[R] Deparse substitute assign with list elements

[soeren.vogel at posteo.ch]

Thanks, Bill, I should have googled more carefully:

http://stackoverflow.com/questions/9561053/assign-values-to-a-list-element-in-r

So, remove

	assign(nm, tmp, parent.frame())

and add

	txt <- paste( nm, '<-', tmp, sep='' )
	eval( parse(text=txt), parent.frame() )

in `foo()` will do the trick.

Bests
Sören

> On 15.05.2015, at 00:02, William Dunlap <wdunlap at tibco.com> wrote:
> 
> 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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