Question

我使用此代码为数字0到49创建一个随机样本。现在，我想为一组自定义值创建一个随机样本。例如：从[1,2,3,4,9,10,11,14,16,22,32,45]中选择5个样本。我该怎么办？

   use rand::{thread_rng, seq};
    use std::time::SystemTime;

    fn main(){
        let mut rng = thread_rng();

        let mut sample = seq::index::sample(&mut rng, 50, 5);
    }

Answer 1

您可以使用IteratorRandom来获得更短的解决方案。这是迭代器的扩展特征，它提供了方便的功能，例如choose_multiple和choose_multiple_fill：

use rand::{seq::IteratorRandom, thread_rng}; // 0.6.1

fn main() {
    let mut rng = thread_rng();
    let v = vec![1, 2, 3, 4, 5];
    let sample = v.iter().choose_multiple(&mut rng, 2);

    println!("{:?}", sample);
}

Answer 2

permutation之类的声音

extern crate permutate;
use permutate::Permutator;
use std::io::{self, Write};

fn main() {
    let stdout = io::stdout();
    let mut stdout = stdout.lock();
    let list: &[&str] = &["one", "two", "three", "four"];
    let list = [list];
    let mut permutator = Permutator::new(&list[..]);

    if let Some(mut permutation) = permutator.next() {
        for element in &permutation {
            let _ = stdout.write(element.as_bytes());
        }
        let _ = stdout.write(b"\n");
        while permutator.next_with_buffer(&mut permutation) {
            for element in &permutation {
                let _ = stdout.write(element.as_bytes());
            }
            let _ = stdout.write(b"\n");
        }
    }
}

如何从元素向量创建随机样本

2 个答案: