是否有任何haskell函数来连接列表与分隔符？

Question

是否有一个函数来连接列表的元素与分隔符？例如:

> foobar " " ["is","there","such","a","function","?"]
["is there such a function ?"]

谢谢你的回复!

Answer 1

是的,有:

Prelude> import Data.List
Prelude Data.List> intercalate " " ["is","there","such","a","function","?"]
"is there such a function ?"

intersperse 有点笼统:

Prelude> import Data.List
Prelude Data.List> concat (intersperse " " ["is","there","such","a","function","?"])
"is there such a function ?"

此外,对于您想要与空格字符连接的特定情况,还有unwords:

Prelude> unwords ["is","there","such","a","function","?"]
"is there such a function ?"

unlines工作方式类似,只是字符串使用换行符进行内爆,并且还会在末尾添加换行符.(这使得序列化文本文件非常有用,每个POSIX标准结尾必须有一个尾随换行符)

@CMCDragonkai不确定你究竟是指什么,但是,这些函数都允许任意字符串作为分隔符和元素.例如,`intercalate',"["some","","string"] ="some ,, string"`和`intercalate""["foo","bar"] ="foobar"` (3认同)
“ unlines”会在每行中添加一个换行符，即“ unlines [“ A”，“ B”] =“ A \ nB \ n”`，因此它与插值不同。 (2认同)

Answer 2

使用foldr写一行字并不难

join sep xs = foldr (\a b-> a ++ if b=="" then b else sep ++ b) "" xs
join " " ["is","there","such","a","function","?"]