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如何从 std::sample 中提取剩余元素

[英]How to extract the remaining elements from std::sample

In the sample code here it shows cases how to use std::sample as shown below此处的示例代码中它显示了如何使用 std::sample 的情况,如下所示

std::string in = "hgfedcba", out;
std::sample(in.begin(), in.end(), std::back_inserter(out),
                5, std::mt19937{std::random_device{}()});
std::cout << "five random letters out of " << in << " : " << out << '\n';

Possible output:可能的输出:

five random letters out of hgfedcba: gfcba

My question is not only I want gfcba I also want to extract the remaining elements that are the not sampled, eg hed .我的问题不仅是我想要gfcba我还想提取未采样的剩余元素,例如hed I know I can write a for loop to compare in and out to extract the remaining elements, but I am wondering if there's a more efficient way to do this.我知道我可以写一个for循环比较inout提取剩余的元素,但我想知道是否有这样做更有效的方式。

If you're not concerned about the order of the characters in the output strings, then you can use std::shuffle to randomise the input string and then copy the first 5 characters of the result to one output string and the last 3 to the other:如果您不关心输出字符串中字符的顺序,那么您可以使用std::shuffle随机化输入字符串,然后将结果的前 5 个字符复制到一个输出字符串,将最后 3 个字符复制到其他:

#include <iostream>
#include <random>
#include <string>
#include <algorithm>

int main ()
{
    std::string in = "hgfedcba";

    std::random_device rd;
    std::mt19937 g (rd ());
    std::shuffle (in.begin(), in.end(), g);

    std::string out5, out3;
    
    for (size_t i = 0; i < 5; ++i)
        out5.push_back (in [i]);
    for (size_t i = 5; i < 8; ++i)
        out3.push_back (in [i]);
    
    std::cout << out5 << " " << out3;
}

Sample output:示例输出:

cbhfd aeg

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