haukex the test you are proposing removes only 1 item from @numbers. If the test removed more than one numbers (or all) one after the other (after a shuffle) could be more realistic (and my hash-based approach could get a chance too :)): 

use warnings;
use strict;
use Benchmark qw/cmpthese/;

use List::Util qw/shuffle/;

my @numbers = ( 0..20 );
my @indices = shuffle 0..$#numbers;

my @expect = ( 0..2,4..20 );
use constant TEST => 0;

cmpthese(-2, {
    splice => sub {
        my @output = @numbers;
        for my $index (@indices){
            splice @output, 0, 1;
        }
        join("\0", @output) eq join("\0", @expect) or die if TEST;
    },
    grep => sub { # https://www.perlmonks.org/?node_id=11123877
        my @output;
        for my $index (@indices){
            @output = @numbers[ grep $_ != $index, 0 .. $#numbers ];
        }
        join("\0", @output) eq join("\0", @expect) or die if TEST;
    },
    hash => sub {
        my %ha; @ha{0..$#numbers} = @numbers;
        for my $index (@indices){
            delete $ha{$index};
        }
        my @output = map { $ha{$_} } sort keys %ha;
        join("\0", @output) eq join("\0", @expect) or die if TEST;
    },
});

__END__

           Rate   grep   hash splice
grep    30916/s     --   -85%   -96%
hash   205775/s   566%     --   -71%
splice 707918/s  2190%   244%     --

[download]

With the above, grep is disadvantaged and splice it just a guess by me (it always removes the first item).

The bottom line: in a loop of removals, hash-based approach can be a contestant too although splice is by far the best but it messes the indices.

bw, bliako

In reply to Re: Fastest way to "pick without replacement" by bliako
in thread Fastest way to "pick without replacement" by haukex

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