使用dateutil package中的relativedelta。这将考虑闰年和其他怪癖。

import datetime
from dateutil.relativedelta import relativedelta

a = '2014-05-06 12:00:56'
b = '2013-03-06 16:08:22'

start = datetime.datetime.strptime(a, '%Y-%m-%d %H:%M:%S')
ends = datetime.datetime.strptime(b, '%Y-%m-%d %H:%M:%S')

diff = relativedelta(start, ends)

>>> print "The difference is %d year %d month %d days %d hours %d minutes" % (diff.years, diff.months, diff.days, diff.hours, diff.minutes)
The difference is 1 year 1 month 29 days 19 hours 52 minutes

您可能需要添加一些逻辑来打印，例如“2 years”而不是“2 year”。

diff是一个timedelta示例。
对于python2，请参见：https://docs.python.org/2/library/datetime.html#timedelta-objects
对于Python 3，请参阅：https://docs.python.org/3/library/datetime.html#timedelta-objects
来自文档：
timdelta示例属性（只读）：

• 天
• 秒
• 微秒

timdelta示例方法：

• 总秒数（）

timdelta类属性为：

• 最小
• 最大
• 解析度

您可以使用days和seconds示例属性来计算所需的内容。
例如：

import datetime

a = '2014-05-06 12:00:56'
b = '2013-03-06 16:08:22'

start = datetime.datetime.strptime(a, '%Y-%m-%d %H:%M:%S')
ends = datetime.datetime.strptime(b, '%Y-%m-%d %H:%M:%S')

diff = start - ends

hours = int(diff.seconds // (60 * 60))
mins = int((diff.seconds // 60) % 60)

要计算时间戳之间的差值：

from time import time

def timestamp_from_seconds(seconds):
    minutes, seconds = divmod(seconds, 60)
    hours, minutes = divmod(minutes, 60)
    days, hours = divmod(hours, 24)
    return days, hours, minutes, seconds

print("\n%d days, %d hours, %d minutes, %d seconds" % timestamp_from_seconds(abs(1680375128- time())))

输出：1天19小时19分钟55秒