[R] Is there a sexy way ...?

[Bert Gunter]

... 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 using 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 using 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 using 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
>>>
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>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>

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