你可以这样做。

data = ['2023-03-15 18:39:42.443+0000', '2023-03-16 08:00:50.500+0000']
df = pd.DataFrame(data, columns=['interactionTime'])
df2 = pd.DataFrame()

df2['InterationTime'] = df['interactionTime'].str[:19]

print(df2)

输出

InterationTime
0  2023-03-15 18:39:42
1  2023-03-16 08:00:50

Process finished with exit code 0

该列似乎是一个完全有效的JSON的列表。
你可以将Json加载到一个有效的Python结构（一个dicts列表）中，获取最后一项（用[-1]索引），然后获取从interactionTime值的开始到19（[:19]）的元素：

import json

import pandas as pd

df = pd.DataFrame({
    "ID": [1],
    "weird_col": [
        '[{"assignedAgentId": "15", "assignedFullName": "Virtual Assistant", "interactionTime": "2023-03-15 18:39:42.443+0000", "interactiveSequence": 1, "dialogId": "fcf12345-e2d-4e2c-be5t-ed2bcf1831gh"}, {"assignedAgentId": "16", "assignedAgentFullName": "S", "interactionTime": "2023-03-16 08:00:50.500+0000", "dialogId": "fcf12345-e2d-4e2c-be5t-ed2bcf1831gh"}]'
    ],
})

df['dates'] = df['weird_col'].apply(
    lambda x: json.loads(x)[-1]['interactionTime'][:19]
)
print(df['dates'])

这就是说...处理完时区，时间偏移等，当你有一个非常好的有效日期时间对象时，你确定你想失去所有有趣的信息吗？当然，你知道得更好，但是...不知道...当你只需要使用pd.to_datetime时，让我很难过：

df['dates'] = df['weird_col'].apply(
    lambda x: pd.to_datetime(json.loads(x)[-1]['interactionTime'])
)
print(df['dates'])

考虑将它们解析为datetime示例！

• 如果以后需要，无需丢弃亚秒级数据
• 处理多个时区（可能非常重要）
• 易于控制显示格式
• 能够进行偏移量数学运算，并使用它进行更多运算

>>> df["interactionTime"] = pd.to_datetime(df["interactionTime"])

自定义显示格式

>>> df["interactionTime"].dt.strftime("%Y-%m-%d %H:%M:%S")  # custom display
0    2023-03-15 18:39:42
1    2023-03-16 08:00:50
Name: interactionTime, dtype: object

从开始日期开始计算秒数的方便性示例

>>> (df["interactionTime"] - pd.to_datetime("2023-03-01 00:00:00+0000")).dt.total_seconds()
0    1276782.443
1    1324850.500
Name: interactionTime, dtype: float64

您应该考虑更改时间格式，而不是提取前19个字符，如下所示：

import pandas as pd
from datetime import datetime

datas = [{"assignedAgentId": "15", "assignedFullName": "Virtual Assistant", 
         "interactionTime": "2023-03-15 18:39:42.443+0000", "interactiveSequence": 1, 
         "dialogId": "fcf12345-e2d-4e2c-be5t-ed2bcf1831gh"}, 
        {"assignedAgentId": "16", "assignedAgentFullName": "S", 
         "interactionTime": "2023-03-16 08:00:50.500+0000", 
         "dialogId": "fcf12345-e2d-4e2c-be5t-ed2bcf1831gh"}]

# Refactoring to 'records' json format
datas = [{k:[v] for k,v in d.items()} for d in datas]
# List of dataframes
dfx = [pd.read_json(str(dat).replace("'", '"'), orient='records') for dat in datas]
# Convert to specified format "%Y-%m-%d %H:%M:%S"
l=[]
for df in dfx:
    df["Interaction Time"] = pd.to_datetime(df.loc[:, "interactionTime"]).dt.strftime('%Y-%m-%d %H:%M:%S')
    l.append(df)

r = pd.concat(l, axis=0).reset_index(drop=True)
# Display the two r-columns
print(r[['interactionTime','Interaction Time']])

interactionTime     Interaction Time
0  2023-03-15 18:39:42.443+0000  2023-03-15 18:39:42
1  2023-03-16 08:00:50.500+0000  2023-03-16 08:00:50

感谢所有提供答案的人，特别是@BorrajaX，他指出了JSON格式。
使用他的方法给了我一个'字符串索引必须是整数'的错误，但它确实帮助我产生了以下内容，实现了我所需要的：

df["interactions"] = df["interactions"].apply(lambda x: json.dumps(x))
df["interactions"] = df["interactions"].apply(lambda x: json.loads(x))

# find the last instance of interactionTime and create a new column with everything after it
new_column_values = []
for interaction in datas2['interactions']:
    interaction_str = json.dumps(interaction)  # convert interaction object to a string
    interaction_time_index = interaction_str.rfind('interactionTime')  # find the last index of interactionTime
    if interaction_time_index != -1:
        new_column_value = interaction_str[interaction_time_index+len('"interactionTime": '):interaction_time_index+len('"interactionTime": ')+19]  # extract everything after interactionTime
        new_column_values.append(new_column_value)
    else:
        new_column_values.append('')  # if interactionTime not found, set value to empty string

# create a new column in the dataframe with the extracted values
df.loc[:, 'new_column'] = new_column_values