由于您标记了pandas，因此可以使用pd.Series构造函数和pd.to_datetime来处理此问题。
试试这个：

import pandas as pd

s = "[Timestamp('2022-08-13 17:25:00'), Timestamp('2022-08-13 18:02:00'), Timestamp('2022-08-13 18:46:00')]"

L = (
        pd.Series(s.strip("[]").split(","))
            .str.extract("'(.*)'", expand=False)
            .apply(pd.to_datetime)
            .to_list()
    )

#输出：

print(L)

[Timestamp('2022-08-13 17:25:00'), Timestamp('2022-08-13 18:02:00'), Timestamp('2022-08-13 18:46:00')]

一个没有Pandas的解决方案，使用regex来识别大字符串中的相关字符串，然后datetime.strptime将它们转换为实际的datetime：

tstring = "[Timestamp('2022-08-13 17:25:00'), Timestamp('2022-08-13 18:02:00'), Timestamp('2022-08-13 18:46:00')]"

import re
from datetime import datetime as dt
dtstrings = re.findall(r"\('([^,]*)'\)",tstring)
datetimes = [dt.strptime(x,'%Y-%m-%d %H:%M:%S') for x in dtstrings]

datetimes
# Out[84]: [datetime.datetime(2022, 8, 13, 17, 25),
# datetime.datetime(2022, 8, 13, 18, 2),
# datetime.datetime(2022, 8, 13, 18, 46)]

一种没有Pandas的方法，生成一系列时间戳：

timestamps="[Timestamp('2022-08-13 17:25:00'), Timestamp('2022-08-13 18:02:00'), 
Timestamp('2022-08-13 18:46:00')]"
# remove square brackets from string and split on "," to create a list from string
timestamps = timestamps[1 :-1 : ].split(", ")
# remove timestamp substrings
dates_ = [elem.replace("Timestamp('","").replace("')","") for elem in 
timestamps]
import datetime
#as datetime
ts_as_datetime = [datetime.datetime.strptime(elem, '%Y-%m-%d %H:%M:%S') for elem in dates_]

#and finally as Timestamp as required:
ts_as_timestamp = [datetime.datetime.timestamp(element) for element in ts_as_datetime]

问候你，乔霍娜。