如何在 python 正则表达式中获取两个子字符串中的特定字符串？

Question

这是示例：review: I love you very much... reviewer:jackson review: I hate you very much... reviewer:madden review: sky is pink and i ... reviewer: tom我想提取字符串review:和之间的字符串...所以以上情况的提取是I love you very muchI hate you very muchsky is pink and i 我使用这种正则表达式但失败了re.findall("review(.*)...",string)它提取了这种结果：I love you very much... reviewer:jackson review: I hate you very much... reviewer:madden review: sky is pink and i

Answer 1

这也可以，而且很简单

str = "review: I love you very much... reviewer:jackson review: I hate you very much... reviewer:madden review: sky is pink and i ... reviewer: tom"

matches = re.findall('review:(.+?)\.\.\.', str)

Answer 2

使用

re.findall(r'\breview:\s*(.*?)\s*\.\.\.', string)

见证明。蟒蛇测试：

import re

regex = r"\breview:\s*(.*?)\s*\.\.\."

string = "review: I love you very much... reviewer:jackson review: I hate you very much... reviewer:madden review: sky is pink and i ... reviewer: tom"

print ( re.findall(regex, string) )

输出：['I love you very much', 'I hate you very much', 'sky is pink and i']

请注意，r"..."表示原始字符串文字的前缀"\b"不是单词边界，而是r"\b"。

解释

NODE                     EXPLANATION

--------------------------------------------------------------------------------

  \b                       the boundary between a word char (\w) and

                           something that is not a word char

--------------------------------------------------------------------------------

  review:                  'review:'

--------------------------------------------------------------------------------

  \s*                      whitespace (\n, \r, \t, \f, and " ") (0 or

                           more times (matching the most amount possible))

--------------------------------------------------------------------------------

  (                        group and capture to \1:

--------------------------------------------------------------------------------

    .*?                      any character except \n (0 or more times

                             (matching the least amount possible))

--------------------------------------------------------------------------------

  )                        end of \1

--------------------------------------------------------------------------------

  \s*                      whitespace (\n, \r, \t, \f, and " ") (0 or

                           more times (matching the most amount possible))

--------------------------------------------------------------------------------

  \.\.\.                   '...'

--------------------------------------------------------------------------------

Answer 3

您可以使用以下利用前瞻的模式：

(?<=review:\s).*?(?=\.\.\.)

inp = "review: I love you very much... reviewer:jackson review: I hate you very much... reviewer:madden review: sky is pink and i ... reviewer: tom"

matches = re.findall(r'(?<=review:\s).*?(?=\.\.\.)', inp)

print(matches)

Answer 4

re.findall与模式一起使用\breview:\s*(.*?)\.\.\.\s*(?=\breviewer:|$)：

inp = "review: I love you very much... reviewer:jackson review: I hate you very much... reviewer:madden review: sky is pink and i ... reviewer: tom"

matches = re.findall(r'\breview:\s*(.*?)\.\.\.\s*(?=\breviewer:|$)', inp)

print(matches)

这打印：

['I love you very much', 'I hate you very much', 'sky is pink and i ']

Answer 5

\对不起，我忘了在前面 添加.

正确的是： re.findall("review:\b?(.*)\.\.\.",string)

而这一次，很重要