Python 从 numpy.timedelta64 值中提取天数
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extracting days from a numpy.timedelta64 value
提问by user7289
I am using pandas/python and I have two date time series s1 and s2, that have been generated using the 'to_datetime' function on a field of the df containing dates/times.
我正在使用 pandas/python 并且我有两个日期时间序列 s1 和 s2，它们是在包含日期/时间的 df 字段上使用“to_datetime”函数生成的。
When I subtract s1 from s2
当我从 s2 中减去 s1
s3 = s2 - s1
s3 = s2 - s1
I get a series, s3, of type
我得到一个系列，s3，类型
timedelta64[ns]
timedelta64[ns]
0 385 days, 04:10:36
1 57 days, 22:54:00
2 642 days, 21:15:23
3 615 days, 00:55:44
4 160 days, 22:13:35
5 196 days, 23:06:49
6 23 days, 22:57:17
7 2 days, 22:17:31
8 622 days, 01:29:25
9 79 days, 20:15:14
10 23 days, 22:46:51
11 268 days, 19:23:04
12 NaT
13 NaT
14 583 days, 03:40:39
How do I look at 1 element of the series:
我如何看待该系列的 1 个元素：
s3[10]
s3[10]
I get something like this:
我得到这样的东西：
numpy.timedelta64(2069211000000000,'ns')
numpy.timedelta64(2069211000000000,'ns')
How do I extract days from s3 and maybe keep them as integers(not so interested in hours/mins etc.)?
我如何从 s3 中提取天数并将它们保留为整数（对小时/分钟等不太感兴趣）？
Thanks in advance for any help.
在此先感谢您的帮助。
采纳答案by Viktor Kerkez
You can convert it to a timedelta with a day precision. To extract the integer value of days you divide it with a timedelta of one day.
您可以将其转换为具有一天精度的 timedelta。要提取天数的整数值，请将其除以一天的时间增量。
>>> x = np.timedelta64(2069211000000000, 'ns')
>>> days = x.astype('timedelta64[D]')
>>> days / np.timedelta64(1, 'D')
23
Or, as @PhillipCloud suggested, just days.astype(int)since the timedeltais just a 64bit integer that is interpreted in various ways depending on the second parameter you passed in ('D', 'ns', ...).
或者，如@PhillipCloud建议，只是days.astype(int)因为timedelta仅仅是一个64位整数，根据你所传递的第二个参数被解释以各种方式（'D'，'ns'，...）。
You can find more about it here.
您可以在此处找到更多相关信息。
回答by mgoldwasser
Suppose you have a timedelta series:
假设您有一个 timedelta 系列：
import pandas as pd
from datetime import datetime
z = pd.DataFrame({'a':[datetime.strptime('20150101', '%Y%m%d')],'b':[datetime.strptime('20140601', '%Y%m%d')]})
td_series = (z['a'] - z['b'])
One way to convert this timedelta column or series is to cast it to a Timedelta object (pandas 0.15.0+) and then extract the days from the object:
转换此 timedelta 列或系列的一种方法是将其转换为 Timedelta 对象（pandas 0.15.0+），然后从对象中提取天数：
td_series.astype(pd.Timedelta).apply(lambda l: l.days)
Another way is to cast the series as a timedelta64 in days, and then cast it as an int:
另一种方法是将系列转换为 timedelta64（以天为单位），然后将其转换为 int：
td_series.astype('timedelta64[D]').astype(int)
回答by Nickil Maveli
Use dt.daysto obtain the days attribute as integers.
用于dt.days以整数形式获取 days 属性。
For eg:
例如：
In [14]: s = pd.Series(pd.timedelta_range(start='1 days', end='12 days', freq='3000T'))
In [15]: s
Out[15]:
0 1 days 00:00:00
1 3 days 02:00:00
2 5 days 04:00:00
3 7 days 06:00:00
4 9 days 08:00:00
5 11 days 10:00:00
dtype: timedelta64[ns]
In [16]: s.dt.days
Out[16]:
0 1
1 3
2 5
3 7
4 9
5 11
dtype: int64
More generally - You can use the .componentsproperty to access a reduced form of timedelta.
更一般地 - 您可以使用该.components属性来访问timedelta.
In [17]: s.dt.components
Out[17]:
days hours minutes seconds milliseconds microseconds nanoseconds
0 1 0 0 0 0 0 0
1 3 2 0 0 0 0 0
2 5 4 0 0 0 0 0
3 7 6 0 0 0 0 0
4 9 8 0 0 0 0 0
5 11 10 0 0 0 0 0
Now, to get the hoursattribute:
现在，要获取hours属性：
In [23]: s.dt.components.hours
Out[23]:
0 0
1 2
2 4
3 6
4 8
5 10
Name: hours, dtype: int64
