Is there a way to drain parts of a vector based on a predicate?

问题

I'm trying to remove some elements from a vector, based on a predicate, and collecting the result. Here's a (not working) example with an expected result:

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]);

This results in the error

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();
  |                                                                ^^^^^

There are several alternatives I looked at, none of them seem to be doing what I want:

Vec::retain removes the elements from the vector, but doesn't give back ownership of the removed elements.
v.drain(..).filter(condition).collect() returns the correct value for drained but empties the whole vector.

回答1:

Not in stable Rust 1.33.0. There's an unstable nightly feature called drain_filter that does exactly what you want:

#![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]);
}

As a stable workaround, you may be able to use Iterator::partition, but it does not reuse the memory:

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]);
}

来源：https://stackoverflow.com/questions/46651354/is-there-a-way-to-drain-parts-of-a-vector-based-on-a-predicate

标签

vector

rust

predicate