将列表中的相邻元素放入元组 [英] Put adjacent elements in List into Tuples

问题描述

给出一个元素列表:

xs = [a, b, c, d, ... z]

其中 a，b，c 等是任意值的占位符.我想实现一个功能 adjacents :: [a]->[(a，a)] 产生

where a, b, c etc are placeholders for arbitrary values. I want to implement a function adjacents :: [a] -> [(a, a)] that produces

adjacentValues = [(a, b), (b, c), (c, d), ... (y, z)]

---

在Haskell中，递归定义相当简洁:

---

In Haskell, a recursive definition is reasonably concise:

adjacents :: [a] -> [(a, a)]
adjacents (x:xs) = (x, head xs) : adjacents xs
adjacents [] = []

Purescript稍微冗长一些:

Purescript is a little more verbose:

adjacents :: forall a. List a -> List (Tuple a a)
adjacents list = case uncons list of 
    Nothing -> []
    Just {head: x, tail: xs} -> case head xs of
                                     Just next -> Tuple x next : adjacents xs
                                     Nothing -> []

有没有一种方法可以表达 adjacents 而无需显式递归(使用折叠)?

Is there a way to express adjacents without explicit recursion (using a fold)?

免责声明:这个问题同时具有Purescript和Haskell标签，因为我想向更多的读者开放.我认为答案不取决于haskells惰性评估语义，因此在两种语言中均有效.

Disclaimer: This question has both Purescript and Haskell tags because I want to open it to a broader audience. I reckon an answer not to depend on haskells lazy-evaluation semantics, and therefore to be valid in both languages.

推荐答案

在Haskell中，无需显式递归，您可以使用列表尾部压缩列表.

In Haskell, without explicit recursion, you can zip a list with its tail.

   let a = [1,2,3,4,5,6,7,8,9,0]

   a `zip` tail a

   => [(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9),(9,0)]

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