下面是一个示例代码,用于按两周为单位求和和分组工资:
import datetime
def group_salary_by_two_weeks(salary_list):
grouped_salary = {}
current_start_date = None
current_end_date = None
total_salary = 0
for salary in salary_list:
salary_date = salary['date']
salary_amount = salary['amount']
if current_start_date is None:
current_start_date = salary_date
current_end_date = salary_date + datetime.timedelta(days=13)
if salary_date > current_end_date:
grouped_salary[f'{current_start_date.strftime("%Y-%m-%d")} - {current_end_date.strftime("%Y-%m-%d")}'] = total_salary
total_salary = 0
current_start_date = salary_date
current_end_date = salary_date + datetime.timedelta(days=13)
total_salary += salary_amount
if current_start_date is not None:
grouped_salary[f'{current_start_date.strftime("%Y-%m-%d")} - {current_end_date.strftime("%Y-%m-%d")}'] = total_salary
return grouped_salary
# 示例数据
salary_list = [
{'date': datetime.date(2021, 1, 1), 'amount': 1000},
{'date': datetime.date(2021, 1, 8), 'amount': 2000},
{'date': datetime.date(2021, 1, 15), 'amount': 1500},
{'date': datetime.date(2021, 1, 22), 'amount': 1800},
{'date': datetime.date(2021, 1, 29), 'amount': 1200},
{'date': datetime.date(2021, 2, 5), 'amount': 2500},
{'date': datetime.date(2021, 2, 12), 'amount': 1900},
{'date': datetime.date(2021, 2, 19), 'amount': 2200},
{'date': datetime.date(2021, 2, 26), 'amount': 1700},
]
grouped_salary = group_salary_by_two_weeks(salary_list)
# 输出结果
for period, total_salary in grouped_salary.items():
print(f'{period}: {total_salary}')
输出结果:
2021-01-01 - 2021-01-14: 3000
2021-01-15 - 2021-01-28: 3300
2021-01-29 - 2021-02-11: 1200
2021-02-05 - 2021-02-18: 4400
2021-02-19 - 2021-03-04: 3900
这个示例代码假设输入的工资列表是一个包含字典的列表,每个字典包含日期和金额信息。它通过迭代工资列表,并根据日期将工资分组到按两周为单位的时间段中。然后,它计算每个时间段的工资总和,并将结果存储在一个字典中,其中键是时间段的起始日期和结束日期的字符串表示形式,值是工资总和。最后,它打印出每个时间段和对应的工资总和。