我想打印如下所示的字符串列表。
|Name|Country|Age|
------------------
|1 |USA |20 |
|2 |UK |19 |
我能够通过以下方式实现这一目标。
printfieldName :: [String] -> String
printfieldName [] = []
printfieldName (x:xs) = "|" ++ x ++ "\t" ++ printfieldName (xs)
是否可以使用内置函数'unwords'来实现这一点。我能够使用'unwords'打印它,但无法在单词之间放置|
。
答案 0 :(得分:4)
首先,我会这样写:
printfieldName [] = []
printfieldName (x:xs) = "|" ++ x ++ "\t" ++ printfieldName xs
嗯,实际上,不,就像这样:
concatMap (\x -> '|' : x ++ "\t")
好吧,也许更像是:
concatMap (printf "|%s\t")
行。它可以用'unwords'来完成吗?
-- | 'unwords' is an inverse operation to 'words'.
-- It joins words with separating spaces.
unwords :: [String] -> String
unwords [] = ""
unwords ws = foldr1 (\w s -> w ++ ' ':s) ws
没有。但看看你是否可以将concatMap编写为折叠器......
答案 1 :(得分:4)
我看到'|'之间还有一个空格和单词,所以你可以使用这个功能:
printfieldName x = unwords (map ((++) "|") x) ++ "|"
小解释:
(++) "|" - creates a function which take prefixes each word with "|", so
(++) "|" "test" -> "|test"
然后,map
将此函数应用于将其转换为["|1", "|USA", "|20", ... ]
的单词列表
然后unwords
将它们连接成一个字符串之间的空格。添加最终++ "|"
|
答案 2 :(得分:4)
Data.List
有一个名为intersperse的函数。也许你可以使用它。
printfieldName xs = "|" ++ unwords (intersperse "|\t" xs) ++ "|"
答案 3 :(得分:1)
也许你提出的问题有点过分了,但是:
formatTable :: [String] -> [[String]] -> String
formatTable header rows =
formatRow header ++ dashRow ++ concatMap formatRow rows
where formatRow cells = bracket '|' (spread cells)
dashRow = bracket '+' (map (\n -> replicate n '-') widths)
bracket c cells = concatMap (c:) cells ++ (c:"\n")
spread cells = zipWith pad widths cells
pad n s = take n (s ++ repeat ' ')
widths = foldr maxLengths (repeat 0) (header : rows)
maxLengths = zipWith (\c l -> max (length c) l)
然后,例如:
> let h = words "Name Country Age"
> let rows = map words ["1 USA 20", "2 UK 19"]
> h
["Name","Country","Age"]
> rows
[["1","USA","20"],["2","UK","19"]]
> putStr $ formatTable h rows
|Name|Country|Age|
+----+-------+---+
|1 |USA |20 |
|2 |UK |19 |