scala 在数组中查找索引的功能代码
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Functional code for finding index in array
提问by Andriy Buday
I would like to refactor this piece of Scalacode in functional style:
我想以函数式的方式重构这段Scala代码：
var k = -1
for (i <- 0 until array.length)
if ((i < array.length - 1) && array(i) < array(i + 1))
k = i
Array in Scalahas indexWhere, which can be used for something like val index = array.indexWhere(c => c == 'a'). I'm looking for something similar, which would take into account two sequential elements of array.
Scala中的数组具有indexWhere，可用于诸如val index = array.indexWhere(c => c == 'a'). 我正在寻找类似的东西，它会考虑数组的两个连续元素。
回答by Travis Brown
When you need to look at adjacent elements in a collection, the usual functional approach is to "zip" the collection with its tail. Consider the following simplified example:
当您需要查看集合中的相邻元素时，通常的功能方法是使用其尾部“压缩”集合。考虑以下简化示例：
scala> val xs = List(5, 4, 2, 3, 1)
xs: List[Int] = List(5, 4, 2, 3, 1)
scala> val tail = xs.tail
tail: List[Int] = List(4, 2, 3, 1)
scala> xs.zip(tail)
res0: List[(Int, Int)] = List((5,4), (4,2), (2,3), (3,1)
Now we can use indexWhere:
现在我们可以使用indexWhere：
scala> res0.indexWhere { case (x, y) => x < y }
res1: Int = 2
In your case, the following is essentially equivalent to your code:
在您的情况下，以下内容基本上等同于您的代码：
val k = (array zip array.tail) lastIndexWhere { case (x, y) => x < y }
I'm using lastIndexWhereinstead of indexWhere, since in your code you don't stop the loop when you hit a pair for which the predicate holds.
我使用的是lastIndexWhere代替indexWhere，因为在您的代码中，当您命中谓词成立的一对时，您不会停止循环。
回答by Adam Rabung
slidinggives you sliding window into the collection, ie
sliding给你滑动窗口进入集合，即
scala> Array(1,2,2,4,5,6, 6).sliding(2).toList
res12: List[Array[Int]] = List(Array(1, 2), Array(2, 2), Array(2, 4), Array(4, 5), Array(5, 6), Array(6, 6))
So that would make finding the index of the first matching pair easy:
这样就可以轻松找到第一个匹配对的索引：
Array(1,2,2,4,5,6, 6).sliding(2).indexWhere { case Array(x1, x2) => x1 == x2 }
That only gives you the first index, use collectto catch em all!
那只给你第一个索引，collect用来捕获它们！
Array(1,2,2,4,5,6, 6)
.sliding(2) //splits each in to pairs
.zipWithIndex //attaches the current index to each pair
.collect { case (Array(x1, x2), index) if (x1 == x2) => index } //collect filters out non-matching pairs AND transforms them to just the inde
