给出以下列表:
List(List(1,2,3), List(4,5))
我想生成所有可能的组合。使用yield
,可以按如下方式完成:
scala> for (x <- l.head; y <- l.last) yield (x,y)
res17: List[(Int, Int)] = List((1,4), (1,5), (2,4), (2,5), (3,4), (3,5))
但我遇到的问题是List [List [Int]]没有修复;它可以增大和缩小,所以我永远不知道我需要提前多少for
个循环。我想要的是能够将该列表传递给一个函数,该函数将动态生成组合而不管我有多少列表,所以:
def generator (x : List[List[Int]) : List[List[Int]]
是否有可以执行此操作的内置库函数。如果不是我该怎么做呢。任何指针和提示都会很棒。
更新:
@DNA的答案使用以下(不是那么大的)嵌套List
结构来打击堆:
List(
List(0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220, 225, 230, 235, 240, 245, 250, 255, 260, 265, 270, 275, 280, 285, 290, 295, 300),
List(0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300),
List(0, 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300),
List(0, 50, 100, 150, 200, 250, 300),
List(0, 100, 200, 300),
List(0, 200),
List(0)
)
调用generator2函数如下:
generator2(
List(
List(0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220, 225, 230, 235, 240, 245, 250, 255, 260, 265, 270, 275, 280, 285, 290, 295, 300),
List(0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300),
List(0, 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240, 260, 280, 300),
List(0, 50, 100, 150, 200, 250, 300),
List(0, 100, 200, 300),
List(0, 200),
List(0)
)
)
有没有办法在不吹堆的情况下生成笛卡尔积?
Exception in thread "main" java.lang.OutOfMemoryError: Java heap space
at scala.LowPriorityImplicits.wrapRefArray(LowPriorityImplicits.scala:73)
at recfun.Main$.recfun$Main$$generator$1(Main.scala:82)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.immutable.List.foreach(List.scala:318)
at scala.collection.TraversableLike$class.flatMap(TraversableLike.scala:251)
at scala.collection.AbstractTraversable.flatMap(Traversable.scala:105)
at recfun.Main$.recfun$Main$$generator$1(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.immutable.List.foreach(List.scala:318)
at scala.collection.TraversableLike$class.flatMap(TraversableLike.scala:251)
at scala.collection.AbstractTraversable.flatMap(Traversable.scala:105)
at recfun.Main$.recfun$Main$$generator$1(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.immutable.List.foreach(List.scala:318)
at scala.collection.TraversableLike$class.flatMap(TraversableLike.scala:251)
at scala.collection.AbstractTraversable.flatMap(Traversable.scala:105)
at recfun.Main$.recfun$Main$$generator$1(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at recfun.Main$$anonfun$recfun$Main$$generator$1$1.apply(Main.scala:83)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.TraversableLike$$anonfun$flatMap$1.apply(TraversableLike.scala:251)
at scala.collection.immutable.List.foreach(List.scala:318)
at scala.collection.TraversableLike$class.flatMap(TraversableLike.scala:251)
答案 0 :(得分:9)
这是一个递归解决方案:
def generator(x: List[List[Int]]): List[List[Int]] = x match {
case Nil => List(Nil)
case h :: _ => h.flatMap(i => generator(x.tail).map(i :: _))
}
产生:
val a = List(List(1, 2, 3), List(4, 5))
val b = List(List(1, 2, 3), List(4, 5), List(6, 7))
generator(a) //> List(List(1, 4), List(1, 5), List(2, 4),
//| List(2, 5), List(3, 4), List(3, 5))
generator(b) //> List(List(1, 4, 6), List(1, 4, 7), List(1, 5, 6),
//| List(1, 5, 7), List(2, 4, 6), List(2, 4, 7),
//| List(2, 5, 6), List(2, 5, 7), Listt(3, 4, 6),
//| List(3, 4, 7), List(3, 5, 6), List(3, 5, 7))
更新:第二个case
也可以写成for
理解,这可能会更清楚一些:
def generator2(x: List[List[Int]]): List[List[Int]] = x match {
case Nil => List(Nil)
case h :: t => for (j <- generator2(t); i <- h) yield i :: j
}
对于较大的数据集,更新2:,如果内存不足,则可以使用Streams(如果有意义地逐步处理结果)。例如:
def generator(x: Stream[Stream[Int]]): Stream[Stream[Int]] =
if (x.isEmpty) Stream(Stream.empty)
else x.head.flatMap(i => generator(x.tail).map(i #:: _))
// NB pass in the data as Stream of Streams, not List of Lists
generator(input).take(3).foreach(x => println(x.toList))
>List(0, 0, 0, 0, 0, 0, 0)
>List(0, 0, 0, 0, 0, 200, 0)
>List(0, 0, 0, 0, 100, 0, 0)
答案 1 :(得分:2)
感觉您的问题可以用递归来描述:
如果你有n个int列表: list1的大小为m和list2,... list n
所以用List(List(1,2),List(3),List(4,5)) 递归调用的结果是List(List(3,4),List(3,5)),并为每个添加2个组合:List(1,3,4),List(2,3,4),List (1,3,5),列表(2,3,5)。
答案 2 :(得分:1)
以西结正是我想要的东西。这只是使其变为通用的一个小调整。
def generateCombinations[T](x: List[List[T]]): List[List[T]] = {
x match {
case Nil => List(Nil)
case h :: _ => h.flatMap(i => generateCombinations(x.tail).map(i :: _))
}
}