对于键在两个枚举对象中都是唯一的情况，我的干净解决方案:

 private static IEnumerable<TResult> FullOuterJoin<Ta, Tb, TKey, TResult>(
            IEnumerable<Ta> a, IEnumerable<Tb> b,
            Func<Ta, TKey> key_a, Func<Tb, TKey> key_b,
            Func<Ta, Tb, TResult> selector)
        {
            var alookup = a.ToLookup(key_a);
            var blookup = b.ToLookup(key_b);
            var keys = new HashSet<TKey>(alookup.Select(p => p.Key));
            keys.UnionWith(blookup.Select(p => p.Key));
            return keys.Select(key => selector(alookup[key].FirstOrDefault(), blookup[key].FirstOrDefault()));
        }

so

    var ax = new[] {
        new { id = 1, first_name = "ali" },
        new { id = 2, first_name = "mohammad" } };
    var bx = new[] {
        new { id = 1, last_name = "rezaei" },
        new { id = 3, last_name = "kazemi" } };

    var list = FullOuterJoin(ax, bx, a => a.id, b => b.id, (a, b) => "f: " + a?.first_name + " l: " + b?.last_name).ToArray();

输出:

f: ali l: rezaei
f: mohammad l:
f:  l: kazemi

我不知道这是否适用于所有情况，但从逻辑上讲，这似乎是正确的。其思想是取一个左外连接和一个右外连接，然后取结果的并集。

var firstNames = new[]
{
    new { ID = 1, Name = "John" },
    new { ID = 2, Name = "Sue" },
};
var lastNames = new[]
{
    new { ID = 1, Name = "Doe" },
    new { ID = 3, Name = "Smith" },
};
var leftOuterJoin =
    from first in firstNames
    join last in lastNames on first.ID equals last.ID into temp
    from last in temp.DefaultIfEmpty()
    select new
    {
        first.ID,
        FirstName = first.Name,
        LastName = last?.Name,
    };
var rightOuterJoin =
    from last in lastNames
    join first in firstNames on last.ID equals first.ID into temp
    from first in temp.DefaultIfEmpty()
    select new
    {
        last.ID,
        FirstName = first?.Name,
        LastName = last.Name,
    };
var fullOuterJoin = leftOuterJoin.Union(rightOuterJoin);

这就像写的一样，因为它是在LINQ to Objects中。如果LINQ to SQL或其他，查询处理器可能不支持安全导航或其他操作。你必须使用条件操作符有条件地获取值。

也就是说,

var leftOuterJoin =
    from first in firstNames
    join last in lastNames on first.ID equals last.ID into temp
    from last in temp.DefaultIfEmpty()
    select new
    {
        first.ID,
        FirstName = first.Name,
        LastName = last != null ? last.Name : default,
    };

这是另一个完整的外部连接

由于对其他命题的简单性和可读性不太满意，我最后得出了这样的结论:

它没有快速的自命(在2020m CPU上加入1000 * 1000大约800毫秒:2.4ghz / 2核)。对我来说，它只是一个紧凑而随意的完全外部连接。

它的工作原理与SQL FULL OUTER JOIN相同(重复保存)

欢呼;-)

using System;
using System.Collections.Generic;
using System.Linq;
namespace NS
{
public static class DataReunion
{
    public static List<Tuple<T1, T2>> FullJoin<T1, T2, TKey>(List<T1> List1, Func<T1, TKey> KeyFunc1, List<T2> List2, Func<T2, TKey> KeyFunc2)
    {
        List<Tuple<T1, T2>> result = new List<Tuple<T1, T2>>();

        Tuple<TKey, T1>[] identifiedList1 = List1.Select(_ => Tuple.Create(KeyFunc1(_), _)).OrderBy(_ => _.Item1).ToArray();
        Tuple<TKey, T2>[] identifiedList2 = List2.Select(_ => Tuple.Create(KeyFunc2(_), _)).OrderBy(_ => _.Item1).ToArray();

        identifiedList1.Where(_ => !identifiedList2.Select(__ => __.Item1).Contains(_.Item1)).ToList().ForEach(_ => {
            result.Add(Tuple.Create<T1, T2>(_.Item2, default(T2)));
        });

        result.AddRange(
            identifiedList1.Join(identifiedList2, left => left.Item1, right => right.Item1, (left, right) => Tuple.Create<T1, T2>(left.Item2, right.Item2)).ToList()
        );

        identifiedList2.Where(_ => !identifiedList1.Select(__ => __.Item1).Contains(_.Item1)).ToList().ForEach(_ => {
            result.Add(Tuple.Create<T1, T2>(default(T1), _.Item2));
        });

        return result;
    }
}
}

这个想法是

基于提供的关键函数生成器构建id 处理仅剩下的项 流程内部连接 只处理正确的项目

下面是一个与之相关的简单测试:

在结束处放置断点，以手动验证它的行为是否符合预期

using System;
using System.Collections.Generic;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using Newtonsoft.Json;
using Newtonsoft.Json.Linq;
using NS;

namespace Tests
{
[TestClass]
public class DataReunionTest
{
    [TestMethod]
    public void Test()
    {
        List<Tuple<Int32, Int32, String>> A = new List<Tuple<Int32, Int32, String>>();
        List<Tuple<Int32, Int32, String>> B = new List<Tuple<Int32, Int32, String>>();

        Random rnd = new Random();

        /* Comment the testing block you do not want to run
        /* Solution to test a wide range of keys*/

        for (int i = 0; i < 500; i += 1)
        {
            A.Add(Tuple.Create(rnd.Next(1, 101), rnd.Next(1, 101), "A"));
            B.Add(Tuple.Create(rnd.Next(1, 101), rnd.Next(1, 101), "B"));
        }

        /* Solution for essential testing*/

        A.Add(Tuple.Create(1, 2, "B11"));
        A.Add(Tuple.Create(1, 2, "B12"));
        A.Add(Tuple.Create(1, 3, "C11"));
        A.Add(Tuple.Create(1, 3, "C12"));
        A.Add(Tuple.Create(1, 3, "C13"));
        A.Add(Tuple.Create(1, 4, "D1"));

        B.Add(Tuple.Create(1, 1, "A21"));
        B.Add(Tuple.Create(1, 1, "A22"));
        B.Add(Tuple.Create(1, 1, "A23"));
        B.Add(Tuple.Create(1, 2, "B21"));
        B.Add(Tuple.Create(1, 2, "B22"));
        B.Add(Tuple.Create(1, 2, "B23"));
        B.Add(Tuple.Create(1, 3, "C2"));
        B.Add(Tuple.Create(1, 5, "E2"));

        Func<Tuple<Int32, Int32, String>, Tuple<Int32, Int32>> key = (_) => Tuple.Create(_.Item1, _.Item2);

        var watch = System.Diagnostics.Stopwatch.StartNew();
        var res = DataReunion.FullJoin(A, key, B, key);
        watch.Stop();
        var elapsedMs = watch.ElapsedMilliseconds;
        String aser = JToken.FromObject(res).ToString(Formatting.Indented);
        Console.Write(elapsedMs);
    }
}

}

我认为LINQ join子句并不是这个问题的正确解决方案，因为join子句的目的并不是按照这个任务解决方案所需的方式来积累数据。合并创建的独立集合的代码变得太复杂了，也许这对于学习目的来说是可以的，但对于真正的应用程序来说不是。解决这个问题的方法之一是下面的代码:

class Program
{
    static void Main(string[] args)
    {
        List<FirstName> firstNames = new List<FirstName>();
        firstNames.Add(new FirstName { ID = 1, Name = "John" });
        firstNames.Add(new FirstName { ID = 2, Name = "Sue" });

        List<LastName> lastNames = new List<LastName>();
        lastNames.Add(new LastName { ID = 1, Name = "Doe" });
        lastNames.Add(new LastName { ID = 3, Name = "Smith" });

        HashSet<int> ids = new HashSet<int>();
        foreach (var name in firstNames)
        {
            ids.Add(name.ID);
        }
        foreach (var name in lastNames)
        {
            ids.Add(name.ID);
        }
        List<FullName> fullNames = new List<FullName>();
        foreach (int id in ids)
        {
            FullName fullName = new FullName();
            fullName.ID = id;
            FirstName firstName = firstNames.Find(f => f.ID == id);
            fullName.FirstName = firstName != null ? firstName.Name : string.Empty;
            LastName lastName = lastNames.Find(l => l.ID == id);
            fullName.LastName = lastName != null ? lastName.Name : string.Empty;
            fullNames.Add(fullName);
        }
    }
}
public class FirstName
{
    public int ID;

    public string Name;
}

public class LastName
{
    public int ID;

    public string Name;
}
class FullName
{
    public int ID;

    public string FirstName;

    public string LastName;
}

如果真正的集合对于HashSet的形成很大，而不是foreach循环可以使用下面的代码:

List<int> firstIds = firstNames.Select(f => f.ID).ToList();
List<int> LastIds = lastNames.Select(l => l.ID).ToList();
HashSet<int> ids = new HashSet<int>(firstIds.Union(LastIds));//Only unique IDs will be included in HashSet

正如您所发现的，Linq没有“外部连接”结构。您所能得到的最接近的是使用您所声明的查询的左外连接。为此，你可以添加任何没有在join中表示的姓氏列表元素:

outerJoin = outerJoin.Concat(lastNames.Select(l=>new
                            {
                                id = l.ID,
                                firstname = String.Empty,
                                surname = l.Name
                            }).Where(l=>!outerJoin.Any(o=>o.id == l.id)));

在两个输入上执行内存流枚举，并为每一行调用选择器。如果在当前迭代中没有相关性，则选择器参数之一将为空。

例子:

   var result = left.FullOuterJoin(
         right, 
         x=>left.Key, 
         x=>right.Key, 
         (l,r) => new { LeftKey = l?.Key, RightKey=r?.Key });

Requires an IComparer for the correlation type, uses the Comparer.Default if not provided. Requires that 'OrderBy' is applied to the input enumerables /// <summary> /// Performs a full outer join on two <see cref="IEnumerable{T}" />. /// </summary> /// <typeparam name="TLeft"></typeparam> /// <typeparam name="TValue"></typeparam> /// <typeparam name="TRight"></typeparam> /// <typeparam name="TResult"></typeparam> /// <param name="left"></param> /// <param name="right"></param> /// <param name="leftKeySelector"></param> /// <param name="rightKeySelector"></param> /// <param name="selector">Expression defining result type</param> /// <param name="keyComparer">A comparer if there is no default for the type</param> /// <returns></returns> [System.Diagnostics.DebuggerStepThrough] public static IEnumerable<TResult> FullOuterJoin<TLeft, TRight, TValue, TResult>( this IEnumerable<TLeft> left, IEnumerable<TRight> right, Func<TLeft, TValue> leftKeySelector, Func<TRight, TValue> rightKeySelector, Func<TLeft, TRight, TResult> selector, IComparer<TValue> keyComparer = null) where TLeft: class where TRight: class where TValue : IComparable { keyComparer = keyComparer ?? Comparer<TValue>.Default; using (var enumLeft = left.OrderBy(leftKeySelector).GetEnumerator()) using (var enumRight = right.OrderBy(rightKeySelector).GetEnumerator()) { var hasLeft = enumLeft.MoveNext(); var hasRight = enumRight.MoveNext(); while (hasLeft || hasRight) { var currentLeft = enumLeft.Current; var valueLeft = hasLeft ? leftKeySelector(currentLeft) : default(TValue); var currentRight = enumRight.Current; var valueRight = hasRight ? rightKeySelector(currentRight) : default(TValue); int compare = !hasLeft ? 1 : !hasRight ? -1 : keyComparer.Compare(valueLeft, valueRight); switch (compare) { case 0: // The selector matches. An inner join is achieved yield return selector(currentLeft, currentRight); hasLeft = enumLeft.MoveNext(); hasRight = enumRight.MoveNext(); break; case -1: yield return selector(currentLeft, default(TRight)); hasLeft = enumLeft.MoveNext(); break; case 1: yield return selector(default(TLeft), currentRight); hasRight = enumRight.MoveNext(); break; } } } }