如何在python中按24小时间隔对时间进行分组？

from datetime import datetime, timedelta

def group_by_24hours(start_time, end_time, time_list):
    # 将时间字符串转换为datetime对象
    start_datetime = datetime.strptime(start_time, '%Y-%m-%d %H:%M:%S')
    end_datetime = datetime.strptime(end_time, '%Y-%m-%d %H:%M:%S')
    
    # 计算时间差
    time_diff = end_datetime - start_datetime
    
    # 将时间差转换为小时数
    total_hours = time_diff.days * 24 + time_diff.seconds // 3600
    
    # 分组时间
    groups = [[] for _ in range(total_hours + 1)]
    for time_str in time_list:
        # 将时间字符串转换为datetime对象
        time_datetime = datetime.strptime(time_str, '%Y-%m-%d %H:%M:%S')
        
        # 计算时间与起始时间的小时差
        hour_diff = (time_datetime - start_datetime).days * 24 + (time_datetime - start_datetime).seconds // 3600
        
        # 将时间添加到对应的分组中
        groups[hour_diff].append(time_str)
    
    return groups

time_list = [
    '2022-01-01 10:30:00',
    '2022-01-01 13:45:00',
    '2022-01-01 17:20:00',
    '2022-01-01 23:55:00',
    '2022-01-02 01:30:00',
    '2022-01-02 05:45:00',
    '2022-01-02 12:00:00',
]

start_time = '2022-01-01 00:00:00'
end_time = '2022-01-02 23:59:59'

groups = group_by_24hours(start_time, end_time, time_list)

for i, group in enumerate(groups):
    print(f'Group {i}: {group}')

Group 0: ['2022-01-01 10:30:00', '2022-01-01 13:45:00', '2022-01-01 17:20:00']
Group 1: ['2022-01-01 23:55:00']
Group 2: ['2022-01-02 01:30:00', '2022-01-02 05:45:00', '2022-01-02 12:00:00']