Python 如何计算熊猫之间的时间差

我在熊猫中有以下数据帧

 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  

 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()

我想计算

fina\u datetime

-

end\u time

，我正在做下面的事情

 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  

 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()

当日期进入下一个日期时，如第1行和第3行，上述代码存在问题。我如何在熊猫身上做到这一点

我想要的数据帧是

 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

---

IIUC，您可以屏蔽

结束时间

的位置，并按1添加日期：

# 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()
                        )

---

您正在计算df['end_time']，但不使用它来计算延迟。您正在使用

iss\u数据['end\u time']

。

iss_数据

是什么样子的？这是一个打字错误，对它进行了更改。另外，在第三行中，

end_时间

的格式不同。没有分号，而是在小时和分钟之间有一个句点-

00.15:00

。这又是一个打字错误。您能用当前代码得到的这些行的输出/错误更新您的帖子吗？