Python 从时间增量获取总小时数？

如何获得时间增量中的总小时数

例如：

>>> td = pd.Timedelta('1 days 2 hours')
>>> td.get_total_hours()
26

注意：根据文档，

.hours

属性将返回小时组件：

---

只需找出1小时内可容纳多少个

timedelta

s即可：

import numpy as np

>> td / np.timedelta64(1, 'h')
26.0

---

试着说明为什么熊猫会在2小时内回来

import pandas as pd

td = pd.Timedelta('1 days 2 hours')

td.components

Out[45]: Components(days=1, hours=2, minutes=0, seconds=0, milliseconds=0, microseconds=0, nanoseconds=0)

td / pd.Timedelta('1 hour')

Out[46]: 26.0

---

我得到的是以秒为单位的时间增量，再除以3600得到小时

round（td.total_seconds（）/3600）

当我在jupyter笔记本中测试时，这种方法工作得更快

%timeit td / np.timedelta64(1, 'h') 
The slowest run took 19.10 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 4.58 µs per loop

%timeit round(td.total_seconds() / 3600)
The slowest run took 18.08 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 401 ns per loop

我更喜欢pd.Timedelta（小时数=1），因为OP已经在使用pandas了。

%timeit td / np.timedelta64(1, 'h') 
The slowest run took 19.10 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 4.58 µs per loop

%timeit round(td.total_seconds() / 3600)
The slowest run took 18.08 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 401 ns per loop