我想从一个不连续的分布中取样而不进行替换(即不重复)。
使用离散函数distribute_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;
}
您是否曾经编码过这种样本而无法替换?可能以更优化的方式?
谢谢。
干杯,
T.A.
最佳答案
该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被禁止。
#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之后,我发现了一种无需替换的加权采样的简单算法,在C++中实现起来有点复杂。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,如果样本大小比基本人口小得多,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;
}
更容易回答
我只是删除了一些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;
}
关于c++ - C++:从离散分布中采样而无需替换,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/53632441/