Binary random array with a specific proportion of ones?

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])

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))>>> arrarray([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)>>> arrarray([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.

If I understand your problem correctly, you might get some help with numpy.random.shuffle

>>> def rand_bin_array(K, N):    arr = np.zeros(N)    arr[:K]  = 1    np.random.shuffle(arr)    return arr>>> rand_bin_array(5,15)array([ 0.,  1.,  0.,  1.,  1.,  1.,  0.,  0.,  0.,  1.,  0.,  0.,  0.,        0.,  0.])