字符串并从 Haskell 的列表中添加特殊字符

Question

在 Haskell 中，我曾经使用过滤器从列表中删除一组数字。但是，我无法使其适用于这种特殊情况。

我有一个字符串列表如下：

["A","B","C","D","E","F","A","B","N"]

我想将[] 和"" 串起来，所以最后的字符串是（带空格）：

A B C D E F A B N

不应该像print filter([) ["A","B","C","D","E","F","A","B","N"] 这样的简单过滤器删除[ 吗？

更新：

我阅读了this document 并且能够得到以下结果：

let example = (concat (intersperse " " ["A","B","C","D","E","F","A","B","N"]))
print example
-- this prints "A B C D E F A B N"

但是，当我使用这个时：

-- where createalphs create a list of strings
-- and userinput is a string entered by the user 

let setofalph = ($ createalphs $ words userinput)
let example = (concat (intersperse " " setofalph))
print example

我收到这个错误

Couldn't match expected type `[[Char]]'
In the second argument of `intersperse', namely `setofalph'
In the first argument of `concat', namely
  `(intersperse " " setofalph)'
In the expression: (concat (intersperse " " setofalph))

Answer 1

unwords 工作正常：

λ> unwords ["A","B","C","D","E","F","A","B","N"]
"A B C D E F A B N"

此外，Data.List.intersperse 和 concat 也会有所帮助。

import Data.List

solution :: [String] -> String
solution =  concat . intersperse " "

这样做是用" " 分隔列表中的每个值，然后将列表连接（连接）在一起。

如果你想和", "分开，你可以很容易地改变上面的函数：

solution :: [String] -> String
solution =  concat . intersperse ", "

这样：

λ> solution ["A","B","C","D","E","F","A","B","N"]
"A, B, C, D, E, F, A, B, N"
λ> putStrLn $ solution ["A","B","C","D","E","F","A","B","N"]
A, B, C, D, E, F, A, B, N

将其放入 IO 的上下文中：

main = do
  x <- getLine
  putStrLn $ solution $ createalphs $ words x