如何从矢量或数组中随机选择一个元素？

Question

我有一个元素是a的向量(String, String).我如何随机选择其中一个元素？

Answer 1

你想要rand箱子,特别是choose方法.

use rand::seq::SliceRandom; // 0.6.5

fn main() {
    let vs = vec![0, 1, 2, 3, 4];
    println!("{:?}", vs.choose(&mut rand::thread_rng()));
}

Answer 2

如果您想选择多个元素，那么random_choice crate 可能适合您：

extern crate random_choice;
use self::random_choice::random_choice;

fn main() {
    let mut samples = vec!["hi", "this", "is", "a", "test!"];
    let weights: Vec<f64> = vec![5.6, 7.8, 9.7, 1.1, 2.0];

    let number_choices = 100;
    let choices = random_choice().random_choice_f64(&samples, &weights, number_choices);

    for choice in choices {
        print!("{}, ", choice);
    }
}

Answer 3

使用choose_multiple：

use rand::seq::SliceRandom; // 0.6.5

fn main() {
    let samples = vec!["hi", "this", "is", "a", "test!"];
    let sample: Vec<_> = samples
        .choose_multiple(&mut rand::thread_rng(), 1)
        .collect();
    println!("{:?}", sample);
}

Answer 4

如果您还想删除所选元素，可以使用以下一种方法（使用 crate rand）：

let mut vec = vec![0,1,2,3,4,5,6,7,8,9];

let index = (rand::random::<f32>() * vec.len() as f32).floor() as usize;
let value = vec.remove( index );

println!("index: {} value: {}", index, value);
println!("{:?}", vec);

铁锈游乐场

remove(index)删除 at 处的值index（将其后面的所有元素向左移动）并返回位于index( docs ) 处的值。

Answer 5

已经包含在randcrate中的另一个加权采样选择是WeightedIndex，它有一个例子：

use rand::prelude::*;
use rand::distributions::WeightedIndex;

let choices = ['a', 'b', 'c'];
let weights = [2,   1,   1];
let dist = WeightedIndex::new(&weights).unwrap();
let mut rng = thread_rng();
for _ in 0..100 {
    // 50% chance to print 'a', 25% chance to print 'b', 25% chance to print 'c'
    println!("{}", choices[dist.sample(&mut rng)]);
}

let items = [('a', 0), ('b', 3), ('c', 7)];
let dist2 = WeightedIndex::new(items.iter().map(|item| item.1)).unwrap();
for _ in 0..100 {
    // 0% chance to print 'a', 30% chance to print 'b', 70% chance to print 'c'
    println!("{}", items[dist2.sample(&mut rng)].0);
}