tried further, still no resolution:
file1 = (space seperated values)
outputfile = (colum1, and 5 onwards) field seperator would be space.
from linux: cut -f1,5- -d" " file |grep -v "^0" | sort -n > to_file;
open(tmpFH,"<","file1");
@tmpAR = <tmpFH>;
close(tmpFH);
foreach $line (@tmpAR) {
@tmpAR2 = split(/ /,"$line");
$line2 = "$tmpAR[0] $tmpAR[5..-1]";
push(@tmpAR3,$line2);
};
@tmpAR3 = sort(@tmpAR3);
print @tmpAR3;
From: Rajeev Prasad <rp.ne...@yahoo.com>
To: Rajeev Prasad <rp.ne...@yahoo.com>; Perl Beginners <beginners@perl.org>
Sent: Thursday, September 1, 2011 6:50 PM
Subject: Re: how to do this in perl (from unix)
I tried this
open(tmpFH,"<","somefile");
@tmpAR = <tmpFH>;
close(tmpFH);
push(my @tmpAR2,$tmpAR[0]);
push (@tmpAR2,$tmpAR[5..-1]);
my @tmpAR3 = grep {!/^0 /} @tmpAR2;
@tmpAR3 = sort(@tmpAR3);
print @tmpAR3;
but getting error:
Argument "" isn't numeric in array element at ./test.pl line 31.
output is not coming correct, only first row of original file is coming.....
From: Rajeev Prasad <rp.ne...@yahoo.com>
To: Perl Beginners <beginners@perl.org>
Sent: Thursday, September 1, 2011 6:38 PM
Subject: how to do this in perl (from unix)
from linux:
cut -f1,5- -d" " file |grep -v "^0" | sort -n > to_file; <======this line:
how to achieve this in perl?
will below work, in perl?
if ( ! -s $sourcedir/somefile ){
open(tmpFH,"<","file2");
@tmpAR = <tmpFH>;
close(tmpFH);
push(@tmpAR2,$tmpAR[0],$tmpAR[5..]); #select on columns 1 and 5 onwards
from the original file
my @tmpAR3 = grep {!^[0 ]} @tmpAR2; #omit line with 0 in the beginning
@tmpAR3 = sort(@tmpAR3); # sort on the first column
}
we can then print tmpAR3 to a file.
plz. advice. thx.
