Hi Gene,
Rather than using R to parse this file, have you considered using either
grep or sed to pre-process the file and then read it in?
It looks like you just want lines starting with numbers, so something like
grep '^[0-9]\+' thefile.csv > otherfile.csv
should be much faster, and then you can just read in otherfile.csv using
read.csv().
Best,
Jim
Gene Leynes 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
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