sorting some data

I am trying to sort some data in bash. Data looks like below.

    2011072开发者_如何转开发4.gz   1347
    20110724.gz   2128
    20110725.gz   1315
    20110725.gz   2334
    20110726.gz   808
    20110726.gz   1088
    -bash-3.2$

After sorting, it should look like

    20110724.gz 3475
    20110725.gz 3649
    20110726.gz 1896

Basically, for a given date, the data are summed up. Can somebody help? Thanks.

hmm, hopefully I figure it out in a few days.

Here's a quick and dirty perl oneliner:

$ perl -e 'my %h = (); while (<>) { chomp; my ($fname, $count) = split; $h{$fname} += $count;} foreach my $k (sort keys %h) {print $k, " ", $h{$k}, "\n"}' < datafile

Here's a perl solution.

Usage: script.pl input.txt > output.txt

Code:

use warnings;
use strict;
use ARGV::readonly;

my %sums;
while (<>) {
    my ($date, $num) = split;
    $sums{$date} += $num;
}
for my $date (sort keys %sums) {
    print "$date $sums{$date}\n";
}

Or as a one-liner:

$ perl -we 'my %h; while(<>) { ($d,$n)=split; $h{$d}+=$n; } print "$_ $h{$_}\n" for sort keys %h;' data2.txt

In case you do need a numerical sort on the dates:

sort { substr($a,0,8) <=> substr($b,0,8) } keys %sums;

You don't need perl for doing that. Some shell trickery will help :)

sort -n -k1,8 <file | while true ; do
    if ! read line ; then
        test -n "$accfile" && echo $accfile $value
        break
    fi
    line=$(echo $line | tr -s ' ' )
    curfile=$(echo $line | cut -d\  -f1)
    curvalue=$(echo $line | cut -d\  -f2)
    if [ $curfile != "$accfile" ] ; then
        # new file, output the last if not empty
        test -n "$accfile" && echo $accfile $value
        accfile=$curfile
        value=$curvalue
    else
        value=$(expr $value \+ $curvalue)
    fi
done

The k parameter tells sort what characters use to sort. As dates are put in number-ordered format, a number sort (-n) works.