python - 使用正则表达式 python 在模式后获取句子

Question

在我的字符串中(示例取自 this turorial )我想获取所有内容，直到第一个 . 之后的通用 (year). 模式:

str = 'purple alice@google.com, (2002).blah monkey. (1991).@abc.com blah dishwasher'

我想我的代码已经差不多完成了，但还没有完成:

test = re.findall(r'[\(\d\d\d\d\).-]+([^.]*)', str)

... 返回:['com, (2002)', 'blah monkey', '(1991)', '@abc', 'com blah dishwasher']

期望的输出是:

['blah monkey', '@abc']

换句话说，我想找到年份模式和下一个点之间的所有内容。

Answer 1

如果你想得到 (year). 和第一个 . 之间的所有东西，你可以使用这个:

\(\d{4}\)\.([^.]*)

参见 Live Demo .

这里解释一下:

"\(\d{4}\)\.([^.]*)"g

\( matches the character ( literally
  \d{4} match a digit [0-9]
    Quantifier: {4} Exactly 4 times
       \) matches the character ) literally
         \. matches the character . literally
1st Capturing group ([^.]*)
    [^.]* match a single character not present in the list below
        Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
        . the literal character .
g modifier: global. All matches (don't return on first match)