Yes. I forget how much difference that can make. Though joost's idea (with several modifications) works out fastest:

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

our $string = join '|', map{
    join rand() < 0.2
        ? 'fred'
        : 'bill',
    'pqr', 'xyz'
} 1 .. 10000;

our $first = 0;
our %counts;
cmpthese -1, {
    orig => q[
        my @arr = split(/\|/, $string);
        my @arr1 = grep { /fred/ } @arr;
        $counts{ orig } = @arr1;
    ],
    Buk1 => q[
        my @arr1 = grep { /fred/ } split /\|/, $string;
        $counts{ Buk1 } = @arr1;
    ],
    jwkrahn => q[
        my @arr1 = grep /fred/, split /\|/, $string;
        $counts{ jwkrahn } = @arr1;
    ],
    Buk2 => q[
        my @arr1 = $string =~ m[(?:^|\|)(.*?fred.*?)(?=\||$)]g;
        $counts{ Buk2 } = @arr1;
    ],
    JOOST => q[
        my @arr1 = $string =~ /(?:^|\|)([^|]*?fred[^|]*?)(?=\||$)/g;
        $counts{ JOOST } = @arr1;
    ],
};

pp \%counts;
__END__
c:\test>junk6
          Rate    orig    Buk1   JOOST jwkrahn    Buk2
orig    20.2/s      --    -28%    -48%    -58%    -84%
Buk1    28.1/s     39%      --    -28%    -42%    -77%
JOOST   39.2/s     94%     39%      --    -19%    -68%
jwkrahn 48.4/s    140%     72%     24%      --    -61%
Buk2     124/s    515%    342%    217%    157%      --
{ Buk1 => 2010, Buk2 => 2010, JOOST => 2010, jwkrahn => 2010, orig => 
+2010 }
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