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合并两个列表的LINQ(复合键上的完全外部连接)

linq c#

Visionary Software Solutions · 技术社区 · 14 年前

我有两张单子

 IEnumerable<Citrus> grapefruit = citrusList.Where(x => x.IsSmall == false);
 IEnumerable<Citrus> tangerines = citrusList.Where(x => x.IsSmall == true);

现在我已经嵌套了foreach循环来检查

 foreach (Citrus fruit in grapefruit)
 {
    foreach (Citrus fruitToo in tangerines)
    {
       PackingContainer container = new PackingContainer();
       if (fruit.Color == fruitToo.Color && 
           fruit.Flavor == fruitToo.Flavor && 
           fruit.Texture == fruitToo.Texture && 
           fruit.State == fruitToo.State)
           { 
              Tangelo tangy = new Tangelo(fruit.Color, fruit.Flavor, fruit.Texture, fruit.State, "A tangelo", new Decimal(0.75);
              container.Add(tangy);
           }
     }
  }

但我相信有更好的办法。我基本上想做一个完全的外部连接(将所有的葡萄柚和橘子联合起来,但要使橘子离开交叉点)。我的最终目标是有一个包装容器,里面有一些葡萄柚,一些橘子,和一些橘子。我相信在林肯有一个更优雅的方法。

……但我无法从 http://msdn.microsoft.com/en-us/library/bb907099.aspx 和 http://msdn.microsoft.com/en-us/library/bb384063.aspx

没什么帮助?

3 回复 | 直到 14 年前

Amy B 14 年前

你不想要一个完整的外部连接,否则你会做一些没有葡萄柚的橘子和一些没有橘子的橘子。

这是一个内部连接。

List<Tangelo> tangelos = (
from fruit in grapefruit
join fruitToo in tangerines
  on new {fruit.Flavor, fruit.Color, fruit.Flavor, fruit.State}
  equals new {fruitToo.Flavor, fruitToo.Color, fruitToo.Flavor, fruitToo.State}
select new Tangelo(fruit.Color, fruit.Flavor, fruit.Texture, fruit.State,
  "A tangelo", new Decimal(0.75))
).ToList()

尝试此筛选以每个橘子仅获取一个橘子:

List<Tangelo> tangelos = (
from fruit in tangerines
where grapefruit.Any(fruitToo => 
  new {fruit.Flavor, fruit.Color, fruit.Flavor, fruit.State}
  == new {fruitToo.Flavor, fruitToo.Color, fruitToo.Flavor, fruitToo.State})
select new Tangelo(fruit.Color, fruit.Flavor, fruit.Texture, fruit.State,
  "A tangelo", new Decimal(0.75))
).ToList()

container.AddRange(tangelos);

diceguyd30 14 年前

实际上听起来你需要一个内部连接,而不是外部连接。嵌套for循环实际上执行的是等效于内部联接的操作。无论如何:

grapefruit
 .Join(
  tangerines,
  x => new { Color = x.Color, Flavor = x.Flavor, Texture = x.Texture, State = x.State },
  x => new { Color = x.Color, Flavor = x.Flavor, Texture = x.Texture, State = x.State },
  (o,i) => new Tangelo(o.Color, o.Flavor, o.Texture, o.State, "A tangelo", new Decimal(0.75))
 ).Map(x => container.Add(x));

其中“Map”是IEnumerables的“ForEach”式扩展方法:

public static void Map<T>(this IEnumerable<T> source, Action<T> func)
{
    foreach (T i in source)
        func(i);
}

编辑:很公平。从这个问题看来,你好像只对坦格洛斯一家感兴趣。这是一个外部连接版本(这是未经测试的,如果有什么不起作用,请告诉我!):

var q =
from fruit in grapefruit.Select(x => new { x.Color, x.Flavor, x.Texture, x.State })
   .Union(tangerines.Select(x => new { x.Color, x.Flavor, x.Texture, x.State }))
join g in grapefruit on fruit equals new { g.Color, g.Flavor, g.Texture, g.State } into jg
from g in jg.DefaultIfEmpty()
join t in tangerines on fruit equals new { t.Color, t.Flavor, t.Texture, t.State } into jt
from t in jt.DefaultIfEmpty()
select  (g == null ? 
   t as Citrus : 
   (t == null ? 
    g as Citrus : 
    new Tangelo(g.Color, g.Flavor, g.Texture, g.State, "A tangelo", new Decimal(0.75)) as Citrus
   )
  );

然后可以使用David B的答案中的map方法或AddRange方法将它们添加到容器中。

Enigmativity 14 年前

我想这就是诀窍:

var cs = from c in citrusList
         group c by new { c.Color, c.Flavor, c.Texture, c.State } into gcs
         let gs = gcs.Where(gc => gc.IsSmall == false)
         let ts = gcs.Where(gc => gc.IsSmall == true)
         let Tangelos = gs
            .Zip(ts, (g, t) =>
                new Tangelo(g.Color, g.Flavor, g.Texture, g.State,
                    "A tangelo", new Decimal(0.75)))
         select new
         {
             gcs.Key,
             Grapefruit = gs.Skip(Tangelos.Count()),
             Tangerines = ts.Skip(Tangelos.Count()),
             Tangelos,
         };

var container = new PackingContainer();

container.AddRange(from c in cs
                   from f in c.Grapefruit
                       .Concat(c.Tangerines)
                       .Concat(c.Tangelos.Cast<Citrus>())
                   select f);