如何在Haskell中使用另一个字符串替换字符串的子字符串而不使用像MissingH这样的外部库？ [英] How can I replace a substring of a string with another in Haskell without using external Libraries like MissingH?

问题描述

我想用Haskell中的一个字符串替换一个子字符串，而不使用外部库，并且如果可能的话，性能良好。

I would like to replace a substring with a string in Haskell, without using external libraries, and, if it is possible, with good performance.

我想过使用 数据。 Text 替换函数，但我不想移植我的整个程序来使用 Text 类型，而不是字符串。将 String 打包成一个 Text 值，然后替换我想要的内容，然后将该文本值解包为 String 对很多字符串很慢。

I thought about using the Data.Text replace functions, but I don't want to port my entire program to use the Text type instead of Strings. Would packing the String into a Text value, then replacing what I wanted to, then unpacking that Text value to a String be slow on a lot of Strings?

推荐答案

试试这个（未经测试）：

Try this (untested):

replace :: Eq a => [a] -> [a] -> [a] -> [a]
replace needle replacement haystack
  = case begins haystack needle of
      Just remains -> replacement ++ remains
      Nothing      -> case haystack of
                        []     -> []
                        x : xs -> x : replace needle replacement xs

begins :: Eq a => [a] -> [a] -> Maybe [a]
begins haystack []                = Just haystack
begins (x : xs) (y : ys) | x == y = begins xs ys
begins _        _                 = Nothing

将改变你的程序使用 Text s而不是 String s。

But in general you will get performance gains from switching your program to use Texts instead of Strings.

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