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def combinations(occurrences: Occurrences): List[Occurrences] = occurrences match {  
    case List() => List(List())
    case occ :: occs =>
        for {
            **occSub <- (0 to occ._2).map((occ._1, _)).toList**
            occsCombination <- combinations(occs)
            } yield (occSub :: occsCombination).filter(x => x._2 != 0)
}
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1 回答 1

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.map((occ._1, _))是 的缩写.map(i => (occ._1, i))

对于 0 和 之间的每个元素occ._2,它创建一个Tuple如上。所以这会返回一个元组列表,其中第一个元素固定,第二个元素从 0 到occ._2.

例如:

scala> val occ = (42,5)
occ: (Int, Int) = (42,5)

scala> (0 to occ._2).map(i => (occ._1, i)).toList
res0: List[(Int, Int)] = List((42,0), (42,1), (42,2), (42,3), (42,4), (42,5))

scala> (0 to occ._2).map((occ._1, _)).toList
res1: List[(Int, Int)] = List((42,0), (42,1), (42,2), (42,3), (42,4), (42,5))
于 2013-05-17T11:40:18.477 回答