展平列表清单 [英] Flatten a list of lists

问题描述

例如 flatten [] = [] 或 flatten [1,2,3,4] = [1,2,3,4] 或 flatten [[1,2] ，[3]，4,5]] = [1,2,3,4,5]

我遇到问题可以根据扁平函数赋予什么类型来匹配类型。

以下是我的：

pre> 数据A a = B a | C [a]导出（Show，Eq，Ord）

flatten：:(显示a，等式a，ord a）=> A a - > a
flatten（C []）=（C []）
flatten（C（x：xs））=（C flatten x）++（C flatten xs）
flatten B我）可以说这个问题是，这个 ++ 操作符期望它的两个参数都有一个列表，我试图给它一些 A 类型的东西。我添加了 A 类型，因此该函数可以获取单个元素或元素列表。

有谁知道以不同的方式来做到这一点，或解释我可以做些什么来解决类型错误？

解决方案

这个问题有点不清楚，但是列出一个列表是一个名为<$ c $的标准函数c> concat 在类型签名的前奏中 [[a]] - > [α] 。

如果您从上面开始创建嵌套列表的数据类型，则可能需要将数据类型调整为如下所示：

 数据列出a = List [a] | ListOfLists [列表a] 
  

然后你可以把它们压扁成一个列表;

  flatten ::列出a  - > [a] 
 flatten（List xs）= xs 
 flatten（ListOfLists xss）= concatMap flatten xss 
  

作为一项测试，

 > flatten（ListOfLists [List [1,2]，List [3]，ListOfLists [List [4]，List [5]]]）
 [1,2,3,4,5] 
  

I have to write a function that flattens a list of lists.

For example flatten [] = [] or flatten [1,2,3,4] = [1,2,3,4] or flatten [[1,2],[3],4,5]] = [1,2,3,4,5]

I'm having trouble with the being able to match the type depending on what is given to the flatten function.

Here's what I have:

data A a = B a | C [a] deriving (Show, Eq, Ord)

flatten::(Show a, Eq a, Ord a)=>A a -> A a
flatten (C []) = (C [])
flatten (C (x:xs) ) = (C flatten x) ++ (C flatten xs)
flatten (B a) = (C [a])

From what I can tell the issue is that the ++ operator is expecting a list for both of its arguments and I'm trying to give it something of type A. I've added the A type so the function can either get a single element or a list of elements.

Does anyone know a different way to do this differently, or explain what I can do to fix the type error?

解决方案

It's a bit unclear what you are asking for, but flattening a list of list is a standard function called concat in the prelude with type signature [[a]] -> [a].

If you make a data type of nested lists as you have started above, maybe you want to adjust your data type to something like this:

 data Lists a = List [a] | ListOfLists [Lists a]

Then you can flatten these to a list;

 flatten :: Lists a -> [a]
 flatten (List xs) = xs
 flatten (ListOfLists xss) = concatMap flatten xss

As a test,

 > flatten (ListOfLists [List [1,2],List [3],ListOfLists [List [4],List[5]]])
 [1,2,3,4,5]

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