Re^2: minimum, maximum and average of a list of numbers at the same time

The logical conclusion:

use List::Util qw( sum );

sub min_max_avg {
    my ( $min, $max ) = ( sort { $a <=> $b } @_ )[ 0, -1 ];
    return ( $min, $max, sum @_ );
}
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Makeshifts last the longest.

Comment on Re^2: minimum, maximum and average of a list of numbers at the same time Download Code

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Re^3: minimum, maximum and average of a list of numbers at the same time by BrowserUk (Patriarch) on Nov 10, 2005 at 20:56 UTC
How is a O(NlogN) substitute for a O(2N) operation the "logical conclusion" of a thread that started out attempting to reduce the latter to O(3N/2)? Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal? "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice.	[reply] [d/l]
Re^4: minimum, maximum and average of a list of numbers at the same time by Aristotle (Chancellor) on Nov 10, 2005 at 21:57 UTC
It’s not. Imagine an “of your approach” at the end of that sentence. Besides, it may well be that this O(N + N long N) algorithm implemented in C is so much faster per step than the O(3N/2) approach implemented in Perl that you’ll need very long arrays before the “fast” algorithm beats the “slow” one. But I haven’t bothered to benchmark, so this guess is probably wrong (as a lot of my guesses about performance have been in the past). Not that I care too much. If I had a desire to do this quickly, I’d patch List::MoreUtils. It’s a numbercrunch-ish problem, and I find that trying to solve those in Perl is rarely intuitive or particularly didactic (other than in learning about the particular quirks of the implementation of `perl` that we happen to have). In contrast, doing them in C is often straightforwardly optimisable on the abstract algorithmic level without surprises in implementation. At most, I have a vague desire to see is where the breakeven point between the O(N log N) sort vs the O(2N) `min`+`max` approaches is. Makeshifts last the longest.	[reply]