Question

我在一本书中找到了一个很好的例子，我正试图解决这个问题。我正在尝试编写一个名为“pointer”的函数，其签名为pointer :: String -> Int。它将带有看起来像[Int]的“指针”文本，然后返回找到的指针总数。

指针函数将检查的文本如下所示：

txt :: String
txt = "[1] and [2] are friends who grew up together who " ++
      "went to the same school and got the same degrees." ++
      "They eventually opened up a store named [2] which was pretty successful."

在命令行中，我们将按如下方式运行代码：

> pointer txt 
3

3表示找到的指针数。

我了解的是：

我知道“单词”会将字符串分解为带有单词的列表。示例：

单词“所有这些苹果都在哪里？”

[ “哪里”， “是”， “所有”， “的”， “这些”， “苹果？”]
我得到“过滤器”将选择列表中的特定元素。例如：

过滤器（＆gt; 3）[1,5,6,4,3]
```
     [5,6,4]
```
我得到“长度”将返回列表的长度

我认为我需要做什么：

Step 1) look at txt and then break it down into single words until you have a long list of words.
Step 2) use filter to examine the list for [1] or [2]. Once found, filter will place these pointers into an list.
Step 3) call the length function on the resulting list.

面临的问题：

我正在努力抓住我所知道的一切并实施它。

Answer 1

这是一个假设的ghci会议：

ghci> words txt
[ "[1]", "and", "[2]", "are", "friends", "who", ...]

ghci> filter (\w -> w == "[1]" || w == "[2]") (words txt)
[ "[1]", "[2]", "[2]" ]

ghci> length ( filter (\w -> w == "[1]" || w == "[2]") (words txt) )
3

您可以使用$运算符使最后一个表达式更具可读性：

length $ filter (\w -> w == "[1]" || w == "[2]") $ words txt

Answer 2

如果您希望能够在字符串中找到[Int]类型的所有模式 - 例如[3]，[465]等，而不仅仅是[1]和[2]，最简单的方法是使用正则表达式：

{-# LANGUAGE NoOverloadedStrings #-}

import Text.Regex.Posix

txt :: String
txt = "[1] and [2] are friends who grew up together who " ++
      "went to the same school and got the same degrees." ++
      "They eventually opened up a store named [2] which was pretty successful."

pointer :: String -> Int
pointer source = source =~ "\\[[0-9]{1,}\\]"

我们现在可以运行：

pointer txt
> 3

Answer 3

这适用于单个数字“指针”：

pointer :: String -> Int
pointer ('[':_:']':xs) = 1 + pointer xs
pointer (_:        xs) = pointer xs
pointer _              = 0

使用ie提供的解析器组合器可以更好地处理这个问题。 Parsec，但这可能有点矫枉过正。

通过字符串搜索

3 个答案: