如何將數字列表拆分為一組包含 Haskell 中所有數字的列表

Question

我如何將 Haskell 中的列表（例如，“222 33244”）拆分為 ["222","33","2","444"] 僅通過前奏的遞歸和函數？

我目前的嘗試是：

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，由於我不太了解您的方法，並且 ghci 在您的代碼中列出了 18 個編譯錯誤，恐怕我無法幫助您嘗試解決方案。

正如評論中所指出的，可能的解決方案是：

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. 過濾字符串中不是數字的每個元素（Data.Char.isNumber 也會這樣做，但前提是只使用 Prelude 函數）並在過濾后的列表上調用listSplit' 。

2. (ws ++ [x : takeWhile (==x) xs])收集xs所有內容，直到遇到一個不等於x的字母，將其包裝在一個列表中並將其附加到ws 。

3. (dropWhile (==x) xs)刪除了每個在信xs ，直到它到達一個字母不等於x 。

4. 最后，該函數使用更新的ws和減少的xs調用自身

5. 如果沒有更多剩余元素，則函數返回ws

Answer 2

如果您的目標是使用很少的預定義函數，這可能會給您一些想法：

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])

當然，您也可以用自己的遞歸替換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

我有時間來實現這一點，但我認為這就是你正在尋找的。

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)