... And, in fact, I just realized that
c(do.call(rbind, x))
is even better.
-- Bert
On Thu, Sep 26, 2024 at 9:26?PM Bert Gunter <bgunter.4567 at gmail.com>
wrote:
> Sorry, hit send by accident.
> The 2-line version is:
>
> x <- do.call(rbind, x)
> dim(x) <- NULL
>
> Cheers,
> Bert
>
> On Thu, Sep 26, 2024 at 9:23?PM Bert Gunter <bgunter.4567 at
gmail.com>
> wrote:
>
>> How about:
>> as.vector(do.call(rbind,x))
>>
>> Cheers,
>> Bert
>>
>>
>>
>>
>> However, I much prefer a 2 line version:
>>
>> On Thu, Sep 26, 2024 at 8:56?PM Rolf Turner <rolfturner at
posteo.net>
>> wrote:
>>
>>>
>>> I have (toy example):
>>>
>>> x <- list(`1` = c(7, 13, 1, 4, 10),
>>> `2` = c(2, 5, 14, 8, 11),
>>> `3` = c(6, 9, 15, 12, 3))
>>> and
>>>
>>> f <- factor(rep(1:3,5))
>>>
>>> I want to create a vector v of length 15 such that the entries of
v,
>>> corresponding to level l of f are the entries of x[[l]]. I.e. I
want
>>> v to equal
>>>
>>> c(7, 2, 6, 13, 5, 9, 1, 14, 15, 4, 8, 12, 10, 11, 3)
>>>
>>> I can create v "easily enough", using say, a for-loop.
It seems to me,
>>> though, that there should be sexier (single command) way of
achieving
>>> the desired result. However I cannot devise one.
>>>
>>> Can anyone point me in the right direction? Thanks.
>>>
>>> cheers,
>>>
>>> Rolf Turner
>>>
>>> --
>>> Honorary Research Fellow
>>> Department of Statistics
>>> University of Auckland
>>> Stats. Dep't. (secretaries) phone:
>>> +64-9-373-7599 ext. 89622
>>> Home phone: +64-9-480-4619
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> https://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
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