[SOLVED] Generate random integer without an upper bound

Generate random integer without an upper bound

I want to generate a random seed in a predictable way.

I was hoping to do this

seed = 12345
prng_0 = random.Random(seed)
prng_1 = random.Random(prng_0.rand_int(0))

There, 0 is the lower bound, but it turns out I need to give it an upper bound as well. I don't want to set a fixed upper bound.

I'm doing this because I need reproducibility when testing. Namely, this is a function receiving a seed and building its prng, prng_0, then calling multiple times another function that needs to receive a different seed every time.

def funct_a(seed=None):
    prng_1 = random.Random(seed)
    prng_2 = numpy.random.RandomState(prng_1.randint(0, 4294967296))
    print(prng_1.random())
    print(prng_2.random())

def funct_b(seed=None):
    prng_0 = random.Random(seed)
    for i in range(0, 5):
        seed = prng_0.randint(0)  # not working, needs upper bound
        funct_a(seed)

funct_b(12345)  # test call

Interestingly enough, numpy has a definite upper seed value, as testified by the doc and by this error

ValueError: Seed must be between 0 and 4294967295

Solution

When I don't want an upper bound I'll often use sys.maxint for the upper bound as an approximation