如何使用C ++ Uniform_int_distribution进行采样而不进行替换 [英] How to sample without replacement using c++ uniform_int_distribution

问题描述

我想在c ++随机库中使用uniform_int_distribution.但是，仅在替换时进行采样，如下例所示.如何在不更换的情况下取样?

I want to use the uniform_int_distribution in the c++ random library. However, it only does the sampling with replacement, as in the example below. How can I sample without replacement?

#include <iostream>
#include <random>

int main()
{
  std::default_random_engine generator;
  std::uniform_int_distribution<int> distribution(1,4);

  for(int i=0; i<4; ++i)
    std::cout << distribution(generator) << std::endl;

  return 0;
}

推荐答案

在初始化为{1, 2, 3, 4}的std::array<int>或std::vector<int>上使用std::shuffle.

Use std::shuffle on, say, a std::array<int> or a std::vector<int>, initialised to {1, 2, 3, 4}.

然后依次读取容器中的内容.

Then read back the container contents in order.

与绘制随机数并仅在以前没有绘制它时才接受它相比，它具有更好的统计属性.

This will have better statistical properties than drawing a random number and accepting it only if it hasn't been drawn before.

参考 http://en.cppreference.com/w/cpp/algorithm/random_shuffle

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