在Haskell中從列表中添加字符串並添加特殊字符

Question

在Haskell中，我曾經使用過濾器從列表中剝離數字集。 但是，我無法使其適用於這種特殊情況。

我有一個字符串列表，如下所示：

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

我想將[]和""字符串化，所以最后一個字符串是（帶空格）：

A B C D E F A B N

不應使用像print filter([) ["A","B","C","D","E","F","A","B","N"]這樣的簡單過濾print filter([) ["A","B","C","D","E","F","A","B","N"]刪除[ ？

更新：

我閱讀了這份文件 ，並得到以下結果：

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，當我使用這個：

-- 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

我得到這個錯誤

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正常工作：

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

另外， Data.List.intersperse和concat也會有所幫助。

import Data.List

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

這是用" "分隔列表中的每個值，然后將列表連接（連接）在一起。

如果要用", "分隔，則可以輕松更改上述功能：

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

以便：

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

要將其放入您的IO上下文中：

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