 5)
pop(w)
#[1] 5
pop(w)
#[1] 2 4 6
pop(w)
#[1] 1
pop(w)
#NULL
#Warning message:
#In pop(w) : ‘w’ length is not > 0
You can do it using pryr::promise_info(l)$env, but it's a very un-R-like thing to do. Functions shouldn't have side effects.
For example,
pop <- function(l) {
info <- pryr::promise_info(l)
if (!is.name(info$code))
stop("Argument expression should be a name.")
result <- l[[1]] # work on lists too
assign(as.character(info$code), l[-1], envir = info$env)
result
}
l <- c(1, 3, 5)
pop(l)
#> Registered S3 method overwritten by 'pryr':
#> method from
#> print.bytes Rcpp
#> [1] 1
l
#> [1] 3 5
Created on 2020-08-15 by the reprex package (v0.3.0)
Edited to add: Interestingly, none of the three answers so far works in complicated situations like this one:
f <- function(x) {
cat("The pop(x) result is", pop(x), "\n")
cat("Now x is ", x, "\n")
cat("Now l is ", l, "\n")
}
l <- c(1, 3, 5)
f(l)
#RuiBarradas's answer gives
The pop(x) result is 5
Now x is 1 3 5
Now l is 1 3 5
(He pops the last value rather than the first which is not a big deal, but neither x nor l is modified.)
#AllanCameron's answer gives
The pop(x) result is 1
Now x is 3 5
Now l is 1 3 5
This is arguably correct (x got popped), but I think it would be nice to have l being popped, and that seems tricky.
My answer dies with this message:
Error in pop(x) : Argument expression should be a name.
which seems like a bug: obviously whether it's getting x or l, it really is a name. The problem seems to be in pryr::promise_info, which returns the compiled code that would return the value of x, rather than just the code for x. If I turn off JIT compiling by compiler::enableJIT(0), I get the same result as #AllanCameron. It's not clear to me how to unwind back the right amount to pop l instead of just x.

How create an R function that take a FUN as parameter

To extend the usability of a R function, we need to pass an argument of type function (FUN), Could you please demonstrate how to declare a function parameter inside in another function and how to call it. Like
MyOperation <- function(x, y, FUN){
int sum <- x+y
if (!missing(FUN)) sum<-FUN(sum)}
return sum
}
Res <- MyOperation(x=1, y=2, FUN=function(n){...})
You don't declare variables in R. Also you can specify a default value right in the formal argument list. You don't need to use missing in this situation.
This runs FUN(x + y) or returns x+y if FUN is not specified.
myOp2 <- function(x, y, FUN = identity) FUN(x + y)
myOp2(1, 2)
## [1] 3
myOp2(1, 3, sqrt)
## [1] 2
One enhancement might be to allow the function to be specified either as a function or as a character string:
myOp2a <- function(x, y, FUN = identity) {
FUN <- match.fun(FUN)
FUN(x + y)
}
myOp2a(1, 3, "sqrt")
## [1] 2
myOp2a(1, 3, sqrt)
## [1] 2
This sums x and y if FUN is not specified; otherwise, it runs FUN with the arguments x and y.
myOp3 <- function(x, y, FUN = sum) FUN(x, y)
myOp3(1, 2)
## [1] 3
myOp3(1, 2, min)
## [1] 1
You just have some basic R syntax problems there. There's no int in R, your function closing bracket was in the wrong place, return() is a function in R -- not a keyword. Check out
MyOperation<-function(x,y,FUN){
sum <- x+y
if (!missing(FUN)) sum<-FUN(sum)
return(sum)
}
MyOperation(x=1,y=2)
# [1] 3
MyOperation(x=1,y=2,FUN=function(n){n+100})
# [1] 103

Hiding output when saving into variable

I would like to hide printed output when saving output of my own function.
f2 <- function(x) {
cat("x + 5 = ", x + 5)
invisible(x + 5)
}
f2(1) # prints
a <- f2(1) # also prints
In other words I would like to make my function print
x + 5 = 6
when calling f2(1) but in case of calling a <- f2(1) I dont want to show any printed output. Is there any easy way how to do that?
You can use a class system for this. Here's a simple S3 example:
f2 <- function(x) {
names(x) <- paste(x, "+ 5")
class(x) <- c(class(x), 'foo')
x + 5
}
print.foo <- function(x) { cat(names(x), "=", x)}
In practice:
> x <- 3
> f2(x)
3 + 5 = 8
> y <- f2(x)
>
Note that the print.foo function does not handle vectors of length > 1 gracefully. That could be fixed, if desired.

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

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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