有没有一种方法可以创建一个表示某个月的对象，并且可以进行迭代？

Question

在工作中，我们需要填写一份月薪表，上面写着我们教过的所有课程。每个老师一周中的每一天都教授完全相同的课程

我试图做的是允许用户输入一个月，然后创建一个表示该月的对象，然后按天循环对象。如果这一天是星期一，请填写该老师在星期一所教的课程

到目前为止，我已经尝试使用datetime模块。我可以想出一种方法来输入开始和结束日期，但不能创建一个月，我可以通过一个用户输入来处理我想要的方式

我希望能够设置它，其他员工只需输入他们的姓名和月份，程序就可以完成剩下的工作，因为这不是一个技术熟练的环境。如果有人知道一个方法，这可以通过datetime或其他模块来完成，我会很高兴

Answer 1

您