[R] Extracting elements from a nested list

[Erich Neuwirth]

mapply(function(x1,x2)x1[[x2]],all.predicted.values,max.growth,SIMPLIFY=FALSE)

gives a list of factors.


On 10/18/2010 8:40 PM, Gregory Ryslik wrote:
> Hi Everyone,
> 
> This is closer to what I need but this returns me a matrix where each
> element is a factor. Instead I would want a list of lists. The first
> entry of the list should equal the first column of the matrix that
> mapply makes, the second entry to the second column etc...
> 
> I've attached the two files that have all.predicted.values and
> max.growth from dput to make for easy testing. Thanks again!
> 
> 
> 
> 
> Kind regards,
> Greg
> 
> On Oct 18, 2010, at 1:33 PM, Erich Neuwirth wrote:
> 
>> You probably need mapply since you have 2 list of arguments which you
>> want to use "in sync"
>>
>> mapply(function(x1,x2)x1[[x2]],all.predicted.values,max.growth)
>>
>> might be what you want.
>>
>>
>>
>> On Oct 18, 2010, at 5:17 PM, Gregory Ryslik wrote:
>>
>>> Unfortunately, that gives me null everywhere. Here's the data I have
>>> for all.predicted.values and max.growth. Perhaps this will help. Thus
>>> I want all.predicted.values[[1]][[4]] then
>>> all.predicted.values[[2]][3]] and then all.predicted.values[[3]][[4]].
>>>
>>> I've attached what your statement outputs at the end.
>>>
>>> Thanks again!
>>>
>>> Browse[2]> max.growth
>>> [[1]]
>>> [1] 4
>>>
>>> [[2]]
>>> [1] 3
>>>
>>> [[3]]
>>> [1] 4
>>>
>>> Browse[2]> all.predicted.values
>>> [[1]]
>>> [[1]][[1]]
>>>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> Levels: 0 1 2
>>>
>>> [[1]][[2]]
>>>  [1] 2 2 2 0 2 0 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 0
>>> 0 0 2 2 0 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 0
>>> [55] 0 0 2 0 2 0 0 0 0 2 2 2 2 0 2 2 2 0 2 2 0 0 2 2 2 2 2 2 2 0 0 0
>>> 2 0 2 2 2 2 0 2 2 2 0 2 0 0
>>> Levels: 0 1 2
>>>
>>> [[1]][[3]]
>>>  [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0
>>> 0 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0
>>> [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 2 2 0 0 0 0 0 0 2 0 0
>>> Levels: 0 1 2
>>>
>>> [[1]][[4]]
>>>  [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0
>>> 0 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0
>>> [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 2 2 0 0 0 0 0 0 2 0 0
>>> Levels: 0 1 2
>>>
>>>
>>> [[2]]
>>> [[2]][[1]]
>>>  [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>> 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>> [55] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>> 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>> Levels: 0 1 2
>>>
>>> [[2]][[2]]
>>>  [1] 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2
>>> 2 2 1 2 2 2 1 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2
>>> [55] 2 2 2 2 1 2 2 2 2 1 2 2 1 1 1 2 2 2 1 2 1 2 1 2 1 2 2 2 1 1 2 2
>>> 1 2 2 1 1 2 1 1 1 2 2 1 2 2
>>> Levels: 0 1 2
>>>
>>> [[2]][[3]]
>>>  [1] 2 2 2 0 1 2 2 2 2 2 1 2 2 2 0 1 2 1 2 2 2 2 2 2 2 0 0 2 1 2 2 2
>>> 0 0 1 2 0 0 1 2 0 1 1 2 2 2 0 2 2 2 0 1 2 2
>>> [55] 0 2 2 2 1 0 0 0 0 1 2 2 1 1 1 2 2 0 1 2 1 0 1 2 1 2 2 2 1 1 2 2
>>> 1 2 2 1 1 2 1 1 1 2 2 1 0 2
>>> Levels: 0 1 2
>>>
>>>
>>> [[3]]
>>> [[3]][[1]]
>>>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>> Levels: 0 1 2
>>>
>>> [[3]][[2]]
>>>  [1] 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 2
>>> 0 0 2 2 2 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 2
>>> [55] 0 2 2 2 2 2 0 0 2 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>> 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>> Levels: 0 1 2
>>>
>>> [[3]][[3]]
>>>  [1] 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0
>>> 0 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0
>>> [55] 0 0 0 0 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 0
>>> 1 0 0 1 1 0 1 1 1 0 0 1 1 0
>>> Levels: 0 1 2
>>>
>>> [[3]][[4]]
>>>  [1] 2 2 2 0 1 0 2 2 0 2 1 2 2 0 0 1 1 1 1 0 2 0 0 0 2 0 0 0 1 2 0 0
>>> 0 0 1 2 0 0 1 2 0 1 1 2 0 0 0 2 2 0 0 1 2 0
>>> [55] 0 0 0 0 1 0 0 0 0 1 0 2 1 1 1 2 0 0 1 2 1 1 1 2 1 2 2 2 1 1 0 0
>>> 1 0 2 1 1 2 1 1 1 2 0 1 1 0
>>> Levels: 0 1 2
>>>
>>>
>>> Browse[2]>
>>> predicted.values.for.max.growth<-diag(sapply(all.predicted.values,'[[','max.growth'))
>>> Browse[2]> predicted.values.for.max.growth
>>> [[1]]
>>> NULL
>>>
>>> [[2]]
>>> [1] 0
>>>
>>> [[3]]
>>> [1] 0
>>>
>>> [[4]]
>>> [1] 0
>>>
>>> [[5]]
>>> NULL
>>>
>>> [[6]]
>>> [1] 0
>>>
>>> [[7]]
>>> [1] 0
>>>
>>> [[8]]
>>> [1] 0
>>>
>>> [[9]]
>>> NULL
>>>
>>>
>>>
>>> On Oct 18, 2010, at 11:08 AM, Henrique Dallazuanna wrote:
>>>
>>>> Try this:
>>>>
>>>> diag(sapply(all.predicted.values, '[[', 'max.growth'))
>>>>
>>>>
>>>> On Mon, Oct 18, 2010 at 12:59 PM, Gregory Ryslik <rsaber at comcast.net
>>>> <mailto:rsaber at comcast.net>> wrote:
>>>> Hi,
>>>>
>>>> I have a list of n items and the ith element has m_i elements within it.
>>>>
>>>> I want to do something like:
>>>>
>>>> predicted.values<- lapply(all.predicted.values,'[[',max.growth[[i]])
>>>>
>>>> Where max.growth[[i]] is the element I want to extract from each of
>>>> the ith predicted elements. Thus, for example, I want to extract the
>>>> max.growth[[1]] element from  all.predicted.values[[1]] (which is
>>>> itself a list). Then I want to extract max.growth[[2]] element from
>>>> all.predicted.values[[2]].
>>>>
>>>> I realize I can do this with a for loop but then if I can do this as
>>>> one line that would be preferable.
>>>>
>>>> Thanks!
>>>>
>>>> Greg
>>>>       [[alternative HTML version deleted]]
>>>>
>>>> ______________________________________________
>>>> R-help at r-project.org <mailto: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.
>>>>
>>>>
>>>>
>>>> -- 
>>>> Henrique Dallazuanna
>>>> Curitiba-Paraná-Brasil
>>>> 25° 25' 40" S 49° 16' 22" O
>>>
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help at r-project.org <mailto: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.
>>
>> --
>> Erich Neuwirth
>> Didactic Center for Computer Science and Institute for Scientific
>> Computing
>> University of Vienna
>>
>>
>>
>>
>