如何将数字列表拆分为一组包含 Haskell 中所有数字的列表

Question

如何仅通过前奏的递归和函数将Haskell中的列表（例如“222 33244”）拆分为[“222”，“33”，“2”，“444”]？

我目前的尝试是：

list xs
  |length xs == 0 = ""
  |otherwise = listSplit xs

 listSplit (x:xs)
  |x == head xs = x : ListSplitNext x xs
  |otherwise = x:[]

 listSplitNext a (x:xs)
  |a == x = a : listSplitNext x xs
  |otherwise = listSplit xs

Answer 1

因此，由于我不太了解您的方法，并且 ghci 在您的代码中列出了 18 个编译错误，因此恐怕无法帮助您尝试解决方案。

正如评论中指出的，一个可能的解决方案是：

listSplit xs = listSplit' [] (filter (`elem` ['0'..'9']) xs)

listSplit' ws (x:xs) = listSplit' (ws ++ [x : takeWhile (==x) xs]) (dropWhile (==x) xs)
listSplit' ws [] = ws

1. 过滤字符串中不是数字的每个元素（Data.Char.isNumber 也会这样做，但前提是只使用 Prelude 函数）并在过滤后的列表上调用 listSplit'。

2. (ws ++ [x : takeWhile (==x) xs]) 收集 xs 中的所有内容，直到遇到不等于 x 的字母，将其包装在列表中并将其附加到 ws。

3. (dropWhile (==x) xs) 删除 xs 中的每个字母，直到它到达不等于 x 的字母。

4. 最后，函数使用更新后的ws 和缩减后的xs 调用自身

5. 如果没有剩余元素，函数返回ws

Answer 2

如果您的目标是使用很少的预定义函数，这可能会给您一些想法：

listSplit :: String -> [String]
listSplit xs =
  let (as, a) = foldr go ([], []) numbers
  in  a : as
 where
  isNumber x = x `elem` ['0'..'9']
  numbers    = filter isNumber xs

  go cur (res, []) = (res, [cur])
  go cur (res, lst@(a:_))
    | a == cur  = (res, a : lst)
    | otherwise = (lst : res, [cur])

当然你也可以用你自己的递归替换foldr：

numberSplit :: String -> [String]
numberSplit xs =
  let numbers = filter (`elem` ['0'..'9']) xs
  in  listSplit numbers


listSplit :: Eq a => [a] -> [[a]]
listSplit =
  reverse . go [] []
 where
  go acc as [] = as : acc
  go acc [] (x:xs) = go acc [x] xs
  go acc as@(a:_) (x:xs)
    | a == x    = go acc (a : as) xs
    | otherwise = go (as : acc) [x] xs

Answer 3

我有时间来实现它，但我认为这就是您正在寻找的。​​p>

listSplit s = go filtered
  where filtered = [c | c <- s, elem c ['0'..'9']]
        go [] = []
        go (x:xs) = (x : takeWhile (== x) xs) : (go $ dropWhile (== x) xs)