C++: Sampling from discrete distribution without replacement - c++

Question

I'd like to sample from a discrete distribution without replacement (i.e., without repetition).
With the function discrete_distribution, it is possible to sample with replacement. And, with this function, I implemented sampling without replacement in a very rough way:
#include <iostream>
#include <random>
#include <vector>
#include <array>
int main()
{
const int sampleSize = 8; // Size of the sample
std::vector<double> weights = {2,2,1,1,2,2,1,1,2,2}; // 10 possible outcome with different weights
std::random_device rd;
std::mt19937 generator(rd());
/// WITH REPLACEMENT
std::discrete_distribution<int> distribution(weights.begin(), weights.end());
std::array<int, 10> p ={};
for(int i=0; i<sampleSize; ++i){
int number = distribution(generator);
++p[number];
}
std::cout << "Discrete_distribution with replacement:" << std::endl;
for (int i=0; i<10; ++i)
std::cout << i << ": " << std::string(p[i],'*') << std::endl;
/// WITHOUT REPLACEMENT
p = {};
for(int i=0; i<sampleSize; ++i){
std::discrete_distribution<int> distribution(weights.begin(), weights.end());
int number = distribution(generator);
weights[number] = 0; // the weight associate to the sampled value is set to 0
++p[number];
}
std::cout << "Discrete_distribution without replacement:" << std::endl;
for (int i=0; i<10; ++i)
std::cout << i << ": " << std::string(p[i],'*') << std::endl;
return 0;
}
Have you ever coded such sampling without replacement? Probably in a more optimized way?
Thank you.
Cheers,
T.A.

Answer 1

This solution might be a bit shorter. Unfortunately, it needs to create a discrete_distribution<> object in every step, which might be prohibitive when drawing a lot of samples.
#include <iostream>
#include <boost/random/discrete_distribution.hpp>
#include <boost/random/mersenne_twister.hpp>
using namespace boost::random;
int main(int, char**) {
std::vector<double> w = { 2, 2, 1, 1, 2, 2, 1, 1, 2, 2 };
discrete_distribution<> dist(w);
int n = 10;
boost::random::mt19937 gen;
std::vector<int> samples;
for (auto i = 0; i < n; i++) {
samples.push_back(dist(gen));
w[*samples.rbegin()] = 0;
dist = discrete_distribution<>(w);
}
for (auto iter : samples) {
std::cout << iter << " ";
}
return 0;
}
Improved answer:
After carefully looking for a similar question on this site (Faster weighted sampling without replacement), I found a stunningly simple algorithm for weighted sampling without replacement, it is just a bit complicated to implement in C++. Note, that this is not the most efficient algorithm, but it seems to me the simplest one to implement.
In https://doi.org/10.1016/j.ipl.2005.11.003 the method is described in detail.
Especially, it is not efficient if the sample size is much smaller than the basic population.
#include <iostream>
#include <iterator>
#include <boost/random/uniform_01.hpp>
#include <boost/random/mersenne_twister.hpp>
using namespace boost::random;
int main(int, char**) {
std::vector<double> w = { 2, 2, 1, 1, 2, 2, 1, 1, 2, 10 };
uniform_01<> dist;
boost::random::mt19937 gen;
std::vector<double> vals;
std::generate_n(std::back_inserter(vals), w.size(), [&dist,&gen]() { return dist(gen); });
std::transform(vals.begin(), vals.end(), w.begin(), vals.begin(), [&](auto r, auto w) { return std::pow(r, 1. / w); });
std::vector<std::pair<double, int>> valIndices;
size_t index = 0;
std::transform(vals.begin(), vals.end(), std::back_inserter(valIndices), [&index](auto v) { return std::pair<double,size_t>(v,index++); });
std::sort(valIndices.begin(), valIndices.end(), [](auto x, auto y) { return x.first > y.first; });
std::vector<int> samples;
std::transform(valIndices.begin(), valIndices.end(), std::back_inserter(samples), [](auto v) { return v.second; });
for (auto iter : samples) {
std::cout << iter << " ";
}
return 0;
}
Easier answer
I just removed some of the STL functions and replaced it with simple for loops.
#include <iostream>
#include <iterator>
#include <boost/random/uniform_01.hpp>
#include <boost/random/mersenne_twister.hpp>
#include <algorithm>
using namespace boost::random;
int main(int, char**) {
std::vector<double> w = { 2, 2, 1, 1, 2, 2, 1, 1, 2, 1000 };
uniform_01<> dist;
boost::random::mt19937 gen(342575235);
std::vector<double> vals;
for (auto iter : w) {
vals.push_back(std::pow(dist(gen), 1. / iter));
}
// Sorting vals, but retain the indices.
// There is unfortunately no easy way to do this with STL.
std::vector<std::pair<int, double>> valsWithIndices;
for (size_t iter = 0; iter < vals.size(); iter++) {
valsWithIndices.emplace_back(iter, vals[iter]);
}
std::sort(valsWithIndices.begin(), valsWithIndices.end(), [](auto x, auto y) {return x.second > y.second; });
std::vector<size_t> samples;
int sampleSize = 8;
for (auto iter = 0; iter < sampleSize; iter++) {
samples.push_back(valsWithIndices[iter].first);
}
for (auto iter : samples) {
std::cout << iter << " ";
}
return 0;
}

