在 Rust 中对 Rayon 并行化 for 循环中的成功求和-解网

问：

我很难理解 monty hall 问题，所以我把一些 Rust 放在一起。我有一个功能可以确定参与者是否赢得游戏节目。如果他们赢得了游戏节目，则该函数返回 true，否则返回 false。我想测试非常大的数字，所以我在 for 循环中添加了一个 Rayon 并行迭代器。代码如下：

fn gameshow(num_doors: usize, num_runs: usize, change: bool) -> usize {
    let mut won_games:usize = 0;
    let runs: Vec<Vec<bool>> = vec![stagehand(num_doors); num_runs]; // Creates common solution 
    if change {
        ParallelIterator::for_each(IntoParallelIterator::into_par_iter(runs), |game| {
            let winner: bool = run_game_no_change(game); // Each door is chosen randomly when tested
            if winner { won_games += 1;}
        });
    }
    else {
        ParallelIterator::for_each(IntoParallelIterator::into_par_iter(runs), |game| {
            let winner: bool = run_game_no_change(game); // Each door is chosen randomly when tested
            if winner { won_games += 1;}
        });     
    }
    won_games   
}

编译时，它会抛出“无法赋值给，因为它是闭包中捕获的变量”错误。won_gamesFn

锈色人造丝

fn gameshow(num_doors: usize, num_runs: usize, change: bool) -> usize {
    let runs: Vec<Vec<bool>> = vec![stagehand(num_doors); num_runs]; // Creates common solution 
    let won_games = if change {
        runs.into_par_iter().map(|game| {
            let winner: bool = run_game_no_change(game); // Each door is chosen randomly when tested
            if winner { 1 } else { 0 }
        }).sum()
    } else {
        runs.into_par_iter().map(|game| {
            let winner: bool = run_game_no_change(game); // Each door is chosen randomly when tested
            if winner { 1 } else { 0 }
        }).sum()    
    };

    won_games   
}

在 Rust 中对 Rayon 并行化 for 循环中的成功求和

Summing successes inside of a Rayon Parallelized for loop in Rust

评论

评论