使用python时间处理方法

在实际中遇到一个时间处理问题，需要将 Sep 06, 2014 19:30 （UTC 时间） 和 当前时间比较早晚，知道 此 2014-09-06 19:30 格式时间的运算。因此，在处理时，就想

先将sep 格式时间转化成 后面一个格式的时间样子，没有找到相关函数，于是自己简单的写了个程序，仅记于此，以备查询

代码如下：

# -*- coding: utf-8 -*- 
from datetime import date
from datetime import datetime
from datetime import timedelta
 
# # #如果是返回当前时间，可以简单的写成
# # time.localtime()
# # #这个返回UTC时间
# # time.gmtime()
# lt = time.localtime()
# tm = time.gmtime()
# ft = time.strftime('%Y-%m-%d %H-%M',lt)
# ft2 = time.strftime('%Y-%m-%d %H:%M',tm)
# print ft, ft2
# print '--------------------------------------------------------'
# now = datetime.datetime.now()
# now = now.replace(day = 1)
# print now
# print now.time()
#
# mytime = ['2014-09-06 20:19']
# #mytime2 = '2014-10-09 14:32'
# str = "".join(mytime)
# print str
# retime  = datetime.strptime(str,'%Y-%m-%d %H:%M')
# print retime
# retime = retime +timedelta(hours = 8)
# print retime
#
# tdtime = datetime.now()
# print tdtime
# if retime <= tdtime - timedelta(days = 7):
#     print "too early"
#
# #Sep 06, 2014 19:30
 
monthdic = {'Jan':'01', 'Feb':'02', 'Mar':'03', 'Apr':'04', 'May':'05', 'Jun':'06', 'Jul':'07', 'Aug':'08', 'Sep':'09', 'Oct':'10', 'Nov':'11', 'Dec':'12'}
def time_format(timestr):
    timestr = timestr.replace(',','')
    #print timestr
    timelist = timestr.split()
    #print timelist
    mon = "".join(timelist[0])
    #print  mon
    timelist[0] = monthdic[mon]
    #print timelist
    mytime = "".join(timelist[2])+'-'+"".join(timelist[0])+'-'+"".join(timelist[1])+' '+ "".join(timelist[3])
    return mytime
if name == 'main':
    timestr = 'Sep 06, 2014 19:30'
    str = time_format(timestr)
    print str      
    mytime  = datetime.strptime(str,'%Y-%m-%d %H:%M') 
    print mytime
    mytime = mytime +timedelta(hours = 8)
    print mytime  
    tdtime = datetime.now()
    print tdtime
    if mytime <= tdtime - timedelta(days = 7):
        print "too early"

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