Re^5: Benchmark, -s versus schwartzian (vs hash)

in reply to Re^4: Benchmark, -s versus schwartzian (vs hash)
in thread Benchmark, -s versus schwartzian

The node you've mentioned actually uses an even faster form of the Schwartzian transform I've completely forgot about.

Thus, I've added it to my benchmark too:

sub sort_arr {
        my @h = map -s, @all;
        @results = @all[sort {$h[$a]<=>$h[$b]} 0..@h-1];
}
[download]

The results are amazing: this form is about 1.5 times as efficent than any of the other ones.

Comment on Re^5: Benchmark, -s versus schwartzian (vs hash) Download Code

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Re^6: Benchmark, -s versus schwartzian (vs hash) by Aristotle (Chancellor) on Aug 23, 2004 at 15:58 UTC
Yes, because it uses an array, and only one array. What's even more important to me is that it's also decidely easier on memory. Every single array in Perl has a minimum overhead of about 100 bytes without even counting the individual scalars inside it (and the one holding the reference to it..). When you're trying to sort a couple dozen million elements with a Schwartzian transform, as I've had to, that adds up alarmingly. In comparison, this method is basically free in terms of memory cost. Nitpick: you should really say `0 .. $#h` when that's what you mean, as is the case here. Makeshifts last the longest.	[reply]

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Re^6: Benchmark, -s versus schwartzian (vs hash)
by Aristotle (Chancellor) on Aug 23, 2004 at 15:58 UTC

Yes, because it uses an array, and only one array.

What's even more important to me is that it's also decidely easier on memory. Every single array in Perl has a minimum overhead of about 100 bytes without even counting the individual scalars inside it (and the one holding the reference to it..). When you're trying to sort a couple dozen million elements with a Schwartzian transform, as I've had to, that adds up alarmingly. In comparison, this method is basically free in terms of memory cost.

Nitpick: you should really say 0 .. $#h when that's what you mean, as is the case here.

Makeshifts last the longest.

[reply]

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