Question

是否有一个函数可以使用分隔符连接列表的元素？例如：

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

感谢您的回复！

Answer 1

是的，there is：

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，对于您想要加入空格字符的特定情况，还有unwords：

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

unlines的工作方式类似，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很有用，每个POSIX标准必须以尾随换行符结束）

Answer 2

如果有人感兴趣，还可以使用一些其他的散布和插入实现方案：

myIntersperse :: a -> [a] -> [a]
myIntersperse _ [] = []
myIntersperse e xs = init $ xs >>= (:[e])

myIntercalate :: [a] -> [[a]] -> [a]
myIntercalate e xs = concat $ myIntersperse e xs

xs >>= f等效于concat (map f xs)。

Answer 3

joinBy sep cont = drop (length sep) $ concat $ map (\w -> sep ++ w) cont

Answer 4

使用foldr编写单行并不难。

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

Answer 5

如果您想编写自己的intercalate和intersperse版本：

intercalate :: [a] -> [[a]] -> [a]
intercalate s [] = []
intercalate s [x] = x
intercalate s (x:xs) = x ++ s ++ (intercalate s xs)

intersperse :: a -> [a] -> [a]
intersperse s [] = []
intersperse s [x] = [x]
intersperse s (x:xs) = x : s : (intersperse s xs)

是否有任何haskell函数来连接列表与分隔符？

5 个答案: