Haskell 提取字符串中的子字符串答案

【问题标题】：Haskell extract substring within a stringHaskell 提取字符串中的子字符串
【发布时间】：2014-02-13 09:45:08
【问题描述】：

我的目标是找出子字符串在字符串中存在的次数。我要查找的子字符串类型为“[n]”，其中 n 可以是任何变量。

我的尝试涉及使用 words 函数拆分字符串，如果字符串的“头”是“[”，则创建一个新的字符串列表同一字符串的“最后一个”是“]”

我遇到的问题是我输入了一个字符串，当使用它进行拆分时功能词，创建了一个看起来像这样“[2]”的字符串现在，我仍然希望这算作“[n]”类型的出现

一个例子是我想要这个字符串，

asdf[1]jkl[2]asdf[1]jkl

返回 3。

这是我的代码：

-- String that will be tested on references function
txt :: String
txt = "[1] and [2] both feature characters who will do whatever it takes to " ++
  "get to their goal, and in the end the thing they want the most ends " ++
  "up destroying them.  In case of [2], this is a whale..."

-- Function that will take a list of Strings and return a list that contains
-- any String of the type [n], where n is an variable
ref :: [String] -> [String]
ref [] = []
ref xs = [x | x <- xs, head x == '[', last x == ']']

-- Function takes a text with references in the format [n] and returns
-- the total number of references.
-- Example :  ghci> references txt -- -> 3
references :: String -> Integer   
references txt = len (ref (words txt))

如果有人能告诉我如何在字符串中搜索子字符串或如何解析给定子字符串的字符串，将不胜感激。

【问题讨论】：

标签： string parsing haskell substring extract

【解决方案1】：

我会用一个正则表达式，这样写：

import Text.Regex.Posix

txt :: String
txt = "[1] and [2] both feature characters who will do whatever it takes to " ++
  "get to their goal, and in the end the thing they want the most ends " ++
  "up destroying them.  In case of [2], this is a whale..."


-- references counts the number of references in the input string
references :: String -> Int
references str = str =~ "\\[[0-9]*\\]"

main = putStrLn $ show $ references txt -- outputs 3

【讨论】：

感谢 jcarpenter！您介意解释一下 =~ 运算符的作用吗？那是导入库的一部分吗？我更希望弄清楚每当 [n] 发生时如何解析，因为我想最终用一个字符串替换每个 [n] ，该字符串位于由任何 n 索引的列表中。
我不知道 =~ 在内部是如何工作的。它将正则表达式与字符串匹配，并且可以返回各种不同的类型。谷歌或其他人可以比我更详细地阐述它。

【解决方案2】：

正则表达式对于这样一个简单的问题来说太过分了。

references = length . consume

consume []       = []
consume ('[':xs) = let (v,rest) = consume' xs in v:consume rest
consume (_  :xs) = consume xs

consume' []       = ([], []) 
consume' (']':xs) = ([], xs)
consume' (x  :xs) = let (v,rest) = consume' xs in (x:v, rest)

consume 等待[，然后调用consume'，它会收集所有内容，直到]。

【讨论】：

我更喜欢这个而不是其他答案，因为 A）它是 Haskell 中的简洁解决方案，而不是正则表达式；和 B）这可能更容易理解和修改 OP 的用例。

【解决方案3】：

这是一个解决方案 sepCap.

import Replace.Megaparsec
import Text.Megaparsec
import Text.Megaparsec.Char
import Data.Either
import Data.Maybe

txt = "[1] and [2] both feature characters who will do whatever it takes to " ++
  "get to their goal, and in the end the thing they want the most ends " ++
  "up destroying them.  In case of [2], this is a whale..."

pattern = single '[' *> anySingle <* single ']' :: Parsec Void String Char
length $ rights $ fromJust $ parseMaybe (sepCap pattern) txt

【讨论】：