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’]
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)