Is this odd behavior a floating point problem?

[wickedjester]

Hello!

I've got an array with 40 elements with each element having a value of '0.001'. If I add them all together and divide by 40 to get the average, I get something like:

$Avg = 0.001025

rather than 0.001, which is what it should really return.

Now, the script I'm righting is a chemical diffusion model dealing with very small numbers and this kind of inaccuracy is causing me problems. If this is a floating point issue, can anyone give me a recommendation on how to deal with this?

Many thanks!

[Eliya]

For reasons described in excruciating detail in the document already cited, there will be errors, but according to a quick test on my system, they are nowhere near as large as you claim:

$ perl -e '$x=0.001; $sum += $x for 1..40; printf "%.20f", $sum/40'
0.00100000000000000067

Looks more like a "one off" error to me (i.e. summing over one more than you divide by):

$ perl -e '$x=0.001; $sum += $x for 0..40; printf "%.20f", $sum/40'
0.00102500000000000074

[Anonymous Monk]

:D Looks like two separate off-by-one error (OBOE) errors to me :)

First you start with non-zero and add 40 times (one too many), then you start with non-zero and add 41 times (one too many twice).

If you start with non-zero you need to add only 39 times, or start with zero and add 40 times :)

In short

perl -MData::Dump -e " @f = map { 0.001 } 1 .. 40; dd\@f; $o = 0; for(
+@f){ dd $o+=$_; } dd int @f; dd $o/int(@f); "

perl -MData::Dump -e "  $o = 0; for(1 .. 40){ dd $o+= 0.001; } dd $o/4
+0; "

It didn't dawn on me to check wickedjesters (or your) math until ww raised the quesiton

[Eliya]

... you start with non-zero ...

Not sure what you're talking about.

n times adding x to zero is mathematically (but not necessarily numerically) the same as n * x.

$ perl -le '$sum += 1 for 1..40; print $sum'
40

So where is the problem?  I think you overlooked that $sum is initially undef/zero.

[roboticus]

wickedjester:

You don't show your code, but I'm pretty sure you're not doing what you think you're doing. Specifically, I believe you're adding 41 copies of 0.001, as that's the only way I can reproduce your results:

$ cat t.pl
#!/usr/bin/perl
my @a = (0.001) x 41;
my $sum=0;
$sum += $_ for @a;
print "Avg: ", $sum/40, "\n";

$ perl t.pl
Avg: 0.001025

...roboticus

When your only tool is a hammer, all problems look like your thumb.

[Anonymous Monk]

What Every Computer Scientist Should Know About Floating-Point Arithmetic

[toolic]

• sprintf
• perlfaq4 Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?

[Khen1950fx]

As I see it, you're performing addition, division, and averaging. You can dispense with the addition, division, and since you have an array, just do an average of the elements.

#!/usr/bin/perl -l

use strict;
use warnings;
use Array::Average;

print average(
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001, 0.001,
);

Returns: 0.001

[Eliya]

You can dispense with the addition, division,...

And how do you think the module arrives at its result?  What it does is exactly addition and division — which of course suffers from the same floating point issues.  Here's the relevant code snippet:

  if (@data) {
    my $sum=0;
    $sum+=$_ foreach @data;
    return $sum/scalar(@data);
  } else {
    return undef;
  }

Anyhow, as has already been pointed out, the OP's problem has likely nothing whatsoever to do with those general floating point issues, but is presumably simply the result of having computed the sum incorrectly.