[R] lapply getting names of the list
dwinsemius at comcast.net
Thu Dec 9 20:21:32 CET 2010
On Dec 9, 2010, at 12:44 PM, Sashi Challa 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) ## outputs “A1”
> ## now A1, B1 are two lists in infile.split as I understand it.
> Correct me if I am wrong.
> filename <- names(x) #### here I expect to see A1 or
> B1, I didn’t, I tried (names(x)) and that gave me “Name” and not
> A1 or B1.
> final_filename <- paste(filename,”toy_set.txt”,sep=”_”)
> write.table(x, file = paste(path,
> final_filename,sep=”/”, row.names=FALSE, quote=FALSE,sep=”\t”)
> } )
> In lapply I wanted to give a unique filename to all the split Sample
> Ids, i.e. name them here as <dragged to the c() construct>.
> How do I get those names, i.e. A1, B1 to a create a filename like
names(file.split) <- c("A1_toy_set.txt", "B1_toy_set_txt")
> When I write each of the element in the list obtained after split
> into a file,
How are you proposing do do this "writing"?
> the column names would have names like A1.Name, A1.SampleID, A1.Al1,
Are you sure? Why would you think that?
> Can I get rid of “A1” in the column names within the lapply (other
> than reading in the file again and changing the names) ?
> Thanks for your time,
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David Winsemius, MD
West Hartford, CT
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