On Jun 9, 8:45 pm, Mensanator <mensana...@aol.com> wrote: > On Jun 9, 6:05 pm, "Gabriel Genellina" <gagsl-...@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 -- http://mail.python.org/mailman/listinfo/python-list
