如果我正确理解了您的问题，您正在寻找的是time结构，而不是datetime。
考虑pandas，您可以使用dt.time修改或创建新列。
可重现的例子：

import pandas as pd
df = pd.DataFrame({'foo': pd.date_range("2018-01-01", periods=5, freq="H")})

df.assign(bar = df['foo'].dt.time)
print(df)

import pandas as pd
import array
from pandas import Timestamp
from datetime import datetime as dt

test = ([[Timestamp('2010-01-01 11:30:00'), Timestamp('2010-01-01 13:30:00')], 
[Timestamp('2010-01-02 11:30:00'), Timestamp('2010-01-02 12:10:00')], 
[Timestamp('2010-01-05 16:40:00'), Timestamp('2010-01-05 18:10:00')],
[Timestamp('2010-12-30 14:30:00'), Timestamp('2010-12-30 15:20:00')], 
[Timestamp('2010-12-31 01:40:00'), Timestamp('2010-12-31 02:40:00')], 
[Timestamp('2010-12-31 14:40:00'), Timestamp('2010-12-31 15:40:00')]]) 

# assuming the date is always the same just subtract the two dates

pd.Series([ v2 - v1 for v1, v2 in test])

0   0 days 02:00:00
1   0 days 00:40:00
2   0 days 01:30:00
3   0 days 00:50:00
4   0 days 01:00:00
5   0 days 01:00:00
dtype: timedelta64[ns]

# if we only care about the time difference and wish to ignore date then set date to common value 1900-01-01

v1 = pd.Series([ dt.strptime('1900-01-01 ' + str(pd.Timestamp.time(v1)), '%Y-%m-%d %H:%M:%S') for v1, v2 in test])
v2 = pd.Series([ dt.strptime('1900-01-01 ' + str(pd.Timestamp.time(v2)), '%Y-%m-%d %H:%M:%S') for v1, v2 in test])

# now subtract first timestamp in tuple from first to find elapsed time

v2 - v1

0   0 days 02:00:00
1   0 days 00:40:00
2   0 days 01:30:00
3   0 days 00:50:00
4   0 days 01:00:00
5   0 days 01:00:00
dtype: timedelta64[ns]

使用您提供的样本数据，无论您是否使用通用基线日期，结果都是相同的