How to change p-value of Numpy.random.choice based on position in array? - numpy

Question

I have a 3D array of size NxNxN. I would like to fill this array with random booleans, which I can do with:
a = np.random.choice([False,True],size=(N,N,N))
However, I would like the likelihood (or p-value) of choosing either True or False to be based on the element's position in the array. I thought maybe I could do this with the p-value parameter, but that only then works for selecting how often True/False is chosen for the entire array.
Is there any way to set specific p-values for the entire (N,N,N) array? I guess that would amount to an (N,N,N,2) array then, with the extra 2 being for the p-value for False and p-value for True (though p_True = 1 - p_False). I feel like there's a simpler way to do this that I'm not thinking of.
Edit:
So say I want to create a simple array, a, of shape (1,2) (just two elements, but multidimensional on purpose). I want to fill these two elements with True/False. I have another array filled with the likelihood or p-value with which I want those elements to be False, say p_False, where p_False.shape = (1,2). Let's say I want the first element to have a 25% chance of being False, but the second element to have a 50% chance of being false, so then p_False = np.array([0.25,0.5]).
I tried something along the lines of:
a = np.random.choice([[False,True],[False,True]],p=[[.25,.75],[.5,.5]])
but I got a ValueError: a must be 1-dimensional.

Answer 1

To generate an array with different probabilities, you can use the following code:
# define an initial value of N
N = 512
# generate an array of probabilities. You can eventually build your own, since the size is respected
prob_array = np.array((range(0,N*N*N)))
# rescale the probabilities between 0 and 1
prob_array = (prob_array - np.min(prob_array)) / (np.max(prob_array) - np.min(prob_array))
# generate the random based on the probabilities, cast to booleans and reshape
np.reshape(np.array(np.random.binomial(1, p=prob_array, size=N*N*N), dtype=bool), (N,N,N))
This generates an array with lots of Falses in the beginning and lots of Trues in the end:
array([[[False, False, False, ..., False, False, False],
[False, False, False, ..., False, False, False],
[False, False, False, ..., False, False, False],
...,
[False, False, False, ..., False, False, False],
[False, False, False, ..., False, False, False],
[False, False, False, ..., False, False, False]],
...,
[[ True, True, True, ..., True, True, True],
[ True, True, True, ..., True, True, True],
[ True, True, True, ..., True, True, True],
...,
[ True, True, True, ..., True, True, True],
[ True, True, True, ..., True, True, True],
[ True, True, True, ..., True, True, True]]])

Answer 2

Use the binomial method with an array of numbers in [0, 1]. Here is an example, which sets each element to 0 or 1 depending on a randomly chosen probability:
import numpy
gen=numpy.random.Generator(numpy.random.PCG64())
ret=gen.binomial(1, gen.uniform(size=(3, 3, 3)))
If you want each item to be True or False rather than 0 or 1, I'm afraid I don't know how to do so.
Note that numpy.random.Generator was introduced in NumPy 1.7. You are recommended to use the latest version of NumPy; in the meantime, you can use the following:
import numpy
ret=numpy.random.binomial(1, numpy.random.uniform(size=(3, 3, 3)))