|
in reply to Re: Re: reversing a sort...
in thread reversing a sort...
reverse sort { arbitrary code } @list will always be less efficient than sort { arbitrary code with $a and $b swapped } @list
Not always, the number of comparisons and swaps performed by sort will depend on the order of the list. On my machine, reverse sort {$a <=> $b} (1..100000); is faster than sort {$b <=> $a} (1..100000);
For almost all cases, you are correct, but it is not _always_ the case. |
Re: Re: Re: Re: reversing a sort...
by chipmunk (Parson) on May 09, 2001 at 05:09 UTC
|
I quite intentionally did not say one would be always be "faster". I said reverse sort is less efficient. sort and reverse sort are both O(n log n), but the one with reverse has a larger constant factor and is less efficient.
By the way, if you want to know the fastest way to sort the list (1..100_000) in descending order, it's:
reverse (1 .. 100_000);
;)
Update: Doh. As Tilly explains below, I totally messed up the Big O analysis. Anyway, there is an added inefficiency with reverse sort, that's my point. :) | [reply]
[d/l] |
|
|
| [reply] |
|
