时间读入为时间增量,前面有大量天

datetime.timdelta 对象将天、秒和微秒存储为属性。我们可以通过 pandas.DataFrame 具有 dt 以下内容:

设置一些虚拟数据

import datetime as dt
import pandas as pd

df = pd.DataFrame(
    data=(
        dt.timedelta(days=17711, hours=i, minutes=i, seconds=i) for i in range(0, 10)
    ), 
    columns=['Duration']
)

print(df['Duration'])

             Duration
0 17711 days 00:00:00
1 17711 days 01:01:01
2 17711 days 02:02:02
3 17711 days 03:03:03
4 17711 days 04:04:04
5 17711 days 05:05:05
6 17711 days 06:06:06
7 17711 days 07:07:07
8 17711 days 08:08:08
9 17711 days 09:09:09
Name: Duration, dtype: timedelta64[ns]

计算秒数并将其转换为小时

print(df['Duration'].dt.seconds / 3600)

0    0.000000
1    1.016944
2    2.033889
3    3.050833
4    4.067778
5    5.084722
6    6.101667
7    7.118611
8    8.135556
9    9.152500
Name: Duration, dtype: float64    

只有几个小时

print(df['Duration'].dt.seconds // 3600)

0    0
1    1
2    2
3    3
4    4
5    5
6    6
7    7
8    8
9    9
Name: Duration, dtype: int64

使用 split() 使用regex应该做你想做的事我想:

df[['Days', 'Time']] = df['Duration'].str.split('.* days', expand=True)

这将把列分成两部分,然后您可以使用“time”键调用它。

代码 以下内容:

>>> import pandas as pd
>>> d = ['17711 days 08:27:00', 
...      '17711 days 07:56:00',
...      '17711 days 04:22:00',
...      '17711 days 07:29:00',
...      '17711 days 06:46:00']
>>> df = pd.DataFrame({'Duration': d})
>>> df[['Days', 'Time']] = df['Duration'].str.split('.* days', expand=True)
>>> df.Time = pd.to_timedelta(df.Time)
>>> df.Time.head()
0   08:27:00
1   07:56:00
2   04:22:00
3   07:29:00
4   06:46:00
Name: Time, dtype: timedelta64[ns]