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>>>import datetime
>>>d1=datetime.datetime(2015,7,5)
>>>d2=datetime.datetime(2005,7,4)
>>>(d1-d2).seconds
//结果86400s
# 计算两个时间的间隔天数
# 注意：输入月份和天时，不能输入08,09等，会被识别为 8 进制而出错！（8进制是不超过07的）
# 解决办法：要把月份和天前面的0去掉。
举例，一个时间偏移后的比较情况：1 #-*-coding=utf-8-*-
2 __author__='zhongtang'
3
4 '''
5 时间戳与字符串的互相转换
6 '''
7
8 import time
9
10 localtime1=time.localtime()
11 time.sleep(5)
12 localtime2=time.localtime(time.time())
13
14 print type(localtime1),localtime1
15 print type(localtime2),localtime2
16
17 gmtime=time.gmtime(time.time())
18 print type(gmtime),gmtime
19
20
21 strtime1='20160518010101'
22 strtime2='20160518020101'
23
24 #字符串变成时间数据结构
25 localtime1=time.strptime(strtime1,'%Y%m%d%H%M%S')
26 localtime2=time.strptime(strtime2,'%Y%m%d%H%M%S')
27
28 print type(localtime1),localtime1
29 print type(localtime2),localtime2
30
31
32 #从时间数据结构转换成时间戳
33 time1= time.mktime(localtime1)
34 time2= time.mktime(localtime2)
35
36 print type(time1),time1
37 print type(time2),time2
38
39 #时间戳可以直接相减，得到以秒为单位的差额
40 print time2-time1
输出结果
1 <type 'time.struct_time'>time.struct_time(tm_year=2016, tm_mon=5, tm_mday=19, tm_hour=9, tm_min=9, tm_sec=30, tm_wday=3, tm_yday=140, tm_isdst=0)
2 <type 'time.struct_time'>time.struct_time(tm_year=2016, tm_mon=5, tm_mday=19, tm_hour=9, tm_min=9, tm_sec=35, tm_wday=3, tm_yday=140, tm_isdst=0)
3 <type 'time.struct_time'>time.struct_time(tm_year=2016, tm_mon=5, tm_mday=19, tm_hour=1, tm_min=9, tm_sec=35, tm_wday=3, tm_yday=140, tm_isdst=0)
4 <type 'time.struct_time'>time.struct_time(tm_year=2016, tm_mon=5, tm_mday=18, tm_hour=1, tm_min=1, tm_sec=1, tm_wday=2, tm_yday=139, tm_isdst=-1)
5 <type 'time.struct_time'>time.struct_time(tm_year=2016, tm_mon=5, tm_mday=18, tm_hour=2, tm_min=1, tm_sec=1, tm_wday=2, tm_yday=139, tm_isdst=-1)
6 <type 'float'>1463504461.0
7 <type 'float'>1463508061.0
8 3600.0
python中的最小时间单位是毫秒，没办法精确到微秒用time包的time()函数可以获得当前计算机的挂钟时间，利用它可以获得时间差
import time
time1 = time.time()
#要度量时间的程序
time2 = time.time()
print time2 - time1
