In Python with numpy, for a given natural number x, we'd like to get an array (say, 20×2 or whichever shape you wish) of uniformly distributed floats in the interval [0,x]. Notice that the interval is closed.
Based on http://stackoverflow.com/a/33359758, a completely screwed approach is
x=10
print(np.reshape([int.from_bytes(os.urandom(8))*x/((1<<64)-1) for i in range(40)],(20,2)))
However, the results depend on the operating system. I also don't see whether/why the obtained numbers could be uniformly distributed.
Notice that a (different) approach of somehow obtaining uniform IEEE 754 floats from [0,1] and then multiplying them by x is potentially flawed because, unless x is a power of 2, your numerators (in terms of the representation as numerator * 2^{…}) will always be multiples of x, whereas all non-multiples of x would never show up.
Any help? Pseudo-random numbers would also do (assuming that they are uniformly distributed).
You can also assume that x is small, say, x<2^8, if it helps.
You could adjust the endpoint to the next floating point number after the max:
import numpy as np
def uniform_with_endpoint(low=0, hi=1, size=None):
return np.random.uniform(low, np.nextafter(hi, np.inf), size=size)
This gives a uniform distribution in which hi is included in the possible results.