random number including 1 - i.e. [0,1]
Steven D'Aprano
steven at REMOVE.THIS.cybersource.com.au
Wed Jun 10 01:52:01 EDT 2009
On Tue, 09 Jun 2009 22:21:26 -0700, John Yeung wrote:
> On Jun 9, 11:24 pm, Steven D'Aprano
> <ste... at REMOVE.THIS.cybersource.com.au> wrote:
>> On Tue, 09 Jun 2009 18:28:23 -0700, John Yeung wrote:
>> > The docs are now... sort of correct. For some values of a and b,
>> > uniform() can never return b. Notably, I believe uniform(0, 1) is
>> > equivalent to random(), and will never return 1. However, uniform(1,
>> > 2) CAN return 2, if this is any indication:
>>
>> >>>> a=0.0
>> >>>> b=1.0
>> >>>> a+(b-a)*z < b
>> > True
>> >>>> a=1.0
>> >>>> b=2.0
>> >>>> a+(b-a)*z < b
>> > False
>>
>> But you haven't shown what value z has, so there's no way of
>> interpreting that example.
>
> I'm pretty aggressive about snipping. I left off the quote of z from
> Gabriel. He chose z to be the largest value that random.random() can
> return; namely, the largest float smaller than 1. I've just carried
> over that value into my example.
>
> The point of my example is, with z < 1, uniform(0, 1) is always less
> than 1, but with z < 1, uniform(1, 2) can be 2, according to Gabriel's
> description of uniform().
Ah, that explains it.
And you're right:
>>> import random
>>> class MyRandom(random.Random):
... def random(self):
... return 1.0 - 2**-53
...
>>> r = MyRandom()
>>> r.uniform(1, 2) < 2
False
However:
>>> r.uniform(1, 10) < 10
True
> Therefore, to me the most up-to-date docs (which say that uniform(a, b)
> returns a float in the closed interval [a, b]) is closer to correct than
> before, but still fails to point out the full subtlety of the behavior.
Which is?
--
Steven
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