新答案

将文本转换为Timedeltadf['Origin Time(Local)'] = pd.to_timedelta(df['Origin Time(Local)'])

df['Seconds'] = df['Origin Time(Local)'].dt.total_seconds()

旧答案

考虑数据帧dfdf = pd.DataFrame(dict(Date=pd.date_range('2017-03-01', '2017-03-02', freq='2H')))

Date

0 2017-03-01 00:00:00

1 2017-03-01 02:00:00

2 2017-03-01 04:00:00

3 2017-03-01 06:00:00

4 2017-03-01 08:00:00

5 2017-03-01 10:00:00

6 2017-03-01 12:00:00

7 2017-03-01 14:00:00

8 2017-03-01 16:00:00

9 2017-03-01 18:00:00

10 2017-03-01 20:00:00

11 2017-03-01 22:00:00

12 2017-03-02 00:00:00

从时间戳中减去最近一天并使用total_seconds。total_seconds是Timedelta的属性。通过计算两个Timestamps序列之间的差异，我们得到一个Timedeltas序列。(df.Date - df.Date.dt.floor('D')).dt.total_seconds()

# equivalent to

# (df.Date - pd.to_datetime(df.Date.dt.date)).dt.total_seconds()

0 0.0

1 7200.0

2 14400.0

3 21600.0

4 28800.0

5 36000.0

6 43200.0

7 50400.0

8 57600.0

9 64800.0

10 72000.0

11 79200.0

12 0.0

Name: Date, dtype: float64

把它放到一个新的列中df.assign(seconds=(df.Date - df.Date.dt.floor('D')).dt.total_seconds())

Date seconds

0 2017-03-01 00:00:00 0.0

1 2017-03-01 02:00:00 7200.0

2 2017-03-01 04:00:00 14400.0

3 2017-03-01 06:00:00 21600.0

4 2017-03-01 08:00:00 28800.0

5 2017-03-01 10:00:00 36000.0

6 2017-03-01 12:00:00 43200.0

7 2017-03-01 14:00:00 50400.0

8 2017-03-01 16:00:00 57600.0

9 2017-03-01 18:00:00 64800.0

10 2017-03-01 20:00:00 72000.0

11 2017-03-01 22:00:00 79200.0

12 2017-03-02 00:00:00 0.0