使用LINQ到对象找到一个集合的项目不匹配的另一

我想找到一个集合中的所有项目不匹配的另一个集合。的集合是不同类型的，虽然，我想写一个lambda表达式来指定平等。

I want to find all items in one collection that do not match another collection. The collections are not of the same type, though; I want to write a lambda expression to specify equality.

一个 LINQPad 例子什么我想要做的：

A LINQPad example of what I'm trying to do:

void Main()
{
    var employees = new[]
    {
        new Employee { Id = 20, Name = "Bob" },
        new Employee { Id = 10, Name = "Bill" },
        new Employee { Id = 30, Name = "Frank" }
    };

    var managers = new[]
    {
        new Manager { EmployeeId = 20 },
        new Manager { EmployeeId = 30 }
    };

    var nonManagers =
    from employee in employees
    where !(managers.Any(x => x.EmployeeId == employee.Id))
    select employee;

    nonManagers.Dump();

    // Based on cdonner's answer:

    var nonManagers2 =
    from employee in employees
    join manager in managers
        on employee.Id equals manager.EmployeeId
    into tempManagers
    from manager in tempManagers.DefaultIfEmpty()
    where manager == null
    select employee;

    nonManagers2.Dump();

    // Based on Richard Hein's answer:

    var nonManagers3 =
    employees.Except(
        from employee in employees
        join manager in managers
            on employee.Id equals manager.EmployeeId
        select employee);

    nonManagers3.Dump();
}

public class Employee
{
    public int Id { get; set; }
    public string Name { get; set; }
}

public class Manager
{
    public int EmployeeId { get; set; }
}

以上的作品，并将返回员工比尔（＃10）。它似乎并不优雅，虽然，这可能是低效率的较大集合。在SQL中我可能会做一个LEFT JOIN和查找项目，其中第二ID为NULL。什么是对LINQ这样做的最佳做法

The above works, and will return Employee Bill (#10). It does not seem elegant, though, and it may be inefficient with larger collections. In SQL I'd probably do a LEFT JOIN and find items where the second ID was NULL. What's the best practice for doing this in LINQ?

编辑：？更新以防止取决于标识等于指数的解决方案

Updated to prevent solutions that depend on the Id equaling the index.

编辑：添加cdonner的解决方案 - 任何人有什么简单

Added cdonner's solution - anybody have anything simpler?

编辑：添加一个变种理查德海恩的回答，我目前最喜欢的？ ！感谢大家的一些优秀的答案

Added a variant on Richard Hein's answer, my current favorite. Thanks to everyone for some excellent answers!