问题
Is there a function to concatenate elements of a list with a separator? For example:
> foobar " " ["is","there","such","a","function","?"]
["is there such a function ?"]
Thanks for any reply!
回答1:
Yes, there is:
Prelude> import Data.List
Prelude Data.List> intercalate " " ["is","there","such","a","function","?"]
"is there such a function ?"
intersperse is a bit more general:
Prelude> import Data.List
Prelude Data.List> concat (intersperse " " ["is","there","such","a","function","?"])
"is there such a function ?"
Also, for the specific case where you want to join with a space character, there is unwords:
Prelude> unwords ["is","there","such","a","function","?"]
"is there such a function ?"
unlines works similarly, only that the strings are imploded using the newline character and that a newline character is also added to the end. (This makes it useful for serializing text files, which must per POSIX standard end with a trailing newline)
回答2:
It's not hard to write one-liner using foldr
join sep xs = foldr (\a b-> a ++ if b=="" then b else sep ++ b) "" xs
join " " ["is","there","such","a","function","?"]
回答3:
joinBy sep cont = drop (length sep) $ concat $ map (\w -> sep ++ w) cont
回答4:
If you wanted to write your own versions of intercalate and intersperse:
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)
来源：https://stackoverflow.com/questions/9220986/is-there-any-haskell-function-to-concatenate-list-with-separator