Question

我的目标是找到字符串中存在子字符串的次数。我要查找的子字符串类型为＆＃34; [n]＆＃34;，其中n可以是任何变量。

我的尝试涉及使用单词function分割字符串，然后创建一个新的字符串列表，如果＆＃39; head＆＃39;字符串是＆＃39; [＆＃39;和最后的＆＃39;相同的字符串是＆＃39;]＆＃39;

我遇到的问题是我输入了一个String，当使用时进行拆分功能词，创建了一个看起来像这样的字符串＆＃34; [2]，＆＃34; 现在，我仍然希望将其视为类型＆＃34; [n]＆＃34;

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))

如果有人可以启发我如何在字符串中搜索子字符串或者如何解析给定子字符串的字符串，非常感谢。

Answer 1

我只想使用正则表达式，并按照以下方式编写：

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

Answer 2

对于这样一个简单的问题，正则表达式是一个巨大的过度杀伤力。

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[，然后拨打consume'，收集所有内容，直到]。

Answer 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

Haskell提取字符串中的子字符串

3 个答案: