python - 将纪元时间转换为本地时间

Question

我有一个 pandas 数据框，其中一列是表示为纪元时间的时间。数据框如下所示:

0    1539340322842
1    1539340426841
2    1539340438482
3    1539340485658
4    1539340495920
Name: Time, dtype: int64

我试过这个 df["local_time"] = df.epoch_time.dt.tz_localize("UTC") 但这并没有给我本地时间的结果。这是上面操作的结果:

                        local_time  
0 1970-01-01 00:25:39.340322842+00:00  
1 1970-01-01 00:25:39.340426841+00:00  
2 1970-01-01 00:25:39.340438482+00:00  
3 1970-01-01 00:25:39.340485658+00:00  
4 1970-01-01 00:25:39.340495920+00:00  

给我想要的结果的另一件事是:

def convert_time(x):                                              
    return time.strftime("%Y-%m-%d %H:%M:%S", time.localtime(x/1000)) 

df["local_time"] = df["epoch_time"].apply(convert_time)

无论如何，我可以对上述操作进行矢量化以获得我想要的格式的日期时间吗？

Answer 1

IIUC，我认为你需要 pandas to_datetime 和 units='s':

pd.to_datetime(df.Time/1000,unit='s')
0   2018-10-12 10:32:02.842000008
1   2018-10-12 10:33:46.841000080
2   2018-10-12 10:33:58.482000113
3   2018-10-12 10:34:45.657999992
4   2018-10-12 10:34:55.920000076
Name: Time, dtype: datetime64[ns]

或者使用astype as:

((df.Time)/1000).astype("datetime64[s]")
0   2018-10-12 10:32:02
1   2018-10-12 10:33:46
2   2018-10-12 10:33:58
3   2018-10-12 10:34:45
4   2018-10-12 10:34:55
Name: Time, dtype: datetime64[ns]

或者

pd.to_datetime(df.Time/1000,unit='s',utc=True)
0   2018-10-12 10:32:02.842000008+00:00
1   2018-10-12 10:33:46.841000080+00:00
2   2018-10-12 10:33:58.482000113+00:00
3   2018-10-12 10:34:45.657999992+00:00
4   2018-10-12 10:34:55.920000076+00:00
Name: Time, dtype: datetime64[ns, UTC]

<小时/>

由于'Asia/Kolkata'是05:30:00，只需添加Timedelta:

pd.to_datetime(df.Time/1000,unit='s')+pd.Timedelta("05:30:00")
0   2018-10-12 16:02:02.842000008
1   2018-10-12 16:03:46.841000080
2   2018-10-12 16:03:58.482000113
3   2018-10-12 16:04:45.657999992
4   2018-10-12 16:04:55.920000076
Name: Time, dtype: datetime64[ns]