在 Scala 中得分和求和的最佳方式？
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Best way to score and sum in Scala?
提问by adam77
Is there a better way of doing this:
有没有更好的方法来做到这一点：
val totalScore = set.foldLeft(0)( _ + score(_) )
or this:
或这个：
val totalScore = set.toSeq.map(score(_)).sum
I think it's quite a common operation so was expecting something sleeker like:
我认为这是一个很常见的操作，所以期待更时尚的东西，比如：
val totalScore = set.sum( score(_) )
回答by Daniel C. Sobral
Well, there are alternative ways to write it:
好吧，还有其他的写法：
val totalScore = set.toSeq.map(score(_)).sum
val totalScore = set.toSeq.map(score).sum
val totalScore = set.toSeq map score sum
The last one may require a semi-colon at the end if the next line doesn't start with a keyword. One can also use .viewinstead of .toSeq, which would avoid allocating a temporary collection. However, I'm not sure the .view's present behavior (of showing repeated elements) is the correct one.
如果下一行不以关键字开头，则最后一行可能需要在末尾使用分号。也可以使用.view代替.toSeq，这将避免分配临时集合。但是，我不确定.view当前的行为（显示重复元素）是否正确。
回答by Michel Kr?mer
Seq.sumdoes not take a function which could be used to score the sum. You could define an implicit conversion which "pimps" Traversable:
Seq.sum不采用可用于对总和进行评分的函数。您可以定义“皮条客”的隐式转换Traversable：
implicit def traversableWithSum[A](t: Traversable[A])(implicit m: Numeric[A]) = new {
def sumWith(f: A => A) = t.foldLeft(m.zero)((a, b) => m.plus(a, f(b)))
}
def score(i: Int) = i + 1
val s = Set(1, 2, 3)
val totalScore = s.sumWith(score _)
println(totalScore)
=> 9
Please note that the Numerictrait only exists in Scala 2.8.
请注意，该Numeric特征仅存在于 Scala 2.8 中。
回答by Randall Schulz
Simpler:
更简单：
scala> val is1 = Set(1, 4, 9, 16)
is1: scala.collection.immutable.Set[Int] = Set(1, 4, 9, 16)
scala> is1.reduceLeft(_ + _)
res0: Int = 30
With your score method:
使用您的评分方法：
scoreSet.reduceLeft(_ + score(_))
Warning, though, this fails is the collection being reduced is empty while fold does not:
警告，但是，这失败是减少的集合是空的，而 fold 不是：
scala> val is0 = Set[Int]()
is0: scala.collection.immutable.Set[Int] = Set()
scala> is0.foldLeft(0)(_ + _)
res1: Int = 0
回答by Randall Schulz
Alternately, the Seq#sumoverload that takes an implicit conversion to Numericcould be used if the type in the collection to be scored / summed does not itself have an addition operator. However, because it's an implicit conversion parameter, it won't be applied unless required to make the reduce closure type-check.
或者，如果要评分/求和的集合中的类型本身没有加法运算符，则可以使用采用Seq#sum隐式转换的重载Numeric。但是，因为它是一个隐式转换参数，除非需要进行reduce 闭包类型检查，否则不会应用它。
