ruby drops a small log

bigearsbilly · 02-03-2015, 02:10 AM

I was looking for the log (base 10 ) of 1000

Code:

ruby:
Math::log(1000, 10)
=> 2.9999999999999996

 Math::log(1000)/Math::log(10)
=> 2.9999999999999996

that's a bit rubbish isn't it?

perl

Code:

$ perl -e 'print log(1000)/(log(10))'  
3

lisp

Code:

(log 1000 10)
3

even C

Code:

int main(int argc, char ** argv)
{
    for( ++argv; *argv; ++argv) {

        float x = atof(*argv);
        printf("log10(%f) = %f", x, log10(x));

    }
 return 0;
}
$ ./log 1000
log10(1000.000000) = 3.000000

even a schoolboy should get that right

NevemTeve · 02-03-2015, 03:22 AM

Floating point calculations are never absolutely precise. Please learn this by heart.

Also read this: https://en.wikipedia.org/wiki/Floati...uracy_problems

bigearsbilly · 02-03-2015, 03:26 AM

I am well aware of floating points.

It is nevertheless, still, a bit rubbish.

NevemTeve · 02-03-2015, 03:44 AM

I'm afraid you don't have enough experience in this question, that's why you have exaggerated expectations.

bigearsbilly · 02-03-2015, 03:57 AM

No experience indeed?

I expect the log base 10 of 1000 to be 3

Call me old-fashioned!

bigearsbilly · 02-03-2015, 04:02 AM

even the MS windows calculator gets it right!

grail · 02-03-2015, 07:54 AM

Like many languages ... it is about learning to use the correct tool:

Code:

$ irb
irb(main):001:0> Math.log10(1000)
=> 3.0

There is also log2. The standard log function is to base e, hence when the additional parameter is supplied there is a division done involving e and the new base.
Hence the rounded figure presented in your example.

NevemTeve · 02-03-2015, 08:14 AM

That was a very useful piece of information, even if unrelated to the topic.

grail · 02-03-2015, 09:16 AM

Ok ... so I am lost ... how does using the right function to get the answer be unrelated?

NevemTeve · 02-03-2015, 09:20 AM

Methinks the answer the OP wanted to hear would have been this: 'yes, you're right, Math::log in ruby is lame'

Edit: an example written in C (tested on x86):

Code:

/* acc.c */

#include <stdio.h>

int main (void)
{
    double f;

    f= 0.1;
    printf ("%g\n", 10*f - 1);
    return 0;
}

What result should I expect?

sundialsvcs · 02-03-2015, 12:47 PM

Well, "lame" would be a pretty good word for it, but still it is par for the course. The difference between the implementations is simply the amount of implied rounding that the formatting library built in to the language decided to use. The internal representation of a float-binary number is never exactly as it appears when printed in base-10. It's mildly surprising to see the difference, as well as the obvious kludge in the case of .log10(), but mathematically it's not wrong.

bigearsbilly · 02-04-2015, 01:27 AM

OK well done Grail, that's the answer then.
phew!

In case you were wondering, I was thinking of using log10 for formatting floating points for FX data. I want a 5 digit display.

Truncated this should give me the significand length -1.

log10 (1000.0) = 3.

So I need 1 digit after the point.
I need to format as text anyway I can easily do it the unsubtle way but I thought I'd try it as it's a bit more aesthetically pleasing.