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Compare two hash of arrays in perl

pinky
Novice

Nov 24, 2011, 5:18 AM

Post #1 of 6 (9423 views)

Compare two hash of arrays in perl	Can't Post

Hi,

How do i compare two hash of arrays in perl without using Data::Compare module?

Thanks
Pinky.

Anything wrong with the below code:

foreach my $name (keys %hash_array_cur) {
@cur_array = @{ $hash_array_cur{$name}};
foreach my $name2 (keys %hash_array_gold) {
@gold_array = @{ $hash_array_gold{$name2}};
my $temp = $name2;
$temp =~ s/^\s+//g;
$temp =~ s/\s+$//g;
next if $name !~ /$temp/;
my $temp1 = $hash_array_gold{$name2}[0];
$temp1 =~ s/^\s+//g;
$temp1 =~ s/\s+$//g;
print "golden is for $name : $temp1 \n";
print "generated is for $name : $hash_array_cur{$name}[0] \n";
if ( $hash_array_cur{$name}[0] !~ /$temp1/) {
#print "description in golden file is : $temp1 \n";
#print "description in generated file is : $hash_array_cur{$name}[0] \n";
}
}
}

(This post was edited by pinky on Nov 24, 2011, 5:19 AM)

wickedxter
User

Nov 25, 2011, 6:17 AM

Post #2 of 6 (9394 views)

Re: [pinky] Compare two hash of arrays in perl [In reply to]	Can't Post

Code

my $hash => {Cat => [white, brown, black], 
                           Dog => [unknown, mut, bark],}; 
 
#loop threw hash and keys 
foreach my $ref (keys %hash}){ 
    #loop threw keys and access the array 
    foreach my $array (@{$hash{$ref}}){ 
             #print array items 
              print $array; 
   } 
}

this is the basic idea ....

saurabhsharma
Novice

Jan 13, 2012, 3:15 AM

Post #3 of 6 (8974 views)

Re: [pinky] Compare two hash of arrays in perl [In reply to]	Can't Post

Hello Pinky,

You can try below mentioned code :

my %hash = (a=>1, b=>2, c=>3);
my %hash1 = (d=>4, e=>2, f=>5);
my %hash2 = reverse %hash1;
my %hash3 = reverse %hash;
my @present_only_in_hash1 = grep ! exists $hash2{$_}, values %hash;
my @present_only_in_hash2 = grep ! exists $hash3{$_}, values %hash1;
my @common = grep exists $hash3{$_}, values %hash1;

print "Present only in HASH 1 = @present_only_in_hash1\n";
print "Present only in HASH 2 = @present_only_in_hash2\n";
print "COMMON values = @common\n";

Cheers,

Saurabh

BillKSmith
Veteran

Jan 13, 2012, 9:10 AM

Post #4 of 6 (8952 views)

Re: [pinky] Compare two hash of arrays in perl [In reply to]	Can't Post

It is hard to comment on your code when I do not know what you intend. I can describe what it does do.

It compares two hashes whose values are references to arrays. It reports a difference if both hashes have the same key, but the first element of the corresponding arrays are different. Keys are considered the same if the only difference is that the key from the 'gold' hash contains leading and/or trailing whitespace. Array elements are considered the same if the entire text of the golden one appears anywhere in the other. Keys are ignored if they appear in only one of the hashes.

Even if this is exactly what is intended, it is a poor way to do it. It uses an inner loop rather than exploiting the power of hash look-ups.
Good Luck,
Bill

rovf
Veteran

Jan 16, 2012, 5:14 AM

Post #5 of 6 (8820 views)

Re: [pinky] Compare two hash of arrays in perl [In reply to]	Can't Post

Maybe Set::Scalar from CPAN?

budman
User

Feb 12, 2012, 2:14 PM

Post #6 of 6 (8185 views)

Re: [pinky] Compare two hash of arrays in perl [In reply to]	Can't Post

I use something similar below (more advanced) for comparing incoming data feeds with current known column structure. vendors love to change column orders. :)

Using powers of 2 you can compare multiple arrays.

Code

my %h = ( 
    tab1 => [qw(a b c d)], 
    tab2 => [qw(c d f g h)], 
    tab3 => [qw(b c d a i)], 
); 
my @tables = sort keys %h; 
 
my %matches; 
foreach my $t (1 .. @tables) { 
    $matches{$_} |= 2**$t for @{$h{ $tables[$t-1] }}; 
} 
 
foreach my $m (sort keys %matches) { 
    my @found; 
    foreach my $t (1 .. @tables) { 
        push @found, $tables[$t-1] if $matches{$m} & 2**$t; 
    } 
    print "$m exists in ", (@found? join(",",@found) : 'None' ),"\n"; 
}

Output:

a exists in tab1,tab3
b exists in tab1,tab3
c exists in tab1,tab2,tab3
d exists in tab1,tab2,tab3
f exists in tab2
g exists in tab2
h exists in tab2
i exists in tab3

(This post was edited by budman on Feb 12, 2012, 2:17 PM)

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