您可以使用
regex来代替
st，
nd，
rd，
th用一个空字符串：
import redef solve(s): return re.sub(r'(d)(st|nd|rd|th)', r'1', s)
演示：
>>> datetime.strptime(solve('1st January 2014'), "%d %B %Y")datetime.datetime(2014, 1, 1, 0, 0)>>> datetime.strptime(solve('3rd March 2014'), "%d %B %Y")datetime.datetime(2014, 3, 3, 0, 0)>>> datetime.strptime(solve('2nd June 2014'), "%d %B %Y")datetime.datetime(2014, 6, 2, 0, 0)>>> datetime.strptime(solve('1st August 2014'), "%d %B %Y")datetime.datetime(2014, 8, 1, 0, 0)
