C# - 在匹配键上加入两个 LanguageExt Either<Error, List<T>>

C# - Joining two LanguageExt Either<Error, List<T>> on matching key

在下面的示例中,我有两组具有匹配键的对象。

class Stuff
{
    int Key1 { get; set; }
    ... other props
}

class OtherStuff
{
    int Key2 { get; set; }
    ... other props
}

我想将这些对象的两个列表连接在一起作为一对列表。假设有一个名为 Error 的 class 表示错误状态。我尝试通过以下方式进行;

Either<Error, List<Stuff>> eitherStuff = GetStuff();

Either<Error, List<OtherStuff>> eitherOtherStuff = GetOtherStuff();

Either<Error, List<object>> eitherCombined =
    from stuff in eitherStuff
    select stuff into s1
    from otherStuff in eitherOtherStuff
    select otherStuff into s2
    from s1item in s1
    join s2item in s2
    on s1item.Key1 equals s2item.Key2
    select new {s1item, s2item};

但这失败了

The name 's1' does not exist in the current context.

合并两个的最佳方法是什么Either<Error, List<T>>

Either<Error, List<(Stuff, OtherStuff)>> eitherCombined =
                from stuff in eitherStuff
                from otherStuff in eitherOtherStuff
                select (from s1item in stuff
                        join s2item in otherStuff
                        on s1item.Key1 equals s2item.Key2
                        select (s1item, s2item)).ToList();

外部 LINQ 表达式结合了两个 Either 元素。内部 LINQ 表达式连接列表(如果两个 Either 都是 "right")。

我将您的 return 类型更改为元组以避免对象。你原来的例子就像

Either<Error, List<object>> eitherCombined =
                from stuff in eitherStuff
                from otherStuff in eitherOtherStuff
                select (from s1item in stuff
                        join s2item in otherStuff
                        on s1item.Key1 equals s2item.Key2
                        select new {s1item, s2item} as object).ToList();

您可以像这样使用您的匿名类型(没有不安全 object):

var eitherCombined =
                from stuff in eitherStuff
                from otherStuff in eitherOtherStuff
                select (from s1item in stuff
                        join s2item in otherStuff
                        on s1item.Key1 equals s2item.Key2
                        select new {s1item, s2item}).ToList();

另一个提示:您可以使用 LanguageExt 不可变类型之一来代替 List...