On May 18, 2012, at 12:56 PM, Matthew Ouellette wrote:
Dear R help list,
I am very new to R and I apologize in advance if this has been
answered
before. I have done my best to google/R search what I need but no
luck.
Here is what I am attempting:
I have hundreds of .csv files that I need to sort based on a single
column
of alphanumeric data. All of the files contain matrices that have
identical dimensions and headers, however the data table doesn't begin
until the 74th line in each file. Doing some searching, I have been
able
to create an object with elements consisting of each file in the
folder
containing the targets (please note this is my working directory):
filenames<-list.files()
alldata<-lapply(filenames, read.csv, skip=73, header=TRUE)
At this point I believe I have created an object with N elements
(where N=#
files in the wd), each containing the matrix I am attempting to
sort. I am
completely lost as to how I can sort each matrix
You should learn to use precise terminology to refer to R objects. You
have a list of dataframes (not matrices)
You can loop over then and return a list of transformed (.e.g. sorted)
dataframes:
alldata <- lapply (alldata, function(x) x[order(x[["Name"]], ] )
based on a single column
(say, "Name") and then either overwrite
The above code would overwrite.
the source files or write to a new
directory all of the sorted data.
If you didn't want it overwritten then assign it to a different name.
I half wonder if I should be creating
individual objects for each file that I read in, but I haven't been
able to
figure this out either.
Much better to stick with lists.
Please note that I am trying to sort these files
individually - would a loop be more efficient?
`lapply` is really a loop.
I appreciate the help,
BustedAvi
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David Winsemius, MD
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