使用正则表达式python获取模式后的句子

Question

在我的字符串中(从这个turorial中采用的例子)我希望获得所有内容,直到.通用(year).模式之后的第一个:

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

我想我的代码几乎就在那里但尚未完成:

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

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

所需的输出是:

['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

如果你想要(year).在第一个和第一个之间获得所有东西,.你可以使用它:

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

见Live Demo.

并在此解释:

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