Question

我对scala和编程很新（只是为了好玩）而且我正在尝试理解尾递归和集合，但是debugg真的很难。

我有2个名单：

val quoters = List[Map[String,List[String]]]

val quoted = List[Map[String,List[String]]]

前：

val quoters = List(Map("author1"->List("1","7","8")),Map("author2"->List("2","4","6","3")),Map("author3"->List("5","2","1","3")))
val quoted = List(Map("author1"->List("5","6")),Map("author2"->List("5","8","1")),Map("author3"->List("4")))

“引用”引用“引用”和“引用”也引用“引用”。

在示例中：

 author1 quoted author2 with "1" and "8",
 author2 quoted author3 with "4",
 author3 quoted author1 with "5" & author2 with "5" + "1"

我想找到引用“引用”引用“引用”的“引用”圈子......

输出应该是这样的：

val quotesCircle = List(
 Map("quoter"->"author1","receiver"->"author2","quote"->"4"),
 Map("quoter"->"author2","receiver"->"author3","quote"->"2"),
 Map("quoter"->"author3","receiver"->"author1","quote"->"1")
)

我的问题：

1 /我认为我在滥用收藏品（Json似乎太多了......）

2 /我可以与列表列表交叉：

def getintersect(q1:List[List[String]],q2:List[List[String]])={ 
 for(x<-q1;r<-q2; if (x intersect r) != Nil)yield x intersect r
}

但没有地图列表的结构。

3 /我尝试了这个用于递归，但是它没有用，因为......我真的不知道：

def getintersect(q1:List[List[String]],q2:List[List[String]])= {
    def getQuotedFromIntersect(quoteMatching:List[String],quoted:List[List[String]]):List[List[String]]={
     for(x<-q1;r<-q2; if (x intersect r) != Nil)
       getQuotedFromIntersect(x intersect r,quoted)
    }
}

我希望我足够清楚：/

提前谢谢！

菲利克斯

Answer 1

我认为您的数据结构中有一个额外的图层。我可能会使用“作者”这个名称而不是“引用”，以避免混淆，因为引用/引用之间的共鸣：

val quoters = List(
  Map("author1"->List("1","7","8")),
  Map("author2"->List("2","4","6","3")),
  Map("author3"->List("5","2","1","3")))

val authors = Map(
  "author1" -> List(5, 6),
  "author2" -> List(5, 8, 1),
  "author3" -> List(4))

有了这个，还有一个像这样的小功能：

def findQuoters(article: Int): List[String] = {
  quoters.keys.toList filter { name =>
    quoters(name) contains article
  }
}

然后，您可以开始尝试不同的收集和报告方式，引用哪位作者以及在哪里，例如：

// For each name in the 'authors' map:
//    For each article authored by this name:
//        Find all quoters of this article
//        Report that author 'name' has been quoted for article 'id' by 'whom'...

for (name <- authors.keys) {
  for (article <- authors(name)) {
    val qq = findQuoters(article)
    println("author %s has been quoted %d times for article %d by %s".format(
      name, qq.size, article, qq.mkString(" ")))
  }
}

打印输出如下：

author author1 has been quoted 1 times for article 5 by author3
author author1 has been quoted 1 times for article 6 by author2
author author2 has been quoted 1 times for article 5 by author3
author author2 has been quoted 1 times for article 8 by author1
author author2 has been quoted 2 times for article 1 by author1 author3
author author3 has been quoted 1 times for article 4 by author2

现在，想一想你用这些关联数组映射的内容。您可以有效地构建图形的邻接矩阵。你会如何在有向图中找到所有圆圈？

Scala尾递归在2个列表之间相交

1 个答案: