根据某些条件从 Vec 中删除元素 [英] Removing elements from a Vec based on some condition

问题描述

我的代码如下所示:

struct Bar {
    i: i32,
}

struct Foo {
    v: Vec<Bar>,
}

impl Foo {
    fn foo(&mut self) {
        self.v.drain(self.v.iter().filter(|b| b.i < 10));
    }
}

注意Bar是不允许被复制或克隆的.

Note that Bar is not allowed to be copied or cloned.

我想删除所有不满足该条件的Bar.最初我以为我可以迭代它们并调用 remove 但我不允许有两个可变借用或一个可变借用，如果有一个完全有意义的不可变借用.

I want to delete all Bars that don't satisfy that condition. Initially I thought I could just iterate over them and call remove but I am not allowed to have two mutable borrows or one mutable borrow if there is an immutable borrow which makes total sense.

我想最简单的方法就是clone、filter 和collect，但我不能克隆或复制.

I guess the easiest thing would be to just clone, filter and collect, but I am not allowed to clone or copy.

我该怎么做?

推荐答案

如果你查看 Vec 的接口，你不会发现基于谓词擦除某些元素的方法.相反，您会发现 retain 根据谓词保留元素.

If you look at the interface of Vec, you will not find a method that erases some elements based on a predicate. Instead you will find retain which keeps the elements based on a predicate.

当然，两者都是对称的，只是如果您通过remove"或erase"过滤方法名称(它的描述中确实包含remove")，则更难找到retain.

Of course, both are symmetric, it's just that retain is harder to find if you filter method names by "remove" or "erase" (it does contain "remove" in its description).

提供的示例不言自明:

let mut vec = vec![1, 2, 3, 4];
vec.retain(|&x| x % 2 == 0);
assert_eq!(vec, [2, 4]);

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