Choose list variable given probability of each variable

Question

I\'ve been trying to code a program that uses the softmax activation function in the middle.

Right now, I have a list of probabilities like this:

P[0         


        

Answer 1

You can easily achieve this with numpy. It has a choice function which accepts the parameter of probabilities.

np.random.choice(
  ['pooh', 'rabbit', 'piglet', 'Christopher'], 
  5,
  p=[0.5, 0.1, 0.1, 0.3]
)

Answer 2

This problem is equivalent to sampling from a categorical distribution. This distribution is commonly conflated with the multinomial distribution which models the result of multiple samples from a categorical distribution.

In numpy, it is easy to sample from the multinomial distribution using numpy.random.multinomial, but a specific categorical version of this does not exist. However, it can be accomplished by sampling from the multinomial distribution with a single trial and then returning the non-zero element in the output.

import numpy as np
pvals = [0.10,0.25,0.60,0.05]
ind = np.where(np.random.multinomial(1,pvals))[0][0]

Answer 3

import random

probs = [0.1, 0.25, 0.6, 0.05]
r = random.random()
index = 0
while(r >= 0 and index < len(probs)):
  r -= probs[index]
  index += 1
print index - 1

Answer 4

Hmm interesting, how about...

1. Generate a number between 0 and 1.

2. Walk the list substracting the probability of each item from your number.

3. Pick the item that, after substraction, took your number down to 0 or below.

That's simple, O(n) and should work :)

Answer 5

Basically, make a cumulative probability distribution (CDF) array. Basically, the value of the CDF for a given index is equal to the sum of all values in P equal to or less than that index. Then you generate a random number between 0 and 1 and do a binary search (or linear search if you want). Here's some simple code for it.

from bisect import bisect
from random import random

P = [0.10,0.25,0.60,0.05]

cdf = [P[0]]
for i in xrange(1, len(P)):
    cdf.append(cdf[-1] + P[i])

random_ind = bisect(cdf,random())

of course you can generate a bunch of random indices with something like

rs = [bisect(cdf, random()) for i in xrange(20)]

yielding

[2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2]

(results will, and should vary). Of course, binary search is rather unnecessary for so few of possible indices, but definitely recommended for distributions with more possible indices.