使用Linq和C#创建列表中项目的所有可能组合

时间:2012-05-09 15:52:43

标签: c# linq algorithm list

我有一个类别表:

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

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其显示在以下标签中:

 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:可能会有更多或更少的类别。这不是预先确定的。

有什么建议吗? 谢谢

2 个答案:

答案 0 :(得分:20)

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

使用CartesianProduct Extension Method中的Eric Lippert's Blog

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})); 
}

示例:

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

答案 1 :(得分:12)

尝试交叉加入

var combo = from l1 in LIst1
            from l2    in List2
            select new {l1, l2};