使用 Linq 和 C# 在列表中创建所有可能的项目组合 [重复]

Question

我有一个类别表：

 Catid | Desciption
 1 | Color
 2 | Size
 3 | Material

还有一个类别项目表

 Catid | Name
 1 | Red
 1 | Blue
 1 | Green
 2 | Small
 2 | Med
 2 l Large
 3 | Cotton
 3 | Silk

我需要遍历所有项目并将它们显示在这样的标签中：

 Red Small Cotton
 Red Small Silk
 Red Med Cotton
 Red Med Silk
 Red Large Cotton
 Red Large Silk
 Blue Small Cotton
 Blue Small Silk
 Blue Med Cotton
 Blue Med Silk
 Blue Large Cotton
 Blue Large Silk
 Green Small Cotton
 Green Small Silk
 Green Med Cotton
 Green Med Silk
 Green Large Cotton
 Green Large Silk

请注意：类别可能更多或更少。它不是预先确定的。

有什么建议吗？ 谢谢

Answer 1

var result = list.GroupBy(t => t.Id).CartesianProduct();

使用来自Eric Lippert's Blog 的CartesianProduct Extension Method：

static IEnumerable<IEnumerable<T>> CartesianProduct<T>(
  this IEnumerable<IEnumerable<T>> sequences) 
{ 
  IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() }; 
  return sequences.Aggregate( 
    emptyProduct, 
    (accumulator, sequence) => 
      from accseq in accumulator 
      from item in sequence 
      select accseq.Concat(new[] {item})); 
}

---

例子：

var list = new[]
{
    new { Id = 1, Description = "Red"    },
    new { Id = 1, Description = "Blue"   },
    new { Id = 1, Description = "Green"  },
    new { Id = 2, Description = "Small"  },
    new { Id = 2, Description = "Med"    },
    new { Id = 2, Description = "Large"  },
    new { Id = 3, Description = "Cotton" },
    new { Id = 3, Description = "Silk"   },
};

var result = list.GroupBy(t => t.Id).CartesianProduct();

foreach (var item in result)
{
    Console.WriteLine(string.Join(" ", item.Select(x => x.Description)));
}

输出：

Red Small Cotton
Red Small Silk
Red Med Cotton
Red Med Silk
Red Large Cotton
Red Large Silk
Blue Small Cotton
Blue Small Silk
Blue Med Cotton
Blue Med Silk
Blue Large Cotton
Blue Large Silk
Green Small Cotton
Green Small Silk
Green Med Cotton
Green Med Silk
Green Large Cotton
Green Large Silk

Answer 2

尝试使用交叉连接：

var combo = from l1 in List1
            from l2 in List2
            select new {l1, l2};