After you read in the first line, read the rest of the file with a single scan:
rest <- scan(..., sep="\t", what=0, skip=1)
index <- 1 # used to march through 'rest'
for (i in 1:3000){
for (j in 1:i){
malt[i,j] <- rest[index]
index <- index+1
}
}
There are probably faster ways, but this should go quicker since most
of your previous time was spent in the reading.
On Jan 2, 2008 6:05 PM, Talbot Katz <[EMAIL PROTECTED]> wrote:
>
> Hi.
>
> I have a matrix stored in a large, tab-delimited flat file. The first row
> contains column names. Because the matrix is symmetric, the file has lower
> triangular format, so the second row contains one number, the third row two
> numbers, etc. In general, row k+1 contains k numbers; the matrix has 3000
> rows, so the file has 3001 rows. The file has variable length records, so
> each row ends with its last piece of data. I read in the file and produced
> the full symmetric matrix as follows:
>
> > mana01 <- scan( file = "C:/mat.dat", sep = "\t", nlines = 1, what =
> > "character" )Read 3000 items> nco <- length( mana01 )> malt <- matrix(0,
> > nrow = nco, ncol = nco )> colnames( malt ) <- mana01> rownames( malt ) <-
> > mana01> for ( i in 1:3000 ) { malt[ i, (1:i) ] <- scan( file="C:/mat.dat",
> > skip = i, n = i, quiet = TRUE ) }
> > mat <- malt + t( malt ) - diag( diag( malt ) )>
>
> The for loop took a couple of hours to complete. I suspect there's a much
> faster way to do this. Any suggestions? Thanks!
>
> -- TMK --212-460-5430 home917-656-5351 cell
> [[alternative HTML version deleted]]
>
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>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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