UPDATE

Sorry guys, I have found another way to get the results I wanted

I wanted to stock times inside slices of hours: all the times between, for ex: 11:00 and 12:00 and equally redistribute them inside this slice, but at the same time keeping the old values so as to make the difference between the new and former times Then I realized I had to use a second hash to conserve the minutes

This is the code which was already inspired a lot by BrowserUK but that I have uddated for my purpose. 

#! perl -slw
use strict;
use diagnostics;
use Data::Dump qw[ pp ];

my %in;
my %former_in;

while (<DATA>){
m[(\d\d):(\d\d) (.+)$] and push @{ $in{ $1 } }, $3 and $former_in{$3} 
+= $2;
print STDOUT "1 is $1\n";
print STDOUT "2 is $2\n";
print STDOUT "3 is $3\n";
}

my @keys = keys %in;
my @times = sort @keys;

for my $hr ( sort{ $a <=> $b } keys %in ) {
    my $n = $#{ $in{ $hr } };
    my $step = 60 / $n;
    my $min = 0;
    for my $flight ( @{ $in{ $hr } } ) {
        my $former_time = $former_in{$flight};
        # relative difference
        my $diff = int( $min ) - $former_in{$flight};
        printf( "%02d:%02d %s %s\n", $hr, int( $min ), $flight, $diff 
+);
        $min += $step;
        $min = 59 if $min > 59;
    }
}





__DATA__
11:10 A1
11:30 E4
11:30 Z4
11:50 H5
12:02 H6
12:25 B2
12:25 A8
12:30 F3
12:30 E7
12:50 E15
12:55 E16

[download]

old prog 

sub by_hour{
if ($a =~ /(\d\d):(\d\d)/){
        $h_a = $1;
        $mn_a = $2;
    }
if ($b =~ /(\d\d):(\d\d)/){
        $h_a = $1;
        $mn_a = $2;
    }
if ($h_a < $h_b){
        -1;
    }
if ($h_a > $h_b){
        1;
    }
if ($h_a == $h_b){
    
    if ($mn_a < $mn_b){
        -1;
    }
    if ($mn_a > $mn_b){
        1;
    }
    if ($mn_a == $mn_b){
        0;
    }
}
}

[download]

In reply to sub routine to sort per time by steph_bow

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