Answer 2

The existing answer by Aleph0 works the best of the ones I tested. I tried benchmarking the original solution, the one added by Aleph0, and a new one where you only make a new discrete_distribution when the existing one is over 50% already added items (redrawing when distribution produces an item already in the sample).
I tested with sample size == population size, and weights equal the index. I think the original solution in the question runs in O(n^2), my new one runs in O(n logn) and the one from the paper seems to run in O(n).
-------------------------------------------------------------
Benchmark Time CPU Iterations
-------------------------------------------------------------
BM_Reuse 25252721 ns 25251731 ns 26
BM_NewDistribution 17338706125 ns 17313620000 ns 1
BM_SomePaper 6789525 ns 6779400 ns 100
Code:
#include <array>
#include <benchmark/benchmark.h>
#include <boost/random/mersenne_twister.hpp>
#include <boost/random/uniform_01.hpp>
#include <iostream>
#include <iterator>
#include <random>
#include <vector>
const int sampleSize = 20000;
using namespace boost::random;
static void BM_ReuseDistribution(benchmark::State &state) {
std::vector<double> weights;
weights.resize(sampleSize);
for (auto _ : state) {
for (int i = 0; i < sampleSize; i++) {
weights[i] = i + 1;
}
std::random_device rd;
std::mt19937 generator(rd());
int o[sampleSize];
std::discrete_distribution<int> distribution(weights.begin(),
weights.end());
int numAdded = 0;
int distSize = sampleSize;
for (int i = 0; i < sampleSize; ++i) {
if (numAdded > distSize / 2) {
distSize -= numAdded;
numAdded = 0;
distribution =
std::discrete_distribution<int>(weights.begin(), weights.end());
}
int number = distribution(generator);
if (!weights[number]) {
i -= 1;
continue;
} else {
weights[number] = 0;
o[i] = number;
numAdded += 1;
}
}
}
}
BENCHMARK(BM_ReuseDistribution);
static void BM_NewDistribution(benchmark::State &state) {
std::vector<double> weights;
weights.resize(sampleSize);
for (auto _ : state) {
for (int i = 0; i < sampleSize; i++) {
weights[i] = i + 1;
}
std::random_device rd;
std::mt19937 generator(rd());
int o[sampleSize];
for (int i = 0; i < sampleSize; ++i) {
std::discrete_distribution<int> distribution(weights.begin(),
weights.end());
int number = distribution(generator);
weights[number] = 0;
o[i] = number;
}
}
}
BENCHMARK(BM_NewDistribution);
static void BM_SomePaper(benchmark::State &state) {
std::vector<double> w;
w.resize(sampleSize);
for (auto _ : state) {
for (int i = 0; i < sampleSize; i++) {
w[i] = i + 1;
}
uniform_01<> dist;
boost::random::mt19937 gen;
std::vector<double> vals;
std::generate_n(std::back_inserter(vals), w.size(),
[&dist, &gen]() { return dist(gen); });
std::transform(vals.begin(), vals.end(), w.begin(), vals.begin(),
[&](auto r, auto w) { return std::pow(r, 1. / w); });
std::vector<std::pair<double, int>> valIndices;
size_t index = 0;
std::transform(
vals.begin(), vals.end(), std::back_inserter(valIndices),
[&index](auto v) { return std::pair<double, size_t>(v, index++); });
std::sort(valIndices.begin(), valIndices.end(),
[](auto x, auto y) { return x.first > y.first; });
std::vector<int> samples;
std::transform(valIndices.begin(), valIndices.end(),
std::back_inserter(samples),
[](auto v) { return v.second; });
}
}
BENCHMARK(BM_SomePaper);
BENCHMARK_MAIN();

Answer 3

Thanks for your question and others' nice answer, I meet a same qustion as you. I think you needn't new distribution every time, instead of
dist.param({ wts.begin(), wts.end() });
complete codes are as follows:
//STL改进方案
#include <iostream>
#include <vector>
#include <random>
#include <iomanip>
#include <map>
#include <set>
int main()
{
//随机数引擎采用默认引擎
std::default_random_engine rng;
//随机数引擎采用设备熵值保证随机性
auto gen = std::mt19937{ std::random_device{}() };
std::vector<int> wts(24); //存储权重值
std::vector<int> in(24); //存储总体
std::set<int> out; //存储抽样结果
std::map<int, int> count; //输出计数
int sampleCount = 0; //抽样次数计数
int index = 0; //抽取的下标
int sampleSize = 24; //抽取样本的数量
int sampleTimes = 100000; //抽取样本的次数
//权重赋值
for (int i = 0; i < 24; i++)
{
wts.at(i) = 48 - 2 * i;
}
//总体赋值并输出
std::cout << "总体为24个:" << std::endl;
//赋值
for (int i = 0; i < 24; i++)
{
in.at(i) = i + 1;
std::cout << in.at(i) << " ";
}
std::cout << std::endl;
//产生按照给定权重的离散分布
std::discrete_distribution<size_t> dist{ wts.begin(), wts.end() };
auto probs = dist.probabilities(); // 返回概率计算结果
//输出概率计算结果
std::cout << "总体中各数据的权重为:" << std::endl;
std::copy(probs.begin(), probs.end(), std::ostream_iterator<double>
{ std::cout << std::fixed << std::setprecision(5), “ ”});
std::cout << std::endl << std::endl;
//==========抽样测试==========
for (size_t j = 0; j < sampleTimes; j++)
{
index = dist(gen);
//std::cout << index << “ ”; //输出抽样结果
count[index] += 1; //抽样结果计数
}
double sum = 0.0; //用于概率求和
//输出抽样结果
std::cout << "总共抽样" << sampleTimes << "次，" << "各下标的频数及频率为：" << std::endl;
for (size_t i = 0; i < 24; i++)
{
std::cout << i << "共有" << count[i] << "个 频率为：" << count[i] / double(sampleTimes) << std::endl;
sum += count[i] / double(sampleTimes);
}
std::cout << "总频率为:" << sum << std::endl << std::endl; //输出总概率
//==========抽样测试==========
//从总体中抽样放入集合中，直至集合大小达到样本数
while (out.size() < sampleSize - 1)
{
index = dist(gen); //抽取下标
out.insert(index); //插入集合
sampleCount += 1; //抽样次数增加1
wts.at(index) = 0; //将抽取到的下标索引的权重设置为0
dist.param({ wts.begin(), wts.end() });
probs = dist.probabilities(); // 返回概率计算结果
//输出概率计算结果
std::cout << "总体中各数据的权重为：" << std::endl;
std::copy(probs.begin(), probs.end(), std::ostream_iterator<double>
{ std::cout << std::fixed << std::setprecision(5), “ ”});
std::cout << std::endl << std::endl;
}
//最后一次抽取，单独出来是避免将所有权重都为0，的权重数组赋值给离散分布dist，避免报错
index = dist(gen); //抽取下标
out.insert(index); //插入集合
sampleCount += 1; //抽样次数增加1
//输出抽样结果
std::cout << "从总体中抽取的" << sampleSize << "个样本的下标索引为：" << std::endl;
for (auto iter : out)
{
std::cout << iter << “-”;
}
std::cout << std::endl;
//输出抽样次数
std::cout << "抽样次数为:" << sampleCount << std::endl;
out.clear(); //清空输出集合，为下次抽样做准备
std::cin.get(); //保留控制台窗口
return 0;
}