text = 'yeah, but no, but yeah, but no, but yeah' Search for the location of the first occurrence text.find('no') 10 text = 'Today is 11/27/2012. PyCon starts 3/13/2013.' datepat.findall(text) ['11/27/2012', '3/13/2013']
或者正则
text1 = '11/27/2012' text2 = 'Nov 27, 2012' import re Simple matching: d means match one or more digits if re.match(r'd /d /d ', text1): ... print('yes') ... else: ... print('no') ... yes if re.match(r'd /d /d ', text2): ... print('yes') ... else: ... print('no') ... no
datepat = re.compile(r'(d )/(d )/(d )') m = datepat.match('11/27/2012') m <_sre.SRE_Match object at 0x1005d2750> Extract the contents of each group m.group(0) '11/27/2012' m.group(1) '11' m.group(2) '27' m.group(3) '2012' m.groups() ('11', '27', '2012') month, day, year = m.groups() Find all matches (notice splitting into tuples) text 'Today is 11/27/2012. PyCon starts 3/13/2013.' datepat.findall(text) [('11', '27', '2012'), ('3', '13', '2013')] for month, day, year in datepat.findall(text): ... print('{}-{}-{}'.format(year, month, day)) ... 2012-11-27 2013-3-13
