在我的字符串中(从
this turorial开始采用的例子)我希望获得所有内容,直到第一个.在通用(年)之后.图案：
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洗碗机’]
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.
解决方法
如果你想在(年)之间得到所有东西.和第一个.你可以用这个：
\(\d{4}\)\.([^.]*)
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)
