两个列表的笛卡尔积 [英] Cartesian product of two lists

问题描述

给出一个将数字与多个字符相关联的地图

Given a map where a digit is associated to several characters

scala> val conversion = Map("0" -> List("A", "B"), "1" -> List("C", "D"))
conversion: scala.collection.immutable.Map[java.lang.String,List[java.lang.String]] =
  Map(0 -> List(A, B), 1 -> List(C, D))

我想基于数字序列生成所有可能的字符序列.例子:

I want to generate all possible character sequences based on a sequence of digits. Examples:

"00" -> List("AA", "AB", "BA", "BB")
"01" -> List("AC", "AD", "BC", "BD")

我可以做到这一点

scala> val number = "011"
number: java.lang.String = 011

为每个索引创建可能的字符序列

Create a sequence of possible characters per index

scala> val values = number map { case c => conversion(c.toString) }
values: scala.collection.immutable.IndexedSeq[List[java.lang.String]] =
  Vector(List(A, B), List(C, D), List(C, D))

生成所有可能的字符序列

Generate all the possible character sequences

scala> for {
     | a <- values(0)
     | b <- values(1)
     | c <- values(2)
     | } yield a+b+c
res13: List[java.lang.String] = List(ACC, ACD, ADC, ADD, BCC, BCD, BDC, BDD)

在这里情况变得很丑陋，并且仅适用于三位数的序列.有什么方法可以在任何长度的序列上达到相同的结果吗?

Here things get ugly and it will only work for sequences of three digits. Is there any way to achieve the same result for any sequence length?

推荐答案

以下建议不使用理解.但我认为这毕竟不是一个好主意，因为正如您所注意到的，您将被束缚在一定长度的笛卡尔积上.

The following suggestion is not using a for-comprehension. But I don't think it's a good idea after all, because as you noticed you'd be tied to a certain length of your cartesian product.

scala> def cartesianProduct[T](xss: List[List[T]]): List[List[T]] = xss match {
     |   case Nil => List(Nil)
     |   case h :: t => for(xh <- h; xt <- cartesianProduct(t)) yield xh :: xt
     | }
cartesianProduct: [T](xss: List[List[T]])List[List[T]]

scala> val conversion = Map('0' -> List("A", "B"), '1' -> List("C", "D"))
conversion: scala.collection.immutable.Map[Char,List[java.lang.String]] = Map(0 -> List(A, B), 1 -> List(C, D))

scala> cartesianProduct("01".map(conversion).toList)
res9: List[List[java.lang.String]] = List(List(A, C), List(A, D), List(B, C), List(B, D))

为什么不尾递归?

请注意，上述递归功能不是不是尾递归.这不是问题，因为xss会很短，除非xss中有很多单例列表.之所以如此，是因为结果的大小随xss的非单个元素的数量呈指数增长.

Why not tail-recursive?

Note that above recursive function is not tail-recursive. This isn't a problem, as xss will be short unless you have a lot of singleton lists in xss. This is the case, because the size of the result grows exponentially with the number of non-singleton elements of xss.

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