Select a non uniformely distributed random element from an array

Question

I'm trying to pick numbers from an array at random.

I can easily do this by picking an element using np.random.randint(len(myArray)) - but that gives a uniform distribution.

For my needs I need to pick a random number with higher probability of picking a number near the beginning of the array - so I think something like an exponential probability function would suit better.

Is there a way for me to generate a random integer in a range (1, 1000) using an exponential (or other, non-uniform distribution) to use as an array index?

See stackoverflow.com/questions/40244602/… for starters.

Selcuk
– Selcuk

2019-03-15 00:22:15 +00:00
Commented Mar 15, 2019 at 0:22 — Selcuk
– Selcuk, Commented Mar 15, 2019 at 0:22

Sheldore · Accepted Answer · 2019-03-15 00:37:15Z

3

You can assign an exponential probability vector to choice module from NumPy. The probability vector should add up to 1 therefore you normalize it by the sum of all probabilities.

import numpy as np
from numpy.random import choice

arr = np.arange(0, 1001)
prob = np.exp(arr/1000) # To avoid a large number
rand_draw = choice(arr, 1, p=prob/sum(prob))

To make sure the random distribution follows exponential behavior, you can plot it for 100000 random draws between 0 and 1000.

import matplotlib.pyplot as plt

# above code here

rand_draw = choice(arr, 100000, p=prob/sum(prob))

plt.hist(rand_draw, bins=100)
plt.show()

edited Mar 15, 2019 at 0:37

answered Mar 15, 2019 at 0:24

Sheldore

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1 Comment

Fırat Kıyak Over a year ago

However, if the number of choices is small then numpy.random.choice is significantly slower than a basic implementation of the function. See: stackoverflow.com/q/18622781/6087087

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