On Jun 9, 4:33 pm, Esmail <ebo...@hotmail.com> wrote: > Hi, > > random.random() will generate a random value in the range [0, 1). > > Is there an easy way to generate random values in the range [0, 1]? > I.e., including 1? >
Are you trying to generate a number in the range [0,n] by multiplying a random function that returns [0,1] * n? If so, then you want to do this using: int(random.random()*(n+1)) This will give equal chance of getting any number from 0 to n. If you had a function that returned a random in the range [0,1], then multiplying by n and then truncating would give only the barest sliver of a chance of giving the value n. You could try rounding, but then you get this skew: 0 for values [0, 0.5) (width of 0.5) 1 for value [0.5, 1.5) (width of 1) ... n for value [n-0.5, n] (width of ~0.50000000000000001) Still not a uniform die roll. You have only about 1/2 the probability of getting 0 or n as any other value. If you want to perform a fair roll of a 6-sided die, you would start with int(random.random() * 6). This gives a random number in the range [0,5], with each value of the same probability. How to get our die roll that goes from 1 to 6? Add 1. Thus: die_roll = lambda : int(random.random() * 6) + 1 Or for a n-sided die: die_roll = lambda n : int(random.random() * n) + 1 This is just guessing on my part, but otherwise, I don't know why you would care if random.random() returned values in the range [0,1) or [0,1]. -- Paul -- http://mail.python.org/mailman/listinfo/python-list
