我想从一个不连续的分布中取样而不进行替换（即不重复）。
使用函数 discrete_distribution ，可以进行替换采样。而且，借助此功能，我以非常粗糙的方式实现了无需替换的采样：
#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;
}
您曾经编码过这种采样而无需替换吗？可能以更优化的方式？
谢谢。
干杯
T.A。
答案 0 :(得分：0)
此解决方案可能会短一些。不幸的是，它需要在每个步骤中创建一个discrete_distribution<>对象，这在绘制大量样本时可能会禁止这样做。
#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;
}
改进的答案：
在这个站点（Faster weighted sampling without replacement）上仔细寻找类似问题后，我发现了一种无需替换的加权采样惊人简单的算法，在C ++中实现起来有点复杂。请注意，这不是最有效的算法，但在我看来，这是最简单的算法。
在https://doi.org/10.1016/j.ipl.2005.11.003中详细描述了该方法。
尤其是，如果样本量远小于基本人群，那将是无效的。
#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;
}
更简单的答案
我刚刚删除了一些STL函数，并用简单的for循环代替了它。
#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;
}
答案 1 :(得分：0)
Aleph0 的现有答案在我测试过的答案中效果最好。我尝试对原始解决方案、Aleph0 添加的解决方案和新解决方案进行基准测试，其中只有在现有解决方案超过 50% 的已添加项目时才创建新的 discrete_distribution（当分布产生样本中已有的项目时重绘） ).
我用样本大小==人口大小进行了测试，权重等于指数。我认为问题中的原始解决方案在 O(n^2) 中运行，我的新解决方案在 O(n logn) 中运行，而论文中的解决方案似乎在 O(n) 中运行。
-------------------------------------------------------------
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
代码：
#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();