如何在Haskell中遍历列表时获取当前元素的前后元素？

Haskell标准库中没有直接提供你想要的那种接收三参数lambda的map变体，但可以通过以下几种方式实现需求：

方法1：用zip3组合偏移列表（推荐处理边界情况） #

通过构造三个偏移后的列表，再用zip3将它们组合成包含前序、当前、后序元素的三元组列表。为了处理首尾元素没有前序/后序的情况，我们用Maybe类型标记不存在的邻居：

withNeighbors :: [a] -> [(Maybe a, a, Maybe a)]
withNeighbors [] = []
withNeighbors list = zip3 prevList curList nextList
  where
    prevList = Nothing : map Just (init list)
    curList = list
    nextList = map Just (tail list) ++ [Nothing]

使用示例：

-- 处理[1,2,3,4,5]
processElements :: [(Maybe Int, Int, Maybe Int)] -> [String]
processElements = map (\(prev, cur, next) -> 
  "当前元素: " ++ show cur 
  ++ ", 前序元素: " ++ show prev 
  ++ ", 后序元素: " ++ show next)

main = print $ processElements $ withNeighbors [1,2,3,4,5]

输出结果：

["当前元素: 1, 前序元素: Nothing, 后序元素: Just 2",
 "当前元素: 2, 前序元素: Just 1, 后序元素: Just 3",
 "当前元素: 3, 前序元素: Just 2, 后序元素: Just 4",
 "当前元素: 4, 前序元素: Just 3, 后序元素: Just 5",
 "当前元素: 5, 前序元素: Just 4, 后序元素: Nothing"]

方法2：自定义递归函数（灵活处理边界默认值） #

如果不想用Maybe，而是想给首尾元素指定默认的前序/后序值，可以写递归函数：

withNeighbors' :: a -> a -> [a] -> [(a, a, a)]
withNeighbors' _ _ [] = []
withNeighbors' prevDef nextDef [x] = [(prevDef, x, nextDef)]
withNeighbors' prevDef nextDef (x:y:xs) = (prevDef, x, y) : go x y xs
  where
    go prev cur [] = [(prev, cur, nextDef)]
    go prev cur (n:ns) = (prev, cur, n) : go cur n ns

使用示例：

-- 给首尾指定默认值0
main = print $ withNeighbors' 0 0 [1,2,3,4,5]

输出结果：

[(0,1,2),(1,2,3),(2,3,4),(3,4,5),(4,5,0)]

方法3：滑动窗口方式（利用tails） #

通过Data.List.tails生成所有后缀，再提取每个后缀的前三个元素作为窗口，适合需要批量处理连续三元组的场景：

import Data.List (tails)

slidingWindow :: [a] -> [(Maybe a, Maybe a, Maybe a)]
slidingWindow = map takeWindow . tails
  where
    takeWindow [] = (Nothing, Nothing, Nothing)
    takeWindow [x] = (Nothing, Just x, Nothing)
    takeWindow [x,y] = (Nothing, Just x, Just y)
    takeWindow (x:y:z:_) = (Just x, Just y, Just z)

内容的提问来源于stack exchange，提问作者Joe