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> 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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