迭代迭代矢量的相邻元素时如何避免循环

In Rust, how do I avoid writing these loops? The code takes a vector and multiplies three adjacent elements to a product. Hence, the outer loop goes over all elements that can form a group of three and the inner loop does the multiplication.

在Rust中，如何避免编写这些循环？代码采用向量并将三个相邻元素乘以产品。因此，外部循环遍历可以形成一组三个的所有元素，并且内部循环执行乘法。

The difficulty lies, I think, in the incomplete iteration of the outer loop (from element 0 to last - 3). Further, the inner loop must use a sub-range.

我认为，难点在于外循环的不完整迭代（从元素0到最后-3）。此外，内环必须使用子范围。

Is there a way to avoid writing the loops?

有没有办法避免编写循环？

let v = vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9];
let mut products = Vec::new();
for seq in 0..v.len() - 3 {
    let mut product = 1;
    for offset in 0..3 {
        product *= v[seq + offset];
    }
    products.push(product);
}

1 个解决方案

#1

The function you are searching for is [T]::windows(). You can specify the size of the overlapping windows and it will return an iterator over sub-slices.

您要搜索的函数是[T] :: windows（）。您可以指定重叠窗口的大小，它将在子切片上返回迭代器。

You can obtain the product of all elements within a sub-slice by making an iterator out of it and using Iterator::product().

您可以通过使用Iterator :: product（）创建迭代器来获取子切片中所有元素的乘积。

let v = vec![1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9];
let products: Vec<u64> = v.windows(3)
    .map(|win| win.iter().product())
    .collect();

(Playground)

（操场）

Here we collect all products into a new vector.

在这里，我们将所有产品收集到新的载体中。

A last note: instead of writing down all numbers in the vector manually you could write this instead:

最后一点：不要手动写下向量中的所有数字，而是可以写下来：

let v: Vec<_> = (1..10).chain(1..10).collect();

#1