Re: When does it pay to use the schwartzian transform?

It "pays" when the cost of pre-computing the sort keys for all of your elements is significantly less than re-computing the sort keys each time you make a comparison. It depends on the cost of computing a sort key, and the number of elements.

Comment on Re: When does it pay to use the schwartzian transform?

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Re^2: When does it pay to use the schwartzian transform? by jfroebe (Parson) on Dec 16, 2005 at 18:59 UTC
When in doubt, test it with the Benchmark module Jason L. Froebe Team Sybase member No one has seen what you have seen, and until that happens, we're all going to think that you're nuts. - Jack O'Neil, Stargate SG-1	[reply]
Re^2: When does it pay to use the schwartzian transform? by Perl Mouse (Chaplain) on Dec 18, 2005 at 14:14 UTC
It "pays" when the cost of pre-computing the sort keys for all of your elements is significantly less than re-computing the sort keys each time you make a comparison. Well, in most of those cases, it pays to use the Gutman-Rossler Transform. `Perl --((8:>*`	[reply]


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