Maybe we could make a C implementation of the Fraction module? That would be nice. On Sun, May 31, 2015 at 8:28 PM, <python-ideas-request@python.org> wrote:

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Today's Topics:

1. Re: Python Float Update (Chris Angelico) 2. Re: Python Float Update (random832@fastmail.us) 3. Re: Python Float Update (Jim Witschey) 4. Re: Python Float Update (David Mertz)

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Message: 1 Date: Mon, 1 Jun 2015 12:48:12 +1000 From: Chris Angelico <rosuav@gmail.com> Cc: python-ideas <python-ideas@python.org> Subject: Re: [Python-ideas] Python Float Update Message-ID: <CAPTjJmr=LurRUoKH3KVVYpMFc= W5h6etG5TscV5uU6zWhxVbgQ@mail.gmail.com> Content-Type: text/plain; charset=UTF-8

On Mon, Jun 1, 2015 at 12:25 PM, u8y7541 The Awesome Person <surya.subbarao1@gmail.com> wrote:

...

I will be presenting a modification to the float class, which will

improve its speed and accuracy (reduce floating point errors). This is applicable because Python uses a numerator and denominator rather than a sign and mantissa to represent floats.

...

First, I propose that a float's integer ratio should be accurate. For

example, (1 / 3).as_integer_ratio() should return (1, 3). Instead, it returns(6004799503160661, 18014398509481984).

...

I think you're misunderstanding the as_integer_ratio method. That isn't how Python works internally; that's a service provided for parsing out float internals into something more readable. What you _actually_ are working with is IEEE 754 binary64. (Caveat: I have no idea what Python-the-language stipulates, nor what other Python implementations use, but that's what CPython uses, and you did your initial experiments with CPython. None of this discussion applies *at all* if a Python implementation doesn't use IEEE 754.) So internally, 1/3 is stored as:

0 <-- sign bit (positive) 01111111101 <-- exponent (1021) 0101010101010101010101010101010101010101010101010101 <-- mantissa (52 bits, repeating)

The exponent is offset by 1023, so this means 1.010101.... divided by 2?; the original repeating value is exactly equal to 4/3, so this is correct, but as soon as it's squeezed into a finite-sized mantissa, it gets rounded - in this case, rounded down.

That's where your result comes from. It's been rounded such that it fits inside IEEE 754, and then converted back to a fraction afterwards. You're never going to get an exact result for anything with a denominator that isn't a power of two. Fortunately, Python does offer a solution: store your number as a pair of integers, rather than as a packed floating point value, and all calculations truly will be exact (at the cost of performance):

...

...

...

one_third = fractions.Fraction(1, 3) one_eighth = fractions.Fraction(1, 8) one_third + one_eighth Fraction(11, 24)

This is possibly more what you want to work with.

ChrisA

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Message: 2 Date: Sun, 31 May 2015 23:14:06 -0400 From: random832@fastmail.us To: python-ideas@python.org Subject: Re: [Python-ideas] Python Float Update Message-ID: <1433128446.31560.283106753.6D60F98F@webmail.messagingengine.com> Content-Type: text/plain

On Sun, May 31, 2015, at 22:25, u8y7541 The Awesome Person wrote:

...

First, I propose that a float's integer ratio should be accurate. For example, (1 / 3).as_integer_ratio() should return (1, 3). Instead, it returns(6004799503160661, 18014398509481984).

Even though he's mistaken about the core premise, I do think there's a kernel of a good idea here - it would be nice to have a method (maybe as_integer_ratio, maybe with some parameter added, maybe a different method) to return with the smallest denominator that would result in exactly the original float if divided out, rather than merely the smallest power of two.

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Message: 3 Date: Mon, 01 Jun 2015 03:21:36 +0000 From: Jim Witschey <jim.witschey@gmail.com> To: Chris Angelico <rosuav@gmail.com> Cc: python-ideas <python-ideas@python.org> Subject: Re: [Python-ideas] Python Float Update Message-ID: <CAF+a8-q6kbOwcWk3F47+9PXf2vKgM9ao1uh5=qBw10jqzTC= kg@mail.gmail.com> Content-Type: text/plain; charset="utf-8"

Teachable moments about the implementation of floating-point aside, something in this neighborhood has been considered and rejected before, in PEP 240. However, that was in 2001 - it was apparently created the same day as PEP 237, which introduced transparent conversion of machine ints to bignums in the int type.

I think hiding hardware number implementations has been a success for integers - it's a far superior API. It could be for rationals as well.

Has something like this thread's original proposal - interpeting decimal-number literals as fractional values and using fractions as the result of integer arithmetic - been seriously discussed more recently than PEP 240? If so, why haven't they been implemented? Perhaps enough has changed that it's worth reconsidering.

On Sun, May 31, 2015 at 22:49 Chris Angelico <rosuav@gmail.com> wrote:

...

On Mon, Jun 1, 2015 at 12:25 PM, u8y7541 The Awesome Person <surya.subbarao1@gmail.com> wrote:

...

I will be presenting a modification to the float class, which will

improve its speed and accuracy (reduce floating point errors). This is applicable because Python uses a numerator and denominator rather than a sign and mantissa to represent floats.

...

First, I propose that a float's integer ratio should be accurate. For

example, (1 / 3).as_integer_ratio() should return (1, 3). Instead, it returns(6004799503160661, 18014398509481984).

...

I think you're misunderstanding the as_integer_ratio method. That isn't how Python works internally; that's a service provided for parsing out float internals into something more readable. What you _actually_ are working with is IEEE 754 binary64. (Caveat: I have no idea what Python-the-language stipulates, nor what other Python implementations use, but that's what CPython uses, and you did your initial experiments with CPython. None of this discussion applies *at all* if a Python implementation doesn't use IEEE 754.) So internally, 1/3 is stored as:

0 <-- sign bit (positive) 01111111101 <-- exponent (1021) 0101010101010101010101010101010101010101010101010101 <-- mantissa (52 bits, repeating)

The exponent is offset by 1023, so this means 1.010101.... divided by 2?; the original repeating value is exactly equal to 4/3, so this is correct, but as soon as it's squeezed into a finite-sized mantissa, it gets rounded - in this case, rounded down.

That's where your result comes from. It's been rounded such that it fits inside IEEE 754, and then converted back to a fraction afterwards. You're never going to get an exact result for anything with a denominator that isn't a power of two. Fortunately, Python does offer a solution: store your number as a pair of integers, rather than as a packed floating point value, and all calculations truly will be exact (at the cost of performance):

...

...

...

one_third = fractions.Fraction(1, 3) one_eighth = fractions.Fraction(1, 8) one_third + one_eighth Fraction(11, 24)

This is possibly more what you want to work with.

ChrisA _______________________________________________ Python-ideas mailing list Python-ideas@python.org https://mail.python.org/mailman/listinfo/python-ideas Code of Conduct: http://python.org/psf/codeofconduct/