Message-ID: <AANLkTi=XC6XOT+k58eaGUVtM6YVZsh0nus4LKTvBrPcU@mail.gmail.com>
Date: 2010-12-09T18:06:30Z
From: Joshua Wiley
Subject: lapply getting names of the list
In-Reply-To: <39F0EAC23E9BBF4AB03AF1562B025CC803607E9514@EX-MB05.ohsu.edu>
Hi Sashi,
On Thu, Dec 9, 2010 at 9:44 AM, Sashi Challa <challa at ohsu.edu> wrote:
> Hello All,
>
> I have a toy dataframe like this. It has 8 columns separated by tab.
>
> Name ? ?SampleID ? ? ? ?Al1 ? ? Al2 ? ? X ? ? ? Y ? ? ? R ? ? ? Th
> rs191191 ? ? ? ?A1 ? ? ?A ? ? ? B ? ? ? 0.999 ? 0.09 ? ?0.78 ? ?0.090
> abc928291 ? ? ? A1 ? ? ?B ? ? ? J ? ? ? 0.3838 ?0.3839 ?0.028 ? 0.888
> abcnab ?A1 ? ? ?H ? ? ? K ? ? ? 0.3939 ?0.939 ? 0.3939 ?0.77
> rx82922 B1 ? ? ?J ? ? ? K ? ? ? 0.3838 ?0.393 ? 0.393 ? 0.00
> rcn3939 B1 ? ? ?M ? ? ? O ? ? ? 0.000 ? 0.000 ? 0.000 ? 0.77
> tcn39399 ? ? ? ?B1 ? ? ?P ? ? ? I ? ? ? 0.393 ? 0.393 ? 0.393 ? 0.56
>
> Note that the SampleID is repeating. So I want to be able to split the dataset based on the SampleID and write the splitted dataset of every SampleID into a new file.
> I tried split followed by lapply to do this.
>
> infile <- read.csv("test.txt", sep="\t", as.is = TRUE, header = TRUE)
> infile.split ?<- split(infile, infile$SampleID)
> names(infile.split[1]) ?## outputs ?A1?
correct, names() returns the top level names of infile.split (i.e.,
the two data frame names)
> ## now A1, B1 are two lists in infile.split as I understand it. Correct me if I am wrong.
It is a single, named list containing two data frames (A1 and B1)
(though data frames are built from lists, I think so I suppose in a
way it contains two lists, but that is not really the point).
>
> lapply(infile.split,function(x){
> ? ? ? ? ? ? ?filename <- names(x) #### here I expect to see A1 or B1, I didn?t, I tried (names(x)[1]) and that gave me ?Name? and not A1 or B1.
by using lapply() on the actual object, your function is getting each
element of the list. That is:
infile.split[[1]]
infile.split[[2]]
trying names() on those:
names(infile.split[[1]])
should show what you are getting
> ? ? ? ? ? ? ?final_filename <- paste(filename,?toy_set.txt?,sep=?_?)
> ? ? ? ? ? ? ?write.table(x, file = paste(path, final_filename,sep=?/?, row.names=FALSE, quote=FALSE,sep=?\t?)
FYI I think you are missing a parenthesis in there somewhere
> ?} )
>
> In lapply I wanted to give a unique filename to all the split Sample Ids, i.e. name them here as A1_toy_set.txt, B1_toy_set_txt.
> How do I get those names, i.e. A1, B1 to a create a filename like above.
Try this:
## read your data from the clipboard (obviously you do not need to)
infile <- read.table("clipboard", header = TRUE)
split.infile <- split(dat, dat$SampleID) #split data
path <- "~" # generic path
## rather than applying to the data itself, instead apply to the names
lapply(names(split.infile), function(x) {
write.table(x = split.infile[[x]],
file = paste(path, paste(x, "toy_set.txt", sep = "_"), sep = "/"),
row.names = FALSE, quote = FALSE, sep = "\t")
cat("wrote ", x, fill = TRUE)
})
it will return two NULL lists, but that is fine because it should have
written the files.
> When I write each of the element in the list obtained after split into a file, the column names would have names like A1.Name, A1.SampleID, A1.Al1, ?.. Can I get rid of ?A1? in the column names within the lapply (other than reading in the file again and changing the names) ?
Can you report the results of str(yourdataframe) ? I did not have
that issue just copying and pasting from your email and using the code
I showed above.
Cheers,
Josh
>
> Thanks for your time,
>
> Regards
> Sashi
>
>
> ? ? ? ?[[alternative HTML version deleted]]
>
>
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--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/