如何从元素向量创建随机样本?

Mig*_*uel 3 sample rust

我使用此代码为数字 0 到 49 创建了一个随机样本。现在我想为一组自定义值创建一个随机样本。例如:从 中选择 5 个样本[1, 2, 3, 4, 9, 10, 11, 14, 16, 22, 32, 45]。我怎样才能做到这一点?

use rand::{seq, thread_rng}; // 0.7.3

fn main() {
    let mut rng = thread_rng();
    let sample = seq::index::sample(&mut rng, 50, 5);
}
Run Code Online (Sandbox Code Playgroud)

lhk*_*lhk 5

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性,它提供了方便的功能,例如choose_multiplechoose_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);
}
Run Code Online (Sandbox Code Playgroud)

  • “choose_multiple”这个词有点令人困惑。这个样品有替换还是没有替换?“multiple”表明它正在进行替换采样,但文档没有提及任何相关内容。只是缺乏重载的情况(语义上它与“choose”相同?)强制使用可能被错误解释的函数名称? (4认同)
  • AFAIK,“choose_multiple”不会进行替换采样。 (2认同)

use*_*176 2

听起来您可以使用排列箱中的排列:

extern crate permutate; // 0.3.2

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");
        }
    }
}
Run Code Online (Sandbox Code Playgroud)