[R] name returned by lapply

[Tony Plate]

If you return the value as named list, you get your answer 
using unlist(res, recursive=F):

 > res <- lapply(1:2, function(i) {val <- list(i); 
names(val) <- paste("Hugo", i, sep="_"); return(val)})
 > unlist(res, rec=F)
$Hugo_1
[1] 1

$Hugo_2
[1] 2

 >

Antje wrote:
> Oh true, this would solve the problem too :-)
> Thanks a lot for the suggestions!
> 
> Antje
> 
> 
> 
> Martin Morgan schrieb:
>> Antje <niederlein-rstat at yahoo.de> writes:
>>
>>> Thanks a lot for your help!
>>>
>>> I know that I cannot directly access the list created, I just was not
>>> sure if there is any "format" of the return value which could provide
>>> additionally a name for the returned list.
>>> I tried to return the values as list with the appropriate name but
>>> then I end up with a list entry as list entry...
>>>
>>> Okay, then I'll solve it with a loop and thanks for the hint with the 
>>> article
>>
>> maybe this:
>>
>>> res <- lapply(1:5, function(i) list(key=paste("Hugo", i, sep="_"), 
>>> val=i))
>>> val <- lapply(res, "[[", "val")
>>> names(val) <- lapply(res, "[[", "key")
>>> val
>> $Hugo_1
>> [1] 1
>>
>> $Hugo_2
>> [1] 2
>>
>> $Hugo_3
>> [1] 3
>>
>> $Hugo_4
>> [1] 4
>>
>> $Hugo_5
>> [1] 5
>>
>> Martin
>>
>>> Ciao,
>>> Antje
>>>
>>>
>>>
>>>
>>>
>>> Gavin Simpson schrieb:
>>>> On Fri, 2008-07-18 at 14:19 +0200, Antje wrote:
>>>>> Hi Gavin,
>>>>>
>>>>> thanks a lot for your answer.
>>>>> Maybe I did not explain very well what I want to do and probably
>>>>> chose a bad example. I don't mind spaces or names starting with a
>>>>> number. I could even name it:
>>>>>
>>>>> "Hugo1", "Hugo2", ...
>>>>>
>>>>> My biggest problem is, that not only the values are
>>>>> calculated/estimated within my function but also the names (Yes, in
>>>>> reality my funtion is more complicated).
>>>>> Maybe it's easier to explain like this. the parameter x can be a
>>>>> coordinate position of mountains on earth. Within the funtion the
>>>>> height of the mountain is estimated and it's name.
>>>>> In the end, I'd like to get a list, where the entry is named like
>>>>> the mountain and it contains its height (or other measurements...)
>>>>>
>>>>>
>>>>>> ## now that we have a list, we change the names to what you want
>>>>>> names(ret) <- paste(1:10, "info_within_function")
>>>>> so this would not work, because I don't have the information
>>>>> anymore about the naming...
>>>> OK, so you can't do what you want to do in the manner you tried, via
>>>> lapply as you don't have control of how the list is produced once the
>>>> loop over 1:10 has been performed. At the stage that 'test' is being
>>>> applied, all it knows about is 'x' and it doesn;t have access to the
>>>> list being built up by lapply().
>>>> The *apply family of functions help us to *not* write out formal
>>>> loops
>>>> in R, but here this is causing you a problem. So we can specify an
>>>> explicit loop and fill in information as and when we want from within
>>>> the loop
>>>> ## create list to hold results
>>>> n <- 10
>>>> ret <- vector(mode = "list", length = n)
>>>> ## initialise loop
>>>> for(i in seq_len(n)) {
>>>>     ## do whatever you need to do here, but this line just
>>>>     ## replicates what 'test' did earlier
>>>>     ret[[i]] <- c(1,2,3,4,5)
>>>>     ## now add the name in
>>>>     names(ret)[i] <- paste("Mountain", i, sep = "")
>>>> }
>>>> ret
>>>> Alternatively, collect a vector of names during the loop and then
>>>> once
>>>> the loop is finished do a single call to names(ret) to replace all the
>>>> names at once:
>>>> n <- 10
>>>> ret <- vector(mode = "list", length = n)
>>>> ## new vector to hold vector of names
>>>> name.vec <- character(n)
>>>> for(i in seq_len(n)) {
>>>>     ret[[i]] <- c(1,2,3,4,5)
>>>>     ## now we just fill in this vector as we go
>>>>     name.vec[i] <- paste("Mountain", i, sep = "")
>>>> }
>>>> ## now replace all the names at once
>>>> names(ret) <- name.vec
>>>> ret
>>>> This latter version is likely to more efficient if n is big so you
>>>> don't
>>>> incur the overhead of the repeated calls to names()
>>>> The moral of the story is to not jump to using *apply all the time
>>>> to
>>>> avoid loops. Loops in R are just fine, so use the tool that helps 
>>>> you do
>>>> the job most efficiently *and* most transparently.
>>>> Take a look at the R Help Desk article by Uwe Ligges and John Fox in
>>>> the
>>>> current issue of RNews:
>>>> http://www.r-project.org/doc/Rnews/Rnews_2008-1.pdf
>>>> Which goes into this in much more detail
>>>> HTH
>>>> G
>>>>
>>> ______________________________________________
>>> 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.
>