Drawing from a discrete distribution is directly built into numpy. The function is called random.choice (difficult to find without any reference to discrete distributions in the numpy docs).
elements = [1.1, 2.2, 3.3]
probabilities = [0.2, 0.5, 0.3]
np.random.choice(elements, 10, p=probabilities)
Here is a short, relatively simple function that returns weighted values, it uses NumPy's digitize
, accumulate
, and random_sample
.
import numpy as np
from numpy.random import random_sample
def weighted_values(values, probabilities, size):
bins = np.add.accumulate(probabilities)
return values[np.digitize(random_sample(size), bins)]
values = np.array([1.1, 2.2, 3.3])
probabilities = np.array([0.2, 0.5, 0.3])
print weighted_values(values, probabilities, 10)
#Sample output:
[ 2.2 2.2 1.1 2.2 2.2 3.3 3.3 2.2 3.3 3.3]
It works like this:
accumulate
we create bins.0
, and 1
) using random_sample
digitize
to see which bins these numbers fall into.You were going in a good direction: the built-in scipy.stats.rv_discrete()
quite directly creates a discrete random variable. Here is how it works:
>>> from scipy.stats import rv_discrete
>>> values = numpy.array([1.1, 2.2, 3.3])
>>> probabilities = [0.2, 0.5, 0.3]
>>> distrib = rv_discrete(values=(range(len(values)), probabilities)) # This defines a Scipy probability distribution
>>> distrib.rvs(size=10) # 10 samples from range(len(values))
array([1, 2, 0, 2, 2, 0, 2, 1, 0, 2])
>>> values[_] # Conversion to specific discrete values (the fact that values is a NumPy array is used for the indexing)
[2.2, 3.3, 1.1, 3.3, 3.3, 1.1, 3.3, 2.2, 1.1, 3.3]
The distribution distrib
above thus returns indexes from the values
list.
More generally, rv_discrete()
takes a sequence of integer values in the first elements of its values=(…,…)
argument, and returns these values, in this case; there is no need to convert to specific (float) values. Here is an example:
>>> values = [10, 20, 30]
>>> probabilities = [0.2, 0.5, 0.3]
>>> distrib = rv_discrete(values=(values, probabilities))
>>> distrib.rvs(size=10)
array([20, 20, 20, 20, 20, 20, 20, 30, 20, 20])
where (integer) input values are directly returned with the desired probability.
The simplest DIY way would be to sum up the probabilities into a cumulative distribution. This way, you split the unit interval into sub-intervals of the length equal to your original probabilities. Now generate a single random number uniform on [0,1), and and see to which interval it lands.
You could also use Lea, a pure Python package dedicated to discrete probability distributions.
>>> distrib = Lea.fromValFreqs((1.1,2),(2.2,5),(3.3,3))
>>> distrib
1.1 : 2/10
2.2 : 5/10
3.3 : 3/10
>>> distrib.random(10)
(2.2, 2.2, 1.1, 2.2, 2.2, 2.2, 1.1, 3.3, 1.1, 3.3)
Et voilà!
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