如何在scala中使用fold/reduce来实现这一点?

Sou*_*nta 0 reduce scala fold

我上了课

case class PlaceHolder(time: Long, a: Int, b: Int) 
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我有一个PlaceHolder对象列表,我想创建一个对象,其中包含字段a和所有值的总和b.我不在乎time.我相信它可以通过折叠/缩小操作来完成,但无法用两个字段来计算它是如何做到的.

scala> val range = 1 to 10
range: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> val placeholders = for{ i <- range} yield PlaceHolder(i,i,i) 
placeholders: scala.collection.immutable.IndexedSeq[PlaceHolder] = Vector(PlaceHolder(1,1,1), PlaceHolder(2,2,2), PlaceHolder(3,3,3), PlaceHolder(4,4,4), PlaceHolder(5,5,5), PlaceHolder(6,6,6), PlaceHolder(7,7,7), PlaceHolder(8,8,8), PlaceHolder(9,9,9), PlaceHolder(10,10,10))
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Lee*_*Lee 5

placeholders.foldLeft(PlaceHolder(0,0,0))((acc, p) => PlaceHolder(acc.time + p.time, acc.a + p.a, acc.b + p.b))
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要么

placeholders.reduce((acc, p) => PlaceHolder(acc.time + p.time, acc.a + p.a, acc.b + p.b))
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或者你可以添加一个+运算符PlaceHolder:

case class PlaceHolder(time: Long, a: Int, b: Int) {
    def +(other: PlaceHolder) = PlaceHolder(this.time + other.time, this.a + other.a, this.b + other.b)
}
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然后你可以这样做:

placeholders.reduce(_ + _)
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