我试图从矢量中删除一些元素,基于谓词,并收集结果。这是一个具有预期结果的(不工作)示例:
let mut v: Vec<i32> = vec![1, 2, 3, 4, 5, 6];
let drained: Vec<i32> = v.iter().filter(|e| (*e) % 2 == 0).drain(..).collect();
assert_eq!(v, vec![1, 3, 5]);
assert_eq!(drained, vec![2, 4, 6]);
这导致错误
error[E0599]: no method named `drain` found for type `std::iter::Filter<std::slice::Iter<'_, i32>, [closure@src/main.rs:4:45: 4:62]>` in the current scope
--> src/main.rs:4:64
|
4 | let drained: Vec<i32> = v.iter().filter(|e| (*e) % 2 == 0).drain(..).collect();
| ^^^^^
我看过几种不同的选择,它们似乎都没有做我想做的事情:
Vec::retain
从向量中删除元素,但不会返回已删除元素的所有权。v.drain(..).filter(condition).collect()
返回drained
的正确值,但清空整个向量。不稳定Rust 1.33.0。有一个不稳定的夜间功能叫做drain_filter
,它可以完全满足您的需求:
#![feature(drain_filter)]
fn main() {
let mut v: Vec<i32> = vec![1, 2, 3, 4, 5, 6];
let drained: Vec<i32> = v.drain_filter(|&mut e| e % 2 == 0).collect();
assert_eq!(v, vec![1, 3, 5]);
assert_eq!(drained, vec![2, 4, 6]);
}
作为一个稳定的解决方法,您可以使用Iterator::partition
,但它不会重用内存:
fn main() {
let v: Vec<i32> = vec![1, 2, 3, 4, 5, 6];
let (drained, v): (Vec<_>, Vec<_>) = v.into_iter().partition(|&e| e % 2 == 0);
assert_eq!(v, vec![1, 3, 5]);
assert_eq!(drained, vec![2, 4, 6]);
}