file-io - 如何使多个迭代器可以处理与文件有关的相同数据？

Question

我有一个文件，希望将其读取并将数据过滤成两个不同的集合，并确定每个集合中的项数。

use std::io::{self, BufRead};

fn main() {
    let cursor = io::Cursor::new(b"pillow\nbrick\r\nphone");

    let lines = cursor.lines().map(|l| l.unwrap());

    let soft_count = lines.filter(|line| line.contains("pillow")).count();

    let hard_count = lines.filter(|line| !line.contains("pillow")).count();
}

Playground

GitHub

但是，借阅检查器给我一个错误:

error[E0382]: use of moved value: `lines`
  --> src/main.rs:14:22
   |
8  |     let lines = cursor.lines().map(|l| l.unwrap());
   |         ----- move occurs because `lines` has type `std::iter::Map<std::io::Lines<std::io::Cursor<&[u8; 19]>>, [closure@src/main.rs:8:36: 8:50]>`, which does not implement the `Copy` trait
9  |     
10 |     let soft_count = lines
   |                      ----- value moved here
...
14 |     let hard_count = lines
   |                      ^^^^^ value used here after move

我尝试使用引用计数解决此问题，以允许多个所有权:

use std::io::{self, BufRead};
use std::rc::Rc;

fn main() {
    let cursor = io::Cursor::new(b"pillow\nbrick\r\nphone");

    let lines = Rc::new(cursor.lines().map(|l| l.unwrap()));

    let soft_count = Rc::clone(&lines)
        .filter(|line| line.contains("pillow"))
        .count();

    let hard_count = Rc::clone(&lines)
        .filter(|line| !line.contains("pillow"))
        .count();
}

Playground

Github

我收到类似的错误消息:

error[E0507]: cannot move out of an `Rc`
  --> src/main.rs:11:22
   |
11 |     let soft_count = Rc::clone(&lines)
   |                      ^^^^^^^^^^^^^^^^^ move occurs because value has type `std::iter::Map<std::io::Lines<std::io::Cursor<&[u8; 19]>>, [closure@src/main.rs:9:44: 9:58]>`, which does not implement the `Copy` trait

error[E0507]: cannot move out of an `Rc`
  --> src/main.rs:15:22
   |
15 |     let hard_count = Rc::clone(&lines)
   |                      ^^^^^^^^^^^^^^^^^ move occurs because value has type `std::iter::Map<std::io::Lines<std::io::Cursor<&[u8; 19]>>, [closure@src/main.rs:9:44: 9:58]>`, which does not implement the `Copy` trait

Answer 1

你不能。相反，您将需要克隆迭代器或它的某些构建块。在这种情况下，您可以克隆的最高名称是Cursor:

use std::io::{self, BufRead};

fn main() {
    let cursor = io::Cursor::new(b"pillow\nbrick\r\nphone");

    let lines = cursor.clone().lines().map(|l| l.unwrap());
    let lines2 = cursor.lines().map(|l| l.unwrap());

    let soft_count = lines.filter(|line| line.contains("pillow")).count();

    let hard_count = lines2.filter(|line| !line.contains("pillow")).count();
}

对于实际的 File，您将需要使用 try_clone ，因为它可能会失败。无论哪种情况，您都将两次引用相同的数据，并且仅保留迭代器信息。

对于您的特定情况，您不需要任何这些。实际上，对数据进行两次迭代效率不高。您可以执行的最简单的内置操作是 partition 迭代器:

let (softs, hards): (Vec<_>, Vec<_>) = lines.partition(|line| line.contains("pillow"));

let soft_count = softs.len();
let hard_count = hards.len();

由于您不需要实际值，因此效率仍然有些低下。您可以创建自己的类型来实现 Extend并丢弃值:

#[derive(Debug, Default)]
struct Count(usize);

impl<T> std::iter::Extend<T> for Count {
    fn extend<I>(&mut self, iter: I)
    where
        I: IntoIterator,
    {
        self.0 += iter.into_iter().count();
    }
}

let (softs, hards): (Count, Count) = lines.partition(|line| line.contains("pillow"));

let soft_count = softs.0;
let hard_count = hards.0;

您也可以只使用 for循环或在 fold之上构建一些东西:

let (soft_count, hard_count) = lines.fold((0, 0), |mut state, line| {
    if line.contains("pillow") {
        state.0 += 1;
    } else {
        state.1 += 1;
    }
    state
});