list 展平列表列表
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Flatten a list of lists
提问by captainpirate
I have to write a function that flattens a list of lists.
我必须编写一个函数来展平列表列表。
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]
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]
I'm having trouble with the being able to match the type depending on what is given to the flatten function.
我无法根据提供给 flatten 函数的内容来匹配类型。
Here's what I have:
这是我所拥有的：
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 Atype so the function can either get a single element or a list of elements.
从我可以看出的问题是，++操作员期望它的两个参数都有一个列表，而我正试图给它一些 type A。我添加了A类型，以便函数可以获取单个元素或元素列表。
Does anyone know a different way to do this differently, or explain what I can do to fix the type error?
有谁知道以不同的方式执行此操作的不同方法，或解释我可以做些什么来修复类型错误？
回答by Malin
It's a bit unclear what you are asking for, but flattening a list of list is a standard function called concatin the prelude with type signature [[a]] -> [a].
有点不清楚您在要求什么，但是展平列表列表是concat在 prelude 中调用的标准函数，类型为 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:
如果您按照上面开始的方式创建嵌套列表的数据类型，也许您想将数据类型调整为如下所示：
data Lists a = List [a] | ListOfLists [Lists a]
Then you can flatten these to a list;
然后你可以将这些扁平化为一个列表；
flatten :: Lists a -> [a]
flatten (List xs) = xs
flatten (ListOfLists xss) = concatMap flatten xss
As a test,
作为测试，
> flatten (ListOfLists [List [1,2],List [3],ListOfLists [List [4],List[5]]])
[1,2,3,4,5]
回答by Mike T
Firstly, the A type is on the right track but I don't think it's quite correct. You want it to be able to flatten arbitrarily nested lists, so a value of type "A a" should be able to contain values of type "A a":
首先，A 型走在正确的轨道上，但我认为这不太正确。您希望它能够展平任意嵌套的列表，因此“A a”类型的值应该能够包含“A a”类型的值：
data A a = B a | C [A a]
Secondly, the type of the function should be slightly different. Instead of returning a value of type "A a", you probably want it to return just a list of a, since by definition the function is always returning a flat list. So the type signature is thus:
其次，函数的类型应该略有不同。您可能希望它只返回 a 的列表，而不是返回“A a”类型的值，因为根据定义，该函数始终返回一个平面列表。所以类型签名是这样的：
flatten :: A a -> [a]
Also note that no typeclass constraints are necessary -- this function is completely generic since it does not look at the list's elements at all.
还要注意，不需要类型类约束——这个函数是完全通用的，因为它根本不查看列表的元素。
Here's my implementation:
这是我的实现：
flatten (B a) = [a]
flatten (C []) = []
flatten (C (x:xs)) = flatten x ++ flatten (C xs)
回答by A_P
this one liner will do the job. Although as it was mentioned by Malin the type signature is different:
这一班轮将完成这项工作。尽管正如 Malin 所提到的，类型签名是不同的：
flatten :: [[a]] -> [a]
flatten xs = (\z n -> foldr (\x y -> foldr z y x) n xs) (:) []
simple test
简单的测试
frege> li = [[3,4,2],[1,9,9],[5,8]]
frege> flatten li
[3,4,2,1,9,9,5,8]
