Question

我提前为代码转储道歉。我把它整理得和我一样多可以不丢失我的问题的背景（在粗体下面）。

我有一个结构

use std::rand;
use std::rand::Rng;
use std::rand::distributions::{Weighted, WeightedChoice, Sample, IndependentSample};

struct MarkovChain {
    state: uint,
    weights: Vec<Vec<uint>>,
}

建模马尔可夫链。对尺寸有一些检查我在weights中实施的MarkovChain::new矩阵：

impl MarkovChain {
    fn new(weights: Vec<Vec<uint>>, initial_state: uint) -> MarkovChain {
        let states = weights.len();
        assert!(states > 0);
        assert!(initial_state < states);
        assert!(weights.iter().all(|row| row.len() == states));
        MarkovChain {
            state: initial_state,
            weights: weights,
        }
    }
}

现在我实施Sample：

impl Sample<uint> for MarkovChain {
    fn sample<R: Rng>(&mut self, rng: &mut R) -> uint {
        // I'd like to put the following part in MarkovChain::new
        // instead, but I can't figure out how to store the
        // WeightedChoice inside the MarkovChain struct.
        //BEGIN
        let mut row = self.weights[self.state]
            .iter()
            .enumerate()
            .map(|(i, &wt)| Weighted { item: i, weight: wt })
            .collect::<Vec<Weighted<uint>>>();
        let wc = WeightedChoice::new(row.as_mut_slice());
        //END

        self.state = wc.ind_sample(rng);
        self.state
    }
}

这里的问题是每个都需要构建row和wc 调用时间sample。鉴于典型的用例涉及多次调用sample，这是一个问题。

我想移动row和wc的计算（针对每个州）转而MarkovChain::new，但我似乎无法弄清楚如何在MarkovChain结构中存储WeightedChoice。我该怎么做？

我无法判断这是否真的很难，或者我是否因为多年的垃圾收集语言而遭受脑损伤。

这是马尔可夫链的一个示例用法。如果可能的话，我想保持接口不变：

fn main() {
    // Create the 3-state Markov chain illustrated at
    // https://en.wikipedia.org/w/index.php?title=Markov_chain&oldid=626307401#Example
    let mut mc = MarkovChain::new(vec![vec![900,  75,  25],
                                       vec![150, 800,  50],
                                       vec![250, 250, 500]], 0);

    // Expect around 62.5% 0s, 31.25% 1s, and 6.25% 2s after many iterations.
    let rng = &mut rand::task_rng();
    let mut stats = vec![0u, 0, 0];
    for _ in range(0u, 10000) {
        *stats.get_mut(mc.sample(rng)) += 1;
    }
    println!("Expect approximately [6250, 3125, 625]:");
    println!("{}", stats);
}

这里是Rust playpen中的全部内容。

在结构中存储WeightedChoice

0 个答案: