如何计算 pandas 中的时间差

Question

我在 pandas 中有关注 dataframe

 code     srt_date       srt_time      end_time    fina_datetime
 123      2019-01-01     23:23:00      00:12:00    2019-01-02 00:13:00 
 123      2019-01-02     00:13:00      00:14:00    2019-01-02 00:15:00
 123      2019-01-02     23:00:00      00:15:00    2019-01-03 00:16:00  

我想计算我在 pandas 中做以下事情的fina_datetime - end_time

 df['end_time'] = df['srt_date'].map(str) +" "+ df['end_time'].map(str)
 df['end_time'] = pd.to_datetime(df['end_time'], format = "%Y-%m-%d %H:%M:%S")
 df['latency_in_secs'] = [x-y for x, y in zip(df['final_datetime'] , df['end_time'])]
 df['latency_in_secs'] = df.latency_in_secs.dt.total_seconds()

上面的代码在日期进入下一个日期时出现问题，例如第一行和第三行。 我如何在 pandas 中做到这一点？

我想要的 dataframe 将是

 code     srt_date       srt_time      end_time    fina_datetime        latency_in_secs 
 123      2019-01-01     23:23:00      00:12:00    2019-01-02 00:13:00     60 
 123      2019-01-02     00:13:00      00:14:00    2019-01-02 00:15:00     60
 123      2019-01-02     23:00:00      00.15:00    2019-01-03 00:16:00     60

Answer 1

IIUC，您可以掩盖end_time < srt_time的位置并将日期加一：

# convert to timedelta
df['srt_time'] = pd.to_timedelta(df['srt_time'])
df['end_time'] = pd.to_timedelta(df['end_time'])

# convert to datetime
df['srt_date'] = pd.to_datetime(df['srt_date'])
df['fina_datetime'] = pd.to_datetime(df['fina_datetime'])

# the normal end
end_dates = df['srt_date'] + df['end_time']

# increase the end time with end_time < srt_time by one day
end_dates.loc[df['end_time'].le(df['srt_time'])] += pd.to_timedelta(1, unit='D')

# substract:
df['latency_in_secs'] = (df['fina_datetime'].sub(end_dates)
                             .dt.total_seconds()
                        )

Output：

   code   srt_date srt_time end_time       fina_datetime  latency_in_secs
0   123 2019-01-01 23:23:00 00:12:00 2019-01-02 00:13:00             60.0
1   123 2019-01-02 00:13:00 00:14:00 2019-01-02 00:15:00             60.0
2   123 2019-01-02 23:00:00 00:15:00 2019-01-03 00:16:00             60.0