Question

我有问题。我有一项无法自行解决的任务。这是交易：

extPair :: Eq a => [(a,b)] -> [(a,[b])]

如果我想用以下方式打电话：

extPair [(3,"pear"),(3,"fog"),(4,"dog"),(4,"x"),(3,"y")]

应该给出以下结果：

[(3, ["pear", "fog"]), (4, ["dog", "x"]), (3, ["y"])] :: [(Integer, [[Char]])]

我现在所知道的是：

extPair [] = []
extPair (x:xs)
 | length (x:xs) >= 2 && fst x == fst (head xs) = [(fst x, [snd x, snd (head xs)])] ++ extPair (drop 1 xs)
 | otherwise = [(fst x, [snd x])]

这适用于该示例，但它不适用于以下内容，而且我不知道如何制作它，所以我请求您的帮助。我想添加尽可能多的对：

extPair [(3,"pear"),(3,"fog"),(3,"dark"),(3,"etc"), ... ,(4,"dog"),(4,"x"),(3,"y"),(3,"z")]

所以结果应该是：

    [(3, ["pear", "fog", "dark", "etc", "...", "...", ...]), (4, ["dog", "x"]), (3, ["y", "z"])] :: [(Integer, [[Char]])]

Answer 1

使用foldr函数的另一种解决方案：

extPair :: Eq a => [(a,b)] -> [(a,[b])]
extPair = foldr f []
  where
    f (key, elem) [] = [(key, [elem])]
    f (key1, elem) l@((key2, elems):xs)
      | key1 == key2 = (key2,elem:elems):xs
      | otherwise    = (key1,[elem]):l

Answer 2

您可以先使用groupBy预处理列表：

> import Data.List
> import Data.Function
> groupBy ((==) `on` fst) [(3,"pear"),(3,"fog"),(4,"dog"),(4,"x"),(3,"y")]
[[(3,"pear"),(3,"fog")],[(4,"dog"),(4,"x")],[(3,"y")]]

从那里开始，使用map f获取合适的f。

如果您想编写自己的代码，请考虑递归：

extPair ((n,s):xs) = ???
   where ys = extPair xs

此处的逻辑如下：如果ys以(n,zs)开头，则将s添加到zs，否则将(n,[s])添加到ys }。

Haskell对数和列表

2 个答案: