尝试这个：
def monthdelta(date, delta): m, y = (date.month+delta) % 12, date.year + ((date.month)+delta-1) // 12 if not m: m = 12 d = min(date.day, [31, 29 if y%4==0 and (not y%100==0 or y%400 == 0) else 28, 31,30,31,30,31,31,30,31,30,31][m-1]) return date.replace(day=d,month=m, year=y)>>> for m in range(-12, 12): print(monthdelta(datetime.now(), m))2009-08-06 16:12:27.8230002009-09-06 16:12:27.8550002009-10-06 16:12:27.8700002009-11-06 16:12:27.8700002009-12-06 16:12:27.8700002010-01-06 16:12:27.8700002010-02-06 16:12:27.8700002010-03-06 16:12:27.8860002010-04-06 16:12:27.8860002010-05-06 16:12:27.8860002010-06-06 16:12:27.8860002010-07-06 16:12:27.8860002010-08-06 16:12:27.9010002010-09-06 16:12:27.9010002010-10-06 16:12:27.9010002010-11-06 16:12:27.9010002010-12-06 16:12:27.9010002011-01-06 16:12:27.9170002011-02-06 16:12:27.9170002011-03-06 16:12:27.9170002011-04-06 16:12:27.9170002011-05-06 16:12:27.9170002011-06-06 16:12:27.9330002011-07-06 16:12:27.933000>>> monthdelta(datetime(2010,3,30), -1)datetime.datetime(2010, 2, 28, 0, 0)>>> monthdelta(datetime(2008,3,30), -1)datetime.datetime(2008, 2, 29, 0, 0)
编辑 更正为也处理一天。
编辑 另请参见困惑的答案，指出了以下更简单的计算方法
d：
d = min(date.day, calendar.monthrange(y, m)[1])
