[R] Deparse substitute assign with list elements

[William Dunlap]

You could use a 'replacement function' named 'bar<-', whose last argument
is called 'value', and use bar(variable) <- newValue where you currently
use foo(variable, newValue).

bar <- function(x) {
    x + 3
}
`bar<-` <- function(x, value) {
    bar(value)
}

a <- NA
bar(a) <- 4
a
# [1] 7
b <- list(NA, NA)
bar(b[[1]]) <- 4
b
#[[1]]
#[1] 7
#
#[[2]]
#[1] NA


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Thu, May 14, 2015 at 11:28 AM, <soeren.vogel at posteo.ch> wrote:

> Hello,
>
> When I use function `foo` with list elements (example 2), it defines a new
> object named `b[[1]]`, which is not what I want.  How can I change the
> function code to show the desired behaviour for all data structures passed
> to the function?  Or is there a more appropriate way to sort of "pass by
> references" in a function?
>
> Thanks
> Sören
>
> <src>
> bar <- function(x) {
>         return( x + 3 )
> }
>
> foo <- function(x, value) {
>         nm <- deparse(substitute(x))
>         tmp <- bar( value )
>         assign(nm, tmp, parent.frame())
> }
>
> # 1)
> a <- NA
> foo(a, 4)
> a # 7, fine :-)
>
> # 2)
> b <- list(NA, NA)
> foo(b[[1]], 4) # the first list item should be 7
> b # wanted 7 but still list with two NAs :-(
> </src>
>
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