Thanks!
On 9/22/2017 12:34 PM, Bert Gunter wrote:
> Well, that's a bit like driving from Boston to New York by way of
> Chicago.
>
> See ?structure
>
> test <- list(a=1,b=2,c=3)
> new <- c(4,5,6)
> test.new <- structure(as.list(new), names=names(test))
> test.new
> $a
> [1] 4
>
> $b
> [1] 5
>
> $c
> [1] 6
>
> Cheers,
> Bert
>
>
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
> On Fri, Sep 22, 2017 at 7:51 AM, Evan Cooch <evan.co...@gmail.com
> <mailto:evan.co...@gmail.com>> wrote:
>
> Solved it:
>
> test <- list(a=1,b=2,c=3)
> new <- c(4,5,6)
>
> hold <- as.list(new)
> updated_test <- replace(test,c(1:3),hold)
>
> $a
> [1] 4
>
> $b
> [1] 5
>
> $c
> [1] 6
>
>
>
> mean.parms <- as.list(mean.parms)
>
> mm.parms <- replace(far.parms,c(1:length(far.parms)),mean.parms)
>
>
> On 9/22/2017 10:34 AM, Evan Cooch wrote:
>
> Suppose I have the following:
>
> test <- list(a=1,b=2,c=3)
>
> I also have a vector (or list, or something else...) with new
> numbers
>
> new <- c(4,5,6)
>
> What I'm trying to figure out is how to take the list, and
> update the numbers from {1,2,3} to {4,5,6}
>
> So, in the end,I want the 'update' test list to look like
>
> (a=4,a=5,a=6)
>
> I tried a bunch of obvious things I know about 'replacing'
> things (without success), but the problem in this instance
> seems to be the fact that the list contains elements that are
> expressions (a=1, a=2,...), while the new vector is simply a
> set of numbers.
>
> So, I want to change the numbers in the list, but retain the
> character parts of the expressions in the list (I need to have
> the list of expressions as is for other purposes).
>
> Doable?
>
> Thanks in advance...
>
>
>
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