use warnings;
use strict;
use List::Util qw(shuffle);

my @nums = shuffle(1 .. 10);
my $sample = pop @nums;
print "sample = $sample\n";
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A way-

my @list = ( "a" .. "z" );
while ( @list )
{
    print splice(@list,rand(@list),1), $/;
}
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#!/usr/bin/perl
use strict;
use warnings;
use List::Util qw (shuffle);

my @array = (0 .. 9);
@array = shuffle @array;
print pop(@array), "\n", @array;
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Note this is discussed in perlfaq4's How do I shuffle an array randomly?.

While splice is helpful, shuffle + shift || pop is a very simple and elegant solution. Didn't even twig on that one. Thanks very much.

Shuffleing is O(n) and spliceing is O(1), this may matter if your array is large.

Update: Oops, my understanding of splice was wrong. See JavaFan and ikegami below.

If the array is being used as a grab bag and order doesn't matter, you could use the "pack of cards" algorithm I mentioned in Re: getting random number 8 times never the same and Re: Random data generation..

Regards,

John Davies

$ perl -MDDS -e " @f = 1..3; Dump(\@f); warn splice @f, rand(@f),1; Du
+mp( \@f )"
$ARRAY1 = [
            1,
            2,
            3
          ];
1 at -e line 1.
$ARRAY1 = [
            2,
            3
          ];
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Yet Another Way (for the senile who don't remember shift):

#!/usr/bin/perl
use strict;
use warnings;
# 878281

my @array = ("a", "b", "c", "d", "e", "f",);

my $i = int(rand(6));

my $array = $array[$i];
undef $array[$i];
print "\t\$i: $i and \$array: $array \n";
print "\tRemaining \@array:  \t";
no warnings qq(uninitialized);
for $array(@array) {
    if (defined($array)) { 
        print $array . "\t";
    } else {
        next;
    }
}
print "\n done!\n";
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Looping around this, if one wishes to exhaust the array, is left as an exercise for the interested.