random number including 1 - i.e. [0,1]
John Yeung
gallium.arsenide at gmail.com
Wed Jun 10 03:28:23 CEST 2009
On Jun 9, 8:45 pm, Mensanator <mensana... at aol.com> wrote:
> On Jun 9, 6:05 pm, "Gabriel Genellina" <gagsl-... at yahoo.com.ar> wrote:
> > py> a+(b-a)*z < b # the expression used for uniform(a,b)
> > False
> > py> a+(b-a)*z
> > 11.0
>
> What you do with the number after it's created is not
> random's concern.
Mensanator, you missed Gabriel's point. What he's saying is that,
effectively, random.uniform(a, b) returns a + (b - a) * random.random
(). So z may not be random()'s concern, but it very much is uniform
()'s concern.
> > The docs are already updated to reflect this:http://svn.python.org/view/python/trunk/Doc/library/random.rst?r1=687...
>
> The docs are now wrong. Why would they do that?
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
John
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