Hi Sashi,
On Thu, Dec 9, 2010 at 9:44 AM, Sashi Challa <cha...@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]]
>
>
> ______________________________________________
> R-help@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.
--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/
______________________________________________
R-help@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.
