Beefy Boxes and Bandwidth Generously Provided by pair Networks RobOMonk
Your skill will accomplish
what the force of many cannot
 
PerlMonks  

Is this odd behavior a floating point problem?

by wickedjester (Initiate)
on Mar 23, 2012 at 16:58 UTC ( #961262=perlquestion: print w/ replies, xml ) Need Help??
wickedjester has asked for the wisdom of the Perl Monks concerning the following question:

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!

Comment on Is this odd behavior a floating point problem?
Re: Is this odd behavior a floating point problem?
by Anonymous Monk on Mar 23, 2012 at 17:02 UTC
      Yes, I've seen and skimmed this document, but, not to be rude, but I'm not interested in becoming a computer scientist in order to write a script to do basic math. Adding together 0.001 40 times is pretty basic and if my calculator can do it, I not sure I understand why Perl won't.

        It's not a Perl problem, it's the problem of representing a non-terminating series for base 2 using a finite number of binary digits. You are accustomed to seeing it in base ten when you try to represent 1/3rd, yet I hear no complaints that your ten counting fingers are malfunctioning. Your calculator gets it right by rounding to the eight or ten significant digits that you see on its little LCD display. In other words, it really doesn't get it right; it just covers up the ugliness. And, in fact, I see a nearly identical question every day in reference to C, C++, PHP, and myriad other programming languages over on StackOverflow. It's not a problem unique to Perl.

        I understand that the "What every computer scientist should know..." article is a little beyond what someone who just wants to get the job done might want to digest. That's fine, the article goes into painful details. Try this response (shameless plug, I wrote it), which tries to spell it out in less technical terms: Re: shocking imprecision.


        Dave

        If you don't like the WhatEvery... article -- and I agree that it is written more to impress than inform; and is way over promoted -- then try this. Short, sweet & clear.


        With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
        Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
        "Science is about questioning the status quo. Questioning authority".
        In the absence of evidence, opinion is indistinguishable from prejudice.

        The start of some sanity?

        If I were doing it with a calculator, I'd surely end up with too many or too few entries of 0.001. I might mess it up building an array. Are you sure you have the right number of elements?

        You can check easily enough:

        say scalar(@arr); # 40 say $#arr; # 39
Re: Is this odd behavior a floating point problem?
by Eliya (Vicar) on Mar 23, 2012 at 17:14 UTC

    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

      :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

        ... 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.

Re: Is this odd behavior a floating point problem?
by roboticus (Canon) on Mar 23, 2012 at 17:25 UTC

    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.

Re: Is this odd behavior a floating point problem?
by toolic (Chancellor) on Mar 23, 2012 at 17:29 UTC
    • sprintf
    • perlfaq4 Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?
Re: Is this odd behavior a floating point problem?
by Khen1950fx (Canon) on Mar 23, 2012 at 19:18 UTC
    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
      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.

Log In?
Username:
Password:

What's my password?
Create A New User
Node Status?
node history
Node Type: perlquestion [id://961262]
Approved by lidden
help
Chatterbox?
and the web crawler heard nothing...

How do I use this? | Other CB clients
Other Users?
Others examining the Monastery: (6)
As of 2014-04-18 10:37 GMT
Sections?
Information?
Find Nodes?
Leftovers?
    Voting Booth?

    April first is:







    Results (466 votes), past polls