Pandas 中两个时间戳之间的差异
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Difference between two timestamps in Pandas
提问by Nithin Das
I have the following two time column,"Time1" and "Time2".I have to calculate the "Difference" column,which is (Time2-Time1) in Pandas:
我有以下两个时间列，“Time1”和“Time2”。我必须计算 Pandas 中的“差异”列，即 (Time2-Time1)：
Time1 Time2 Difference
8:59:45 9:27:30 -1 days +23:27:45
9:52:29 10:08:54 -1 days +23:16:26
8:07:15 8:07:53 00:00:38
When Time1 and Time2 are in different hours,I am getting result as"-1 days +" .My desired output for First two values are given below:
当 Time1 和 Time2 在不同的时间时，我得到的结果为“-1 days +”。我想要的前两个值的输出如下：
Time1 Time2 Difference
8:59:45 9:27:30 00:27:45
9:52:29 10:08:54 00:16:26
How can I get this output in Pandas?
如何在 Pandas 中获得此输出？
Both time values are in 'datetime64[ns]' dtype.
两个时间值都在 'datetime64[ns]' dtype 中。
回答by maxymoo
The issue is not that time1and time2are in different hours, it's that time2is before time1so time2-time1is negative, and this is how negative timedeltas are stored. If you just want the difference in minutes as a negative number, you could extract the minutes before calculating the difference:
问题不是那个time1，time2而是在不同的时间，它time2是之前的time1所以time2-time1是负数，这就是负时间增量的存储方式。如果您只想将分钟差异作为负数，则可以在计算差异之前提取分钟：
(df.Time1.dt.minute- df.Time2.dt.minute)
回答by Kamil Sindi
I was not able to reproduce the issue using pandas 17.1:
我无法使用 pandas 17.1 重现该问题：
import pandas as pd
d = {
"start_time": [
"8:59:45",
"9:52:29",
"8:07:15"
],
"end_time": [
"9:27:30",
"10:08:54",
"8:07:53"
]
}
from datetime import datetime
df = pd.DataFrame(data=d)
df['start_time'] = pd.to_datetime(df['start_time'])
df['end_time'] = pd.to_datetime(df['end_time'])
df.end_time - df.start_time
0 00:27:45
1 00:16:25
2 00:00:38
dtype: timedelta64[ns]
