Assign multiple new variables on LHS in a single line

R Variables Tuples Variable Assignment Assign

R Problem Overview

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?

R Solutions

Solution 1 - R

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"

Solution 2 - R

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.

Solution 3 - R

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

Solution 4 - R

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.

Solution 5 - R

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)).

Solution 6 - R

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

Solution 7 - R

list2env(setNames(as.list(rep(2,5)), letters[1:5]), .GlobalEnv)

Served my purpose, i.e., assigning five 2s into first five letters.

Solution 8 - R

Had a similar problem recently and here was my try using purrr::walk2

purrr::walk2(letters,1:26,assign,envir =parent.frame())

Solution 9 - R

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
        )

Solution 10 - R

If your only requirement is to have a single line of code, then how about:

> a<-values[2]; b<-values[4]

Solution 11 - R

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...

Solution 12 - R

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

Solution 13 - R

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()
}