On Sun, 03 Jul 2005 10:56:42 +0100, Tom Anderson wrote:
> On Sun, 3 Jul 2005, Steven D'Aprano wrote:
>
>> On Sun, 03 Jul 2005 02:22:23 +0200, Fredrik Johansson wrote:
>>
>>> On 7/3/05, Tom Anderson <[EMAIL PROTECTED]> wrote:
>>>> That's one way. I'd do:
>>>>
>>>> root = value ** 0.5
>>>>
>>>> Does that mean we can expect Guido to drop math.sqrt in py3k? :)
>>>
>>> I'd rather like to see a well implemented math.nthroot. 64**(1/3.0)
>>> gives 3.9999999999999996, and this error could be avoided.
>>
>> py> math.exp(math.log(64)/3.0)
>> 4.0
>>
>> Success!!!
>
> Eeeeeeenteresting. I have no idea why this works. Given that math.log is
> always going to be approximate for numbers which aren't rational powers of
> e (which, since e is transcendental, is all rational numbers, and
> therefore all python floats, isn't it?), i'd expect to get the same
> roundoff errors here as with exponentiation. Is it just that the errors
> are sufficiently smaller that it looks exact?
I have no idea :-)
Unfortunately, floating point maths is a bit of a black art. Try this
simple calculation:
py> 4.0/3 - 5.0/6
0.49999999999999989
Should be 0.5 exactly.
Many numbers which you can write exactly in decimal cannot be
stored exactly in floating point. Even something as simple as one tenth
0.1 doesn't have an exact representation in binary:
py> 1.0/10
0.10000000000000001
>> Note how much simpler this would be if we could guarantee proper
>> infinities and NaNs in the code. We could turn a 23-line block to a
>> one-liner.
>
> YES! This is something that winds me up no end; as far as i can tell,
> there is no clean programmatic way to make an inf or a NaN; in code i
> write which cares about such things, i have to start:
>
> inf = 1e300 ** 1e300
> nan = inf - inf
>
> Every bloody time. I'm going to be buggered if python ever rolls out some
> sort of bigfloat support.
It fails for me:
py> inf = 1e300 ** 1e300
Traceback (most recent call last):
File "<stdin>", line 1, in ?
OverflowError: (34, 'Numerical result out of range')
But this works:
py> inf = float("inf")
py> inf
inf
> And then god forbid i should actually want to test if a number is NaN,
> since, bizarrely, (x == nan) is true for every x;
Well dip me in chocolate and call me a Tim Tam, you're right. That is
bizarre. No, that's not the word I want... that behaviour is broken.
> instead, i have to
> write:
>
> def isnan(x):
> return (x == 0.0) and (x == 1.0)
>
> The IEEE spec actually says that (x == nan) should be *false* for every x,
> including nan. I'm not sure if this is more or less stupid than what
> python does!
Well, no, the IEEE standard is correct. NaNs aren't equal to anything,
including themselves, since they aren't numbers.
Okay, infinity isn't a number either. But there are rules for arithmetic
for infinity, based on limit calculations, so it makes sense to treat
infinity as a number where possible.
Apple Computer had a maths library that implemented a superset of IEEE
arithmetic. It allowed for 254 different NaNs (twice that if you looked at
the sign bit, which you weren't supposed to do). This wasn't a
deliberate feature as such, it merely fell out naturally from the
bit-patterns that represented NaNs, but it was useful since Apple took
advantage of it by specifying certain particular NaNs for certain failure
modes. Eg NaN(1) might represent the square root of a negative number, but
NaN(2) might be INF-INF.
Furthermore, you could easily query the class of a number (zero,
normalised, denormalised, infinity, NaN), so you would test for a NaN by
doing something like:
if Kind(x) == NanKind: ...
or
if isNaN(x): ...
> And while i'm ranting, how come these expressions aren't the same:
>
> 1e300 * 1e300
> 1e300 ** 2
Because this is floating point, not real maths :-)
I get inf and Overflow respectively. What do you get?
> And finally, does Guido know something about arithmetic that i don't, or
> is this expression:
>
> -1.0 ** 0.5
>
> Evaluated wrongly?
No, it is evaluated according to the rules of precedence. It is equivalent
to -(1.0**0.5). You are confusing it with (-1.0)**0.5 which fails as
expected.
--
Steven.
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