Binary random array with a specific proportion of ones?

A simple way to do this would be to first generate an ndarray with the proportion of zeros and ones you want:

>>> import numpy as np
>>> N = 100
>>> K = 30 # K zeros, N-K ones
>>> arr = np.array([0] * K + [1] * (N-K))
>>> arr
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1])

Then you can just shuffle the array, making the distribution random:

>>> np.random.shuffle(arr)
>>> arr
array([1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0,
       1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1,
       1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1,
       0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1,
       1, 1, 1, 0, 1, 1, 1, 1])

Note that this approach will give you the exact proportion of zeros/ones you request, unlike say the binomial approach. If you don't need the exact proportion, then the binomial approach will work just fine.

Yet another approach, using np.random.choice:

>>> np.random.choice([0, 1], size=(10,), p=[1./3, 2./3])
array([0, 1, 1, 1, 1, 0, 0, 0, 0, 0])