Related
I am fetching bins.txt and saving its data in "data". I tried printing it and it is printing properly.
data <- read.csv("bins.txt", header = FALSE)
for (n in 1:24060)
{
j=(data[n,])
for (i in 1:20)
{
m=(i-1)*80
n=(i*80)-1
if(m<j && j<n)
{
print (i)
}
}
}
I wish to not print(i) but store the values of i in some vector and print it outside the loop and pass it in
obs="vector"
Somewhat like this
No idea what your bins.txt is. Since I really dislike nested loops, here's a suggestion:
(i) define the twenty pairs of min (or m) and max (or j) values in condition check:
m <- lapply(1:20, function(x) (x-1)*80)
n <- lapply(1:20, function(x) (x*80)-1)
(ii) return a list of twenty vectors based against data based on the twenty combinations of m and n:
lapply(1:20, function(x) dat[m[[x]] < dat & dat < n[[x]]])
Assuming that your data is
dat <- seq(0, 1000, length.out=50)
The first six vectors returned are:
[[1]]
[1] 20.40816 40.81633 61.22449
[[2]]
[1] 81.63265 102.04082 122.44898 142.85714
[[3]]
[1] 163.2653 183.6735 204.0816 224.4898
[[4]]
[1] 244.8980 265.3061 285.7143 306.1224
[[5]]
[1] 326.5306 346.9388 367.3469 387.7551
[[6]]
[1] 408.1633 428.5714 448.9796 469.3878
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
I have a list of elements say:
l <- c("x","ya1","xb3","yb3","ab","xc3","y","xa1","yd4")
Out of this list I would like to make a list of the matching x,y pairs, i.e.
(("xa1" "ya1") ("xb3" "yb3") ("x" "y"))
In essence, I need to capture the X elements, the Y elements and then pair them up:
I know how to do the X,Y extraction part:
xelems <- grep("^x", l, perl=TRUE, value=TRUE)
yelems <- grep("^y", l, perl=TRUE, value=TRUE)
An X element pairs up with a Y element when
1. xElem == yElem # if xElem and yElem are one char long, i.e. 'x' and 'y'
2. substr(xElem,1,nchar(xElem)) == substr(yElem,1,nchar(yElem))
There is no order, i.e. matching xElem and yElem can be positioned anywhere.
I am however not very sure about the next part. I am more familiar with the SKILL programming language (SKILL is a LISP derivative) and this is how I write it:
procedure( get_xy_pairs(inputList "l")
let(( yElem (xyPairs nil) xList yList)
xList=setof(i inputList rexMatchp("^x" i))
yList=setof(i inputList rexMatchp("^y" i))
when(xList && yList
unless(length(xList)==length(yList)
warn("xList and yList mismatch : %d vs %d\n" length(xList) length(yList))
)
foreach(xElem xList
if(xElem=="x"
then yElem="y"
else yElem=strcat("y" substring(xElem 2 strlen(xElem)))
)
if(member(yElem yList)
then xyPairs=cons(list(xElem yElem) xyPairs)
else warn("x element %s has no matching y element \n" xElem)
)
)
)
xyPairs
)
)
When run on l, this would return
get_xy_pairs(l)
*WARNING* x element xc3 has no matching y element
(("xa1" "ya1") ("xb3" "yb3") ("x" "y"))
As I am still new to R, I would appreciate if you folks can help. Besides, I do understand the R folks tend to avoid for loops and are more into lapply ?
Maybe something like this would work. (Only tested on your sample data.)
## Remove any item not starting with x or y
l2 <- l[grepl("^x|^y", l)]
## Split into a list of items starting with x
## and items starting with y
L <- split(l2, grepl("^x", l2))
## Give "names" to the "starting with y" group
names(L[[1]]) <- gsub("^y", "x", L[[1]])
## Use match to match the names in the y group with
## the values from the x group. This results in a
## nice named vector with the pairs you want
Matches <- L[[1]][match(L[[2]], names(L[[1]]), nomatch=0)]
Matches
# x xb3 xa1
# "y" "yb3" "ya1"
As a data.frame:
MatchesDF <- data.frame(x = names(Matches), y = unname(Matches))
MatchesDF
# x y
# 1 x y
# 2 xb3 yb3
# 3 xa1 ya1
I would store tuples in a list, i.e:
xypairs
[[1]]
[1] "x" "y"
[[2]]
[2] "xb3" "yb3"
Your procedure can be simplified with match and substring.
xends <- substring(xelems, 2)
yends <- substring(yelems, 2)
ypaired <- match(xends, yends) # Indices of yelems that match xelems
# Now we need to handle the no-matches:
xsorted <- c(xelems, rep(NA, sum(is.na(ypaired))))
ysorted <- yelems[ypaired]
ysorted <- c(ysorted, yelems[!(yelems %in% ysorted)])
# Now we create the list of tuples:
xypairs <- lapply(1:length(ysorted), function(i) {
c(xsorted[i], ysorted[i])
})
Result:
xypairs
[[1]]
[1] "x" "y"
[[2]]
[1] "xb3" "yb3"
[[3]]
[1] "xc3" NA
[[4]]
[1] "xa1" "ya1"
[[5]]
[1] NA "yd4"
Say I have a list:
> fs
[[1]]
NULL
[[2]]
NULL
[[3]]
NULL
[[4]]
[1] 61.90298 58.29699 54.90104 51.70293 48.69110
I want to "reverse fill" the rest of the list by using it's values. Example:
The [[3]] should have the function value of [[4]] pairs:
c( myFunction(fs[[4]][1], fs[[4]][2]), myFunction(fs[[4]][2], fs[[4]][3]), .... )
The [[2]] should have myFunction values of [[3]] etc...
I hope that's clear. What's the right way to do it? For loops? *applys? My last attempt, which leaves 1-3 empty:
n = length(fs)
for (i in rev(1:(n-1)))
child_fs = fs[[i+1]]
res = c()
for (j in 1:(i+1))
up = v(child_fs[j])
do = v(child_fs[j+1])
this_f = myFunction(up, do)
res[j] = this_f
fs[[i]] = res
Make fs easily reproducible
fs <- list(NULL, NULL, NULL, c(61.90298, 58.29699, 54.90104, 51.70293, 48.69110))
To be able to show an example, make a trivial myFunction
myFunction <- function(a, b) {a + b}
You can loop over all but the last positions in fs (in reverse order), and compute each. Just call myFunciton with the vectors which are the next higher position's vectors without the last and without the first element.
for (i in rev(seq_along(fs))[-1]) {
fs[[i]] <- myFunction(head(fs[[i+1]], -1), tail(fs[[i+1]], -1))
}
That assumes myFunction is vectorized (given vectors for inputs, will give a vector for output). If it isn't, you can easily make a version which is.
myFunction <- function(a, b) {a[[1]] + b[[1]]}
for (i in rev(seq_along(fs))[-1]) {
fs[[i]] <- Vectorize(myFunction)(head(fs[[i+1]], -1), tail(fs[[i+1]], -1))
}
In either case, you get
> fs
[[1]]
[1] 453.2 426.8
[[2]]
[1] 233.398 219.802 206.998
[[3]]
[1] 120.200 113.198 106.604 100.394
[[4]]
[1] 61.90298 58.29699 54.90104 51.70293 48.69110
Really, what you have is a starting point
start <- c(61.90298, 58.29699, 54.90104, 51.70293, 48.69110)
a function you want to apply (I made this one up which adds 1 everywhere and deletes the last element)
myFunction <- function(x) head(x + 1, -1L)
and the number of times you want to apply the function (recursively):
n <- 3L
So I would write a function to apply the function n times recursively, then reverse the output list:
apply.n.times <- function(fun, n, x)
if (n == 0L) list(x) else c(list(x), Recall(fun, n - 1L, fun(x)))
rev(apply.n.times(myFunction, n, start))
# [[1]]
# [1] 64.90298 61.29699
#
# [[2]]
# [1] 63.90298 60.29699 56.90104
#
# [[3]]
# [1] 62.90298 59.29699 55.90104 52.70293
#
# [[4]]
# [1] 61.90298 58.29699 54.90104 51.70293 48.69110
Here is a one-line solution (if myFunction can be replaced with something like sum, or in this case rowSums):
Reduce( function(x,y) rowSums( embed(y,2) ), fs, right=TRUE, accumulate=TRUE )
If myFunction needs to accept 2 values and do something with them then this can be expanded a bit to:
Reduce( function(x,y) apply( embed(y,2), 1, function(z) myFunction(z[1],z[2]) ),
fs, right=TRUE, accumulate=TRUE )
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