Question

列表列表包含以下条目。

[
  ["John","Doe","38\r"],
  ["Jane","Doe","35\r"]
]

Person数据类型的构造如下：

data Person = Person { firstName :: String
                     , lastName :: String   
                     , age :: Int  
                     } deriving (Show)

我正在尝试在Haskell中列出人员列表，以便最终的人员列表如下所示：

[
Person {firstName = "John", lastName = "Doe", age = 38},
Person {firstName = "Jane", lastName = "Doe", age = 35}
]

感谢您的帮助！请谢谢！

Answer 1

所以我们需要这种类型的函数：

f :: [[String]] -> [Person]

有一个功能可以帮助我们-map

stringsToPerson :: [String] -> Person
f = map stringsToPerson

现在，一个基本问题变得很清楚了-stringsToPerson不是类型安全的。用户可以给它一个空列表，该函数别无选择，只能返回一个错误。更好的类型是：

stringsToPerson :: (String, String, Int) -> Person
stringsToPerson :: [String] -> Maybe Person

Buuut假设您要坚持选择的类型，则可以通过模式匹配来实现：

stringsToPerson [firstName, lastName, age] = Person firstName lastName (read age)

Answer 2

您可以执行以下操作；

data Person = Person { firstName :: String
                     , lastName :: String
                     , age :: Int
                     } deriving (Show)


mapper :: [[String]] -> [Person]
mapper = map makePerson
         where makePerson [f,l,a] = Person f l $ read a

*Main> mapper [["John","Doe","38\r"], ["Jane","Doe","35\r"]]
[Person {firstName = "John", lastName = "Doe", age = 38},Person {firstName = "Jane", lastName = "Doe", age = 35}]

Haskell-从列表列表中产生人员列表

2 个答案: