Numpy sampling from a 2d numpy array of probabilities

Question

I have a 2d array of probabilities, p that sum to 1.0:

p = np.array([[0.05, 0.05, 0.1 , 0.  ],
              [0.4 , 0.1 , 0.1 , 0.2 ]])

I want to have a function that samples an index from the 2d array p based on the probabilities. For example, I should return [1,0] with a probability of 0.4 or [0,2] with a probability of 0.1.

If I call the function sample(p) which takes the 2d probability array p. Then with the 2d p array as I've defined above I could expect indices anywhere from [0,0] to [1,3] but not including [0,3] as that has a probability of 0 of appearing:

sample(p)
>> [1,0]
sample(p)
>> [0,2]
sample(p)
>> [1,0]
sample(p)
>> [1,3]

Normally I would use numpy.random.choice however it only accepts 1D array probabilities. Is there a way that I can extend this to work for 2d probability arrays of any size .

Would you please elaborate on the example? I am not really following what is happening there. — norok2
– norok2, Commented Apr 5, 2020 at 19:04

norok2 · Accepted Answer · 2020-04-05 19:32:26Z

2

You could use np.unravel_index(), together with np.random.choice() on the result of np.arange() with the .ravel() (or .flatten()) probability.

For example:

p = np.array([[0.05, 0.05, 0.1 , 0.  ],
              [0.4 , 0.1 , 0.1 , 0.2 ]])


def sample_index(p):
    i = np.random.choice(np.arange(p.size), p=p.ravel())
    return np.unravel_index(i, p.shape)

np.random.seed(0)
x = [sample_index(p) for _ in range(10)]
print(x)
# [(1, 0), (1, 2), (1, 1), (1, 0), (1, 0), (1, 1), (1, 0), (1, 3), (1, 3), (1, 0)]

answered Apr 5, 2020 at 19:32

norok2

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Comments

Franz Forstmayr · Accepted Answer · 2020-04-05 19:18:45Z

1

Just make your array flat first :)

p = np.array([[0.05, 0.05, 0.1 , 0.  ],
          [0.4 , 0.1 , 0.1 , 0.2 ]])
r = np.random.choice(np.arange(8), p=p.ravel())

Ravel returns a view of your array in a one dimensional form.

p.ravel().shape
> (8,)

Just checking if its working properly:

r = np.random.choice(np.arange(8), size=10000, p=p.ravel())
np.unique(r, return_counts=True)
> (array([0, 1, 2, 4, 5, 6, 7]), array([ 530,  523,  968, 4014, 1012,  944, 2009]))

edited Apr 5, 2020 at 19:18

answered Apr 5, 2020 at 19:12

Franz Forstmayr

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