Re: Answer: How do I get the root mean square of a list?

Pass-by-reference isn't winning you anything here, since you are explicitly dereferencing and iterating over the entire array anyway. I would rewrite as:

sub std_dev {
  my ($sum, $avg, $variance, $t);

  $sum += $_ for @_;
  $avg = $sum / @_;
  $variance += ($t = ($avg - $_)) * $t for @_;
  $variance /= @_;

  sqrt $variance;
}
[download]

This benchmarks over twice as fast, mostly because it's not calling outside subs for summation and averaging. There's a little intermediate result optimization in there as well.

I wonder, however, if there isn't a more efficient algorithm for calculating standard deviation, that doesn't require iterating twice... tilly?

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

Comment on Re: Answer: How do I get the root mean square of a list? Download Code

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Re (tilly) 2: Answer: How do I get the root mean square of a list? by tilly (Archbishop) on Jun 04, 2001 at 21:15 UTC
You need to iterate twice. However the following version should be slightly faster and I think is numerically more stable: `sub std_dev { my ($sum, $sqr_sum); for (@_) { $sum += $_; $sqr_sum += $_ * $_; } sqrt(($sqr_sum - $sum * $sum / @_)/@_); }` [download] And, of course, if your set of numbers is a sample set from a random distribution, the standard deviation of the numbers is a biased predictor of the true standard deviation, so you may prefer the following: `# Produces an unbiased estimate of a standard deviation # from a sample sub std_dev_samp { my ($sum, $sqr_sum); for (@_) { $sum += $_; $sqr_sum += $_ * $_; } sqrt(($sqr_sum - $sum * $sum / @_)/$#_); }` [download]	[reply] [d/l] [select]
Re: Re: Answer: How do I get the root mean square of a list? by princepawn (Parson) on Jun 04, 2001 at 21:16 UTC
you have heard of PDL havent you? it handles array ops as fast as C within Perl. and it does have rms as part of its stats() function.	[reply]