Question

我希望能够使用类似printf的功能创建一个字符串，从列表中提取变量并插入到模板字符串中。

即

let templateStr = "First comes %s, then comes %s, after which comes %s"
let vars = ["one","two", "three"]

并且某些函数返回：

function returns >>> First comes one, then comes two, after which comes three

即。在Python中我可以做类似的事情：

>>> templateStr = "First comes %s, then comes %s, after which comes %s"
>>> vars = ["one","two", "three"]
>>> outputStr = tempStr % tuple(vars)
>>> print outputStr
First comes one, then comes two, after which comes three

我的尝试

mergeList :: [a] -> [a] -> [a]
mergeList [] ys = ys
mergeList (x:xs) ys = x:mergeList ys xs

-- not actually needed * use: Prelude.concat
listConcat :: [[a]] -> [a]
listConcat [] = []
listConcat (x:xs) = x ++ listConcat xs

-- via @dfeuer list concat is not need because of Prelude.concat
printf' :: String -> [String] -> String
printf' s v = concat $ mergeList (splitOn "%s" s) v

尝试通过@Reid Barton

printf' :: String -> [String] -> String
printf' ('%':'s':rest) (v:vs) = v ++ printf' rest vs
printf' (c:rest)        vs    = c :  printf' rest vs
printf' []              _     = []

两次尝试都给出了

>>> let templateStr = "First comes %s, then comes %s, after which comes %s"
>>> let vars = ["one","two", "three"]
>>> printf' templateStr vars
"First comes one, then comes two, after which comes three"

Answer 1

另一种更直接的方法的概述：

printf' ('%':'s':rest) (v:vs) = ...
printf' (c:rest)       vs     = ...
... -- handle the remaining cases too

Answer 2

好开始！ mergeList很干净。

mergeList :: [a] -> [a] -> [a]
mergeList [] ys = ys
mergeList (x:xs) ys = x:mergeList ys xs



listConcat :: [String] -> String
listConcat [] = []
listConcat (x:xs)
  | null xs   = x
  | otherwise = x ++ listConcat xs

listConcat你可以做得更好。特别是，您目前使用两个基本案例进行递归，但您只需要一个。此外，您可以将类型签名更改为listConcat :: [[a]] -> [a]以使其更通用。一旦你清理了你的版本：标准库中就有这样的功能。你能找到吗？

字符串替换运算符，包含Haskell中的列表

2 个答案: