On Tue, 09 Jun 2009 22:21:26 -0700, John Yeung wrote: > On Jun 9, 11:24 pm, Steven D'Aprano > <ste...@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 -- http://mail.python.org/mailman/listinfo/python-list
