通过将列表项与其邻居进行比较来对列表项进行分组 [英] Grouping list items by comparing them with their neighbors

问题描述

根据相邻值将值列表分组的最优雅方式是什么?

What is the most elegant way of grouping a list of values into groups based on their neighbor values?

我拥有的更广泛的上下文是有一个行列表，需要将其分组为段落.我想说的是，如果两行之间的垂直差异低于阈值，则它们在同一段落中.

The wider context I have is having a list of lines, that need to be grouped into paragraphs. I want to be able to say that if the vertical difference between two lines is lower than threshold, they are in the same paragraph.

我最终以不同的方式解决了这个问题，但我想知道这里的正确解决方案.

I ended up solving this problem differently, but I'm wondering about the correct solution here.

case class Box(y: Int)
val list = List(Box(y=1), Box(y=2), Box(y=5))

def group(list: List[Box], threshold: Int): List[List[Box]] = ???

val grouped = group(list, 2)
> List(List(Box(y=1), Box(y=2)), List(Box(y=5)))

我看过groupBy()，但这一次只能处理一个元素.我还尝试了一种使用 sliding()，但是从原始集合中检索元素变得很尴尬.

I have looked at groupBy(), but that can only work with one element at a time. I have also tried an approach that involved pre-computing differences using sliding(), but then it becomes awkward to retrieve the elements from the original collection.

推荐答案

这是一个单线.概括类型留给读者作为练习.

It's a one liner. Generalising types left as an exercise for the reader.

使用整数和绝对差异而不是行和间距来避免混乱.

Using ints and absolute difference rather than lines and spacing to avoid clutter.

val zs = List(1,2,4,8,9,10,15,16)  
def closeEnough(a:Int, b:Int) = (Math.abs(b -a) <= 2)

zs.drop(1).foldLeft(List(List(zs.head)))
                      ((acc, e)=> if (closeEnough(e, acc.head.head)) 
                                      (e::acc.head)::acc.tail
                                  else
                                       List(e)::acc)
       .map(_.reverse)
       .reverse

// List(List(1, 2, 4), List(8, 9, 10), List(15, 16))

或者两个衬里以获得轻微的效率提升

Or a two liner for a slight efficiency gain

val ys = zs.reverse
ys.drop(1).foldLeft(List(List(ys.head)))
                ((acc, e)=> if (closeEnough(e, acc.head.head)) 
                   (e::acc.head)::acc.tail
                else
                   List(e)::acc)
// List(List(1, 2, 4), List(8, 9, 10), List(15, 16))

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