[R] seq(along= surprise

[baptiste auguie]

Perhaps this is what was intended?


> sims <- list(length=100)

> do.call(seq, sims)

seq by itself does not expect a list, but do.call() can create the  
appropriate call if a list is what you want to pass to the function.

Hope this helps,

baptiste

On 5 Feb 2009, at 19:46, Uwe Ligges wrote:

>
>
> Uwe Ligges wrote:
>>
>>
>> Kjetil Halvorsen wrote:
>>> This surprised me:
>>>
>>>> reps <- 100
>>>> sims <- list(length=reps)
>>>> sims
>>> $length
>>> [1] 100
>>>
>>>> for(i in seq(along=sims))print(i)
>>> [1] 1
>>>
>>> This is R 2.8.1.
>>
>>
>> What is surprising?
>>
>> sims is now a list that contains 1 element called "length" with a
>> numeric value of 100.
>> Then seq(along=sims) is exactly 1, because sims has length 1.
>> Hence i is printed once (1 iteration of the loop) and is 1 in the  
>> first
>> (and only) iteration.
>>
>> Uwe
>
> I should have added that you probably want
>
> sims <- vector(mode="list", length=100)
>
> Uwe
>
>
>>
>>
>>> Kjetil
>>>
>>>    [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

_____________________________

Baptiste Auguié

School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK

Phone: +44 1392 264187

http://newton.ex.ac.uk/research/emag