如何将逗号分隔值的字符串解析为haskell中的字符串列表? [英] how to parse a string of comma-separated values into a list of strings in haskell?

问题描述

因此，如果我在其中的单词列表中有一个字符串(this，is，a，故事，所有关于，如何)" ["this"，"is"，"a"，"story"，全部"，关于"，如何"]作为ReadP String的实例?我尝试了很多不同的方法，其中之一是:

so if I have a string "(this, is, a, story, all, about, how)" into a list of the words inside it ["this", "is", "a", "story", "all", "about", "how"] as an instance of ReadP String? I've tried a bunch of different ways, one of which being this:

parseStr :: ReadP String
parseStr = do
  skipSpaces
  n <- munch1 isAlphaOrDigit
  skipComma
  return $ n

解析除最后一个值以外的所有值.我想如果我将它与以下分析结合起来:

which parses all values but the last. I thought if I combined it with this parse:

parseLast :: ReadP String
parseLast = do
  skipSpaces
  n <- munch1 isAlphaOrDigit
  return $ n

为

parseLet = (many parseStr) +++ parseLast

但是那也不起作用.有小费吗?

but that didn't work either. Any tips?

更多定义

isAlphaOrDigit :: Char -> Bool 
isAlphaOrDigit a = (isDigit a) || (isAlpha a) 
comma = satisfy (','==)
skipComma = const () <$> some comma

推荐答案

解析器 a +++ b 将整个输入字符串发送到 a 到 b ，生成解析器生成的所有结果.相反，您需要一个解析器，将字符串的第一部分发送到 a ，将第二部分发送到 b ，然后让您合并结果.尝试以下方法:

The parser a +++ b sends the entire input string to a and the entire input string to b, producing all the results that either parser produced. You instead want a parser that sends the first part of the string to a and the second part to b, then lets you combine the results. Try this instead:

parseLet = liftA2 (\ss s -> ss ++ [s]) (many parseStr) parseLast

许多解析器库也为这个确切的用例提供了 manySepBy 组合器(也许名称略有不同).您可以考虑在 ReadP 库中查找此类内容.

Many parser libraries also offer a manySepBy combinator (perhaps with a slightly different name) for this exact use case; you might consider looking through the ReadP library for such a thing.

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