你可以创建这样的扩展方法:

public static IEnumerable<TResult> LeftOuterJoin<TSource, TInner, TKey, TResult>(this IEnumerable<TSource> source, IEnumerable<TInner> other, Func<TSource, TKey> func, Func<TInner, TKey> innerkey, Func<TSource, TInner, TResult> res)
    {
        return from f in source
               join b in other on func.Invoke(f) equals innerkey.Invoke(b) into g
               from result in g.DefaultIfEmpty()
               select res.Invoke(f, result);
    }

对于一个表Bar与一个表Foo对Foo的连接(左外)。Foo_Id = Bar。Foo_Id用lambda表示:

var qry = Foo.GroupJoin(
          Bar, 
          foo => foo.Foo_Id,
          bar => bar.Foo_Id,
          (x,y) => new { Foo = x, Bars = y })
       .SelectMany(
           x => x.Bars.DefaultIfEmpty(),
           (x,y) => new { Foo=x.Foo, Bar=y});

要实现两个数据集的连接，不需要使用组连接方法。

内连接:

var qry = Foos.SelectMany
            (
                foo => Bars.Where (bar => foo.Foo_id == bar.Foo_id),
                (foo, bar) => new
                    {
                    Foo = foo,
                    Bar = bar
                    }
            );

对于左连接，只需添加DefaultIfEmpty()

var qry = Foos.SelectMany
            (
                foo => Bars.Where (bar => foo.Foo_id == bar.Foo_id).DefaultIfEmpty(),
                (foo, bar) => new
                    {
                    Foo = foo,
                    Bar = bar
                    }
            );

EF和LINQ to SQL正确转换为SQL。 对于LINQ to Objects，最好使用GroupJoin，因为它在内部使用Lookup。但是如果您正在查询DB，那么跳过GroupJoin是AFAIK作为性能。

Personlay对我来说，这种方式比GroupJoin()更具可读性。

将Marc Gravell的答案转换为一个扩展方法，我做了以下工作。

internal static IEnumerable<Tuple<TLeft, TRight>> LeftJoin<TLeft, TRight, TKey>(
    this IEnumerable<TLeft> left,
    IEnumerable<TRight> right,
    Func<TLeft, TKey> selectKeyLeft,
    Func<TRight, TKey> selectKeyRight,
    TRight defaultRight = default(TRight),
    IEqualityComparer<TKey> cmp = null)
{
    return left.GroupJoin(
            right,
            selectKeyLeft,
            selectKeyRight,
            (x, y) => new Tuple<TLeft, IEnumerable<TRight>>(x, y),
            cmp ?? EqualityComparer<TKey>.Default)
        .SelectMany(
            x => x.Item2.DefaultIfEmpty(defaultRight),
            (x, y) => new Tuple<TLeft, TRight>(x.Item1, y));
}

你可以创建这样的扩展方法:

public static IEnumerable<TResult> LeftOuterJoin<TSource, TInner, TKey, TResult>(this IEnumerable<TSource> source, IEnumerable<TInner> other, Func<TSource, TKey> func, Func<TInner, TKey> innerkey, Func<TSource, TInner, TResult> res)
    {
        return from f in source
               join b in other on func.Invoke(f) equals innerkey.Invoke(b) into g
               from result in g.DefaultIfEmpty()
               select res.Invoke(f, result);
    }

由于这似乎是使用方法(扩展)语法的左外连接的事实上的SO问题，我想我应该在当前选择的答案之外添加一个替代方案(至少在我的经验中)，这是我更常见的答案

// Option 1: Expecting either 0 or 1 matches from the "Right"
// table (Bars in this case):
var qry = Foos.GroupJoin(
          Bars,
          foo => foo.Foo_Id,
          bar => bar.Foo_Id,
          (f,bs) => new { Foo = f, Bar = bs.SingleOrDefault() });

// Option 2: Expecting either 0 or more matches from the "Right" table
// (courtesy of currently selected answer):
var qry = Foos.GroupJoin(
                  Bars, 
                  foo => foo.Foo_Id,
                  bar => bar.Foo_Id,
                  (f,bs) => new { Foo = f, Bars = bs })
              .SelectMany(
                  fooBars => fooBars.Bars.DefaultIfEmpty(),
                  (x,y) => new { Foo = x.Foo, Bar = y });

要使用一个简单的数据集来显示差异(假设我们是根据值本身进行连接):

List<int> tableA = new List<int> { 1, 2, 3 };
List<int?> tableB = new List<int?> { 3, 4, 5 };

// Result using both Option 1 and 2. Option 1 would be a better choice
// if we didn't expect multiple matches in tableB.
{ A = 1, B = null }
{ A = 2, B = null }
{ A = 3, B = 3    }

List<int> tableA = new List<int> { 1, 2, 3 };
List<int?> tableB = new List<int?> { 3, 3, 4 };

// Result using Option 1 would be that an exception gets thrown on
// SingleOrDefault(), but if we use FirstOrDefault() instead to illustrate:
{ A = 1, B = null }
{ A = 2, B = null }
{ A = 3, B = 3    } // Misleading, we had multiple matches.
                    // Which 3 should get selected (not arbitrarily the first)?.

// Result using Option 2:
{ A = 1, B = null }
{ A = 2, B = null }
{ A = 3, B = 3    }
{ A = 3, B = 3    }    

选项2对于典型的左外连接定义是正确的，但正如我前面提到的，根据数据集的不同，它通常是不必要的复杂。