Dru has asked for the wisdom of the Perl Monks concerning the following question:

Monks,

I'm trying to build a complex hash (well, for me anyway) and I have succeeded for the most part. One thing that I am trying to do is count the number of times the source shows up in the file. It doesn't have to be tied to a specific entry, just a total number. I can do this by using two different hashes, but is it possible to do this with one hash?

Thanks,
Dru 

my %count, %hash;

while (<FILE>){
 
 next unless /something/; 
 if (/<LONG REGEX>/){
   
   $hash{$3} = {
        sig => $1,
        src => $3,
        proto => $2,
        dst => $4,
        port => $5,
   };
 
  if ($hash{$3}{src}){
    $count{$3}++;
  }
 }
}

[download]

Replies are listed 'Best First'.
Re: One Hash Instead of Two
by McDarren (Abbot) on Feb 06, 2007 at 16:39 UTC
    I suspect that this code is probably not really doing what you think it is. You are using $3 as a key to your hash on each iteration. Because hash keys must be unique, each time you find a "src" that you've had before, the previous one will simply be overwritten. I suspect this isn't really what you want.

    I could suggest an alternative approach, but first I'd like to see some sample data and the output you would expect to get.

    Cheers,
    Darren :)

[reply]
Re: One Hash Instead of Two
by davorg (Chancellor) on Feb 06, 2007 at 16:30 UTC
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Re: One Hash Instead of Two
by polettix (Vicar) on Feb 06, 2007 at 18:44 UTC
    If the count is all you need, you don't need of %hash at all:
    my %count;

while (<FILE>){
 
 next unless /something/; 
 if (/<LONG REGEX>/){
   
    $count{$3}++ if $3;
 }
}

    [download]

    If McDarren is correct, and you want to keep track of all items, you should probably maintain an array inside $hash{$3} (and a variable rename would be good in this case), and in this case you would have the count "for free", just looking at how many elements each item has: 

    my %hashes_for;  # More sensible name

while (<FILE>){
 
 next unless /something/; 
 if (/<LONG REGEX>/){
   
   push @{$hashes_for{$3}}, {
        sig => $1,
        src => $3,
        proto => $2,
        dst => $4,
        port => $5,
   };
 }
}

# You can now use scalar(@{$hashes_for{whatever}}) instead of $count{w
+hatever}

    [download]

    Note: untested code!

    Flavio
    perl -ple'$_=reverse' <<<ti.xittelop@oivalf

    Don't fool yourself.
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[d/l]
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Re: One Hash Instead of Two
by Dru (Hermit) on Feb 06, 2007 at 21:25 UTC
    All,

    Thank you for your help. This is the code I decided to go with: 
    my %hash;
 
while (<FILE>){
 
 next unless /Something/;
 if (/<LONG REGEX>/){
 
   if ($hash{$3}){
     $hash{$3}{count}++;
     next;
   }

   $hash{$3} = {
        sig => $1,
        src => $3,
        proto => $2,
        dst => $4,
        port => $5,
   };
 }
}

    [download]
    davorg,

    I based my code on yours, just changed it a bit. Thanks for the tip on Data::Dumper, I've heard about it before, just never used it. Very cool

    McDarren,

    Believe it or not, one entry from each source is all I really need. I believe this code will speed things up a bit since it will count the source and go to the next one if we already have it.

    fenLisesi,

    Thanks, I tend to do that more then I wish to admit.

    frodo72,

    Thanks for the code and reply. If I need to build the more complex data structure, I'll probably do something similar to what you gave me. I'll just have to brush up on perlreftut first.

    -Dru
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[d/l]
Re: One Hash Instead of Two
by fenLisesi (Priest) on Feb 06, 2007 at 18:32 UTC
    Side issue: you probably want my (%count, %hash);. Cheers.
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[d/l]

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