Perl Merge file

问题:

I have 3 or multiple files I need to merge, the data looks like this..

file 1
0334.45656
0334.45678
0335.67899
file 2
0334.89765
0335.12346
0335.56789
file 3
0334.12345
0335.45678
0335.98764

Expected output in file 4,

0334.89765
0334.89765
0334.89765
0334.12345
0335.67899
0335.12346
0335.56789
0335.45678
0335.98764

So far I have tried but data in 4rth file does not come in sorted order,

#!/usr/bin/perl
my %hash;
my $outFile = "outFile.txt";
foreach $file(@ARGV)
{
print "$file\n";
open (IN, "$file") || die "cannot open file $!";
open (OUT,">>$outFile") || die "cannot open file $!";
while ( <IN> )
{
    chomp $_;
    ($timestamp,$data) = split (/\./,$_);
    $hash{$timeStamp}{'data'}=$data;
    if (defined $hash{$timeStamp})
    {
    print "$_\n";
    print OUT"$_\n";

        }
}
}
close (IN);
close (OUT);

回答1:

I wouldn't normally suggest this, but unix utilties should be able to handle this just fine.

1. cat the 3 files together.
2. use sort to sort the merged file.

However, using perl, could just do the following:

#!/usr/bin/perl

use strict;
use warnings;

my @data;
push @data, $_ while (<>);

# Because the numbers are all equal length, alpha sort will work here
print for sort @data;

However, as we've discussed, it's possible that the files will be extremely large. Therefore it will be more efficient both in memory and speed if you're able to take advantage of the fact that all the files are already sorted.

The following solution therefore streams the files, pulling out the next one in order each loop of the while:

#!/usr/bin/perl

# Could name this catsort.pl

use strict;
use warnings;
use autodie;

# Initialize File handles
my @fhs = map {open my $fh, '<', $_; $fh} @ARGV;

# First Line of each file
my @data = map {scalar <$_>} @fhs;

# Loop while a next line exists
while (@data) {
    # Pull out the next entry.
    my $index = (sort {$data[$a] cmp $data[$b]} (0..$#data))[0];

    print $data[$index];

    # Fill In next Data at index.
    if (! defined($data[$index] = readline $fhs[$index])) {
        # End of that File
        splice @fhs, $index, 1;
        splice @data, $index, 1;
    }
}

回答2:

Using Miller's idea in a more reusable way,

use strict;
use warnings;

sub get_sort_iterator {
  my @fhs = map {open my $fh, '<', $_ or die $!; $fh} @_;
  my @d;

  return sub {
    for my $i (0 .. $#fhs) {
      # skip to next file handle if it doesn't exists or we have value in $d[$i]
      next if !$fhs[$i] or defined $d[$i];

      # reading from $fhs[$i] file handle was success?
      if ( defined($d[$i] = readline($fhs[$i])) ) { chomp($d[$i]) }
      # file handle at EOF, not needed any more
      else  { undef $fhs[$i] }
    }
    # compare as numbers, return undef if no more data
    my ($index) = sort {$d[$a] <=> $d[$b]} grep { defined $d[$_] } 0..$#d
      or return;

    # return value from $d[$index], and set it to undef
    return delete $d[$index];
  };
}

my $iter = get_sort_iterator(@ARGV);
while (defined(my $x = $iter->())) {
  print "$x\n";
}

output

0334.12345
0334.45656
0334.45678
0334.89765
0335.12346
0335.45678
0335.56789
0335.67899
0335.98764

回答3:

Suppose every input files are already in ascending order and have at least one line in them, this script could merge them in ascending order:

#!/usr/bin/perl

use warnings;
use strict;

use List::Util 'reduce';

sub min_index {
    reduce { $_[$a] < $_[$b] ? $a : $b } 0 .. $#_;
}

my @fhs = map { open my $fh, '<', $_; $fh } @ARGV;
my @data = map { scalar <$_> } @fhs;

while (@data) {
    my $idx = min_index(@data);
    print "$data[$idx]";
    if (! defined($data[$idx] = readline $fhs[$idx])) {
        splice @data, $idx, 1;
        splice @fhs, $idx, 1;
    }
}

---

Note: this is basic the same as the second script offered by @Miller, but a bit clearer and more concise.

回答4:

I suggest this solution, which uses a sorted array of hashes - each hash corresponding to an input file, and containing a file handle fh, the last line read line and the timestamp extracted from the line timestamp.

The hash at the end of the array always corresponds to the input that has the smallest value for the timestamp, so all that is necessary is to repeateedly pop the next value from the array, print its data, read the next line and (if it hasn't reached eof) insert it back into the array in sorted order.

This could produce an appreciable increase in speed over the repeated sorting of all the data for each output line that other answers use.

Note that the program expects the list of input files as parameters on the command line, and sends its merged output to STDOUT. It also assumes that the input files are already sorted.

use strict;
use warnings;
use autodie;

my @data;

for my $file (@ARGV) {
  my $item;
  open $item->{fh}, '<', $file;
  insert_item($item, \@data);
}

while (@data) {
  my $item = pop @data;
  print $item->{line};
  insert_item($item, \@data);
}

sub insert_item {
  my ($item, $array) = @_;
  return if eof $item->{fh};
  $item->{line} = readline $item->{fh};
  ($item->{timestamp}) = $item->{line} =~ /^(\d+)/;
  my $i = 0;
  ++$i while $i < @$array and $item->{timestamp} < $array->[$i]{timestamp};
  splice @$array, $i, 0, $item;
}

output

0334.45656
0334.89765
0334.12345
0334.45678
0335.12346
0335.45678
0335.67899
0335.56789
0335.98764