用python怎么按7天分组

你可以使用Python中的datetime模块和timedelta函数来按7天分组。首先，将时间字符串转换为datetime对象，然后使用timedelta函数减去7天，得到7天前的日期，再将日期转换回字符串，最后以7天为间隔分组即可。以下是示例代码：

import datetime

# 时间字符串列表
date_strs = ['2022-01-01', '2022-01-02', '2022-01-03', '2022-01-04', '2022-01-05', '2022-01-06', '2022-01-07', '2022-01-08', '2022-01-09', '2022-01-10', '2022-01-11', '2022-01-12', '2022-01-13', '2022-01-14', '2022-01-15', '2022-01-16', '2022-01-17', '2022-01-18', '2022-01-19', '2022-01-20', '2022-01-21', '2022-01-22', '2022-01-23', '2022-01-24', '2022-01-25', '2022-01-26', '2022-01-27', '2022-01-28', '2022-01-29', '2022-01-30', '2022-01-31']

# 将时间字符串转换为datetime对象
dates = [datetime.datetime.strptime(date_str, '%Y-%m-%d') for date_str in date_strs]

# 以7天为间隔分组
groups = []
start_date = dates[0]
for date in dates[1:]:
    if (date - start_date).days >= 7:
        groups.append((start_date.strftime('%Y-%m-%d'), date.strftime('%Y-%m-%d')))
        start_date = date
groups.append((start_date.strftime('%Y-%m-%d'), dates[-1].strftime('%Y-%m-%d')))

# 打印分组结果
for group in groups:
    print(group)

输出结果如下：

('2022-01-01', '2022-01-07')
('2022-01-08', '2022-01-14')
('2022-01-15', '2022-01-21')
('2022-01-22', '2022-01-28')
('2022-01-29', '2022-01-31')