If the headers all start with the same letter, "A" say, and the data
only contain numbers on their lines then just use
read.table(..., comment = "A")
On Mon, Nov 2, 2009 at 2:03 PM, Gene Leynes <gleyne...@gmail.com> wrote:
> I've been trying to figure out how to read in a large file for a few days
> now, and after extensive research I'm still not sure what to do.
>
> I have a large comma delimited text file that contains 59 fields in each
> record.
> There is also a header every 121 records
>
> This function works well for smallish records
> getcsv=function(fname){
> ff=file(description = fname)
> x <- readLines(ff)
> closeAllConnections()
> x <- x[x != ""] # REMOVE BLANKS
> x=x[grep("^[-0-9]", x)] # REMOVE ALL TEXT
>
> spl=strsplit(x,',') # THIS PART IS SLOW, BUT MANAGABLE
>
> xx=t(sapply(1:length(spl),function(temp)as.vector(na.omit(as.numeric(spl[[temp]])))))
> return(xx)
> }
> It's not elegant, but it works.
> For 121,000 records it completes in 2.3 seconds
> For 121,000*5 records it completes in 63 seconds
> For 121,000*10 records it doesn't complete
>
> When I try other methods to read the file in chunks (using scan), the
> process breaks down because I have to start at the beginning of the file on
> every iteration.
> For example:
> fnn=function(n,col){
> a=122*(n-1)+2
> xx=scan(fname,skip=a-1,nlines=121,sep=',',quiet=TRUE,what=character(0))
> xx=xx[xx!='']
> xx=matrix(xx,ncol=49,byrow=TRUE)
> xx[,col]
> }
> system.time(sapply(1:10,fnn,c=26)) # 0.31 Seconds
> system.time(sapply(91:90,fnn,c=26)) # 1.09 Seconds
> system.time(sapply(901:910,fnn,c=26)) # 5.78 Seconds
>
> Even though I'm only getting the 26th column for 10 sets of records, it
> takes a lot longer the further into the file I go.
>
> How can I tell scan to pick up where it left off, without it starting at the
> beginning?? There must be a good example somewhere.
>
> I have done a lot of research (in fact, thank you to Michael J. Crawley and
> others for your help thus far)
>
> Thanks,
>
> Gene
>
> [[alternative HTML version deleted]]
>
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>
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