如何压缩两个以上的迭代器？

Question

是否有更直接,更易读的方法来完成以下任务:

fn main() {
    let a = [1, 2, 3];
    let b = [4, 5, 6];
    let c = [7, 8, 9];
    let iter = a.iter()
        .zip(b.iter())
        .zip(c.iter())
        .map(|((x, y), z)| (x, y, z));
}

也就是说,如何从n个迭代中构建迭代器,从而产生n元组？

Answer 1

您可以使用izip!()crate itertools中的宏,它为任意多个迭代器实现这一点:

use itertools::izip;

fn main() {

    let a = [1, 2, 3];
    let b = [4, 5, 6];
    let c = [7, 8, 9];

    // izip!() accepts iterators and/or values with IntoIterator.
    for (x, y, z) in izip!(&a, &b, &c) {

    }
}

你必须在Cargo.toml中添加对itertools的依赖,使用最新的版本.例:

[dependencies]
itertools = "0.8"

Answer 2

您还可以使用提供的命令创建宏.zip，例如

$ cat z.rs
macro_rules! zip {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr), +) => (
        $x.iter().zip(
            zip!($($y), +))
    )
}


fn main() {
    let x = vec![1,2,3];
    let y = vec![4,5,6];
    let z = vec![7,8,9];

    let zipped = zip!(x, y, z);
    println!("{:?}", zipped);
    for (a, (b, c)) in zipped {
        println!("{} {} {}", a, b, c);
    }
}

输出：

$ rustc z.rs && ./z
Zip { a: Iter([1, 2, 3]), b: Zip { a: Iter([4, 5, 6, 67]), b: IntoIter([7, 8, 9]), index: 0, len: 0 }, index: 0, len: 0 }
1 4 7
2 5 8
3 6 9