Weighted random selection with and without replacement

I’d recommend you start by looking at section 3.4.2 of Donald Knuth’s Seminumerical Algorithms.

If your arrays are large, there are more efficient algorithms in chapter 3 of Principles of Random Variate Generation by John Dagpunar. If your arrays are not terribly large or you’re not concerned with squeezing out as much efficiency as possible, the simpler algorithms in Knuth are probably fine.

Here’s what I came up with for weighted selection without replacement:

def WeightedSelectionWithoutReplacement(l, n):
  """Selects without replacement n random elements from a list of (weight, item) tuples."""
  l = sorted((random.random() * x[0], x[1]) for x in l)
  return l[-n:]

This is O(m log m) on the number of items in the list to be selected from. I’m fairly certain this will weight items correctly, though I haven’t verified it in any formal sense.

Here’s what I came up with for weighted selection with replacement:

def WeightedSelectionWithReplacement(l, n):
  """Selects with replacement n random elements from a list of (weight, item) tuples."""
  cuml = []
  total_weight = 0.0
  for weight, item in l:
    total_weight += weight
    cuml.append((total_weight, item))
  return [cuml[bisect.bisect(cuml, random.random()*total_weight)] for x in range(n)]

This is O(m + n log m), where m is the number of items in the input list, and n is the number of items to be selected.

The following is a description of random weighted selection of an element of a
set (or multiset, if repeats are allowed), both with and without replacement in O(n) space
and O(log n) time.

It consists of implementing a binary search tree, sorted by the elements to be
selected, where each node of the tree contains:

1. the element itself (element)
2. the un-normalized weight of the element (elementweight), and
3. the sum of all the un-normalized weights of the left-child node and all of
   its children (leftbranchweight).
4. the sum of all the un-normalized weights of the right-child node and all of
   its chilren (rightbranchweight).

Then we randomly select an element from the BST by descending down the tree. A
rough description of the algorithm follows. The algorithm is given a node of
the tree. Then the values of leftbranchweight, rightbranchweight,
and elementweight of node is summed, and the weights are divided by this
sum, resulting in the values leftbranchprobability,
rightbranchprobability, and elementprobability, respectively. Then a
random number between 0 and 1 (randomnumber) is obtained.

• if the number is less than elementprobability,
  • remove the element from the BST as normal, updating leftbranchweight
    and rightbranchweight of all the necessary nodes, and return the
    element.
• else if the number is less than (elementprobability + leftbranchweight)
  • recurse on leftchild (run the algorithm using leftchild as node)
• else
  • recurse on rightchild

When we finally find, using these weights, which element is to be returned, we either simply return it (with replacement) or we remove it and update relevant weights in the tree (without replacement).

DISCLAIMER: The algorithm is rough, and a treatise on the proper implementation
of a BST is not attempted here; rather, it is hoped that this answer will help
those who really need fast weighted selection without replacement (like I do).

A simple approach that hasn’t been mentioned here is one proposed in Efraimidis and Spirakis. In python you could select m items from n >= m weighted items with strictly positive weights stored in weights, returning the selected indices, with:

import heapq
import math
import random

def WeightedSelectionWithoutReplacement(weights, m):
    elt = [(math.log(random.random()) / weights[i], i) for i in range(len(weights))]
    return [x[1] for x in heapq.nlargest(m, elt)]

This is very similar in structure to the first approach proposed by Nick Johnson. Unfortunately, that approach is biased in selecting the elements (see the comments on the method). Efraimidis and Spirakis proved that their approach is equivalent to random sampling without replacement in the linked paper.

It is possible to do Weighted Random Selection with replacement in O(1) time, after first creating an additional O(N)-sized data structure in O(N) time. The algorithm is based on the Alias Method developed by Walker and Vose, which is well described here.

The essential idea is that each bin in a histogram would be chosen with probability 1/N by a uniform RNG. So we will walk through it, and for any underpopulated bin which would would receive excess hits, assign the excess to an overpopulated bin. For each bin, we store the percentage of hits which belong to it, and the partner bin for the excess. This version tracks small and large bins in place, removing the need for an additional stack. It uses the index of the partner (stored in bucket[1]) as an indicator that they have already been processed.

Here is a minimal python implementation, based on the C implementation here

def prep(weights):
    data_sz = len(weights)
    factor = data_sz/float(sum(weights))
    data = [[w*factor, i] for i,w in enumerate(weights)]
    big=0
    while big<data_sz and data[big][0]<=1.0: big+=1
    for small,bucket in enumerate(data):
        if bucket[1] is not small: continue
        excess = 1.0 - bucket[0]
        while excess > 0:
            if big==data_sz: break
            bucket[1] = big
            bucket = data[big]
            bucket[0] -= excess
            excess = 1.0 - bucket[0]
            if (excess >= 0):
                big+=1
                while big<data_sz and data[big][0]<=1: big+=1
    return data

def sample(data):
    r=random.random()*len(data)
    idx = int(r)
    return data[idx][1] if r-idx > data[idx][0] else idx

Example usage:

TRIALS=1000
weights = [20,1.5,9.8,10,15,10,15.5,10,8,.2];
samples = [0]*len(weights)
data = prep(weights)

for _ in range(int(sum(weights)*TRIALS)):
    samples[sample(data)]+=1

result = [float(s)/TRIALS for s in samples]
err = [a-b for a,b in zip(result,weights)]
print(result)
print([round(e,5) for e in err])
print(sum([e*e for e in err]))

Suppose you want to sample 3 elements without replacement from the list [‘white’,’blue’,’black’,’yellow’,’green’] with a prob. distribution [0.1, 0.2, 0.4, 0.1, 0.2]. Using numpy.random module it is as easy as this:

    import numpy.random as rnd

    sampling_size = 3
    domain = ['white','blue','black','yellow','green']
    probs = [.1, .2, .4, .1, .2]
    sample = rnd.choice(domain, size=sampling_size, replace=False, p=probs)
    # in short: rnd.choice(domain, sampling_size, False, probs)
    print(sample)
    # Possible output: ['white' 'black' 'blue']

Setting the replace flag to True, you have a sampling with replacement.

More info here:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.random.choice.html#numpy.random.choice

We faced a problem to randomly select K validators of N candidates once per epoch proportionally to their stakes. But this gives us the following problem:

Imagine probabilities of each candidate:

0.1
0.1
0.8

Probabilities of each candidate after 1’000’000 selections 2 of 3 without replacement became:

0.254315
0.256755
0.488930

You should know, those original probabilities are not achievable for 2 of 3 selection without replacement.

But we wish initial probabilities to be a profit distribution probabilities. Else it makes small candidate pools more profitable. So we realized that random selection with replacement would help us – to randomly select >K of N and store also weight of each validator for reward distribution:

std::vector<int> validators;
std::vector<int> weights(n);
int totalWeights = 0;

for (int j = 0; validators.size() < m; j++) {
    int value = rand() % likehoodsSum;
    for (int i = 0; i < n; i++) {
        if (value < likehoods[i]) {
            if (weights[i] == 0) {
                validators.push_back(i);
            }
            weights[i]++;
            totalWeights++;
            break;
        }

        value -= likehoods[i];
    }
}

It gives an almost original distribution of rewards on millions of samples:

0.101230
0.099113
0.799657

This is an old question for which numpy now offers an easy solution so I thought I would mention it. Current version of numpy is version 1.2 and numpy.random.choice allows the sampling to be done with or without replacement and with given weights.