The brute force solution:

/bin/perl -w
use strict;
my %h=( k1 => "v1", k2 => "v2", k3 => "v3"); 
my @keys= keys %h; 
my $k= $keys[int rand( @keys)]; 
print "random key: $k\n";'
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If you want to do this on a huge hash and you want to do lots of random access and updates then you should look at Building a Better Hash, a pretty clever way to solve this problem.

$lh=scalar keys %h;
print ((keys %h)[int rand($lh)]);
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In a scalar context, %h returns a string consisting of the number of buckets used, a slash, and the number of buckets allocated. You could have a hash with 1000 keys and only 16 buckets. Thus, your "random key" is limited to one of the first few keys in the hash.

Try $lh = scalar keys %h; instead.

You also don't need int because indexing automatically int's it. {grin}

But you are computing the keys twice, and that's expensive for large hashes. Perhaps this instead:

do { my @k = keys %hash; $k[rand @k] }
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-- Randal L. Schwartz, Perl hacker

scalar keys %hash doesn't actually compute the keys (except for tied hashes).

$lh=(split "/",scalar keys %h)[0];
print ((keys %h)[int rand($lh)]);
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Originally posted as a Categorized Answer.