[R] changing column labels for data frames inside a list

[Sarah Goslee]

On Mon, Mar 30, 2015 at 11:43 AM, Sven E. Templer
<sven.templer at gmail.com> wrote:
>
>
> On 30 March 2015 at 17:31, Bert Gunter <gunter.berton at gene.com> wrote:
>>
>> Sarah's statement is correct.
>>
>> So is yours. They are not contradictory, and I believe Sarah's point
>> was that the OP needed to learn the appropriate syntax.
>>
>
> That's why I pointed to ?return.
> Sarah's statement was not so clear (and might have been misleading) for me
> regarding the R expertise of the OP.

You're right: I totally wasn't clear. I got involved making a working
reproducible example and didn't explain what I was doing. And you are
totally correct, the last expression will be returned. Mea culpa.

Sarah


>>
>> -- Bert
>>
>> Bert Gunter
>> Genentech Nonclinical Biostatistics
>> (650) 467-7374
>>
>> "Data is not information. Information is not knowledge. And knowledge
>> is certainly not wisdom."
>> Clifford Stoll
>>
>>
>>
>>
>> On Mon, Mar 30, 2015 at 7:56 AM, Sven E. Templer <sven.templer at gmail.com>
>> wrote:
>> > On 30 March 2015 at 16:47, Sarah Goslee <sarah.goslee at gmail.com> wrote:
>> >
>> >> colnames(e) <- paste0('pop',1:12)
>> >>
>> >> isn't a function and doesn't return anything.
>> >>
>> >
>> > But
>> > function(e){colnames(e) <- paste0('pop', 1:2)}
>> > is a function and it returns something (the last evaluated expression! -
>> > here the paste0 return):
>> >
>> >> mylist2 <- lapply(mylist, function(e){colnames(e) <- paste0('pop',
>> >> 1:2)})
>> >> mylist2
>> > [[1]]
>> > [1] "pop1" "pop2"
>> >
>> > [[2]]
>> > [1] "pop1" "pop2"
>> >
>> > [[3]]
>> > [1] "pop1" "pop2"
>> >
>> >  from ?return:
>> >
>> > If the end of a function is reached without calling return, the value of
>> > the last evaluated expression is returned.
>> >
>> >>
>> >> > mylist <- list(
>> >> + data.frame(a = runif(10), b = runif(10)),
>> >> + data.frame(c = runif(10), d = runif(10)),
>> >> + data.frame(e = runif(10), f = runif(10)))
>> >> > mylist2 <- lapply(mylist, function(e){colnames(e) <- paste0('pop',
>> >> > 1:2);
>> >> e})
>> >> > colnames(mylist2[[1]])
>> >> [1] "pop1" "pop2"
>> >>
>> >> Sarah
>> >>
>> >> On Mon, Mar 30, 2015 at 9:54 AM, Vikram Chhatre
>> >> <crypticlineage at gmail.com> wrote:
>> >> >> summary(mygenfreqt)
>> >> >                   Length Class  Mode
>> >> > dat1.str 59220  -none- numeric
>> >> > dat2.str 59220  -none- numeric
>> >> > dat3.str 59220  -none- numeric
>> >> >
>> >> >> head(mylist[[1]])
>> >> >            1     2     3     4     5     6     7     8     9    10
>> >> > 11
>> >> >  12
>> >> > L0001.1 0.60 0.500 0.325 0.675 0.600 0.500 0.500 0.375 0.550 0.475
>> >> > 0.350
>> >> > 0.275
>> >> > L0001.2 0.40 0.500 0.675 0.325 0.400 0.500 0.500 0.625 0.450 0.525
>> >> > 0.650
>> >> > 0.725
>> >> >
>> >> > I want to change 1:12 to pop1:pop12
>> >> >
>> >> > mylist<- lapply(mylist, function(e) colnames(e) <-
>> >> > paste0('pop',1:12))
>> >> >
>> >> > What this is doing is replacing the data frames with just names
>> >> > pop1:pop12.  I just want to replace the column labels.
>> >> >
>> >> > Thanks for any suggestions.
>> >> >
>> >>
>> >> --
>> >> Sarah Goslee
>> >> http://www.functionaldiversity.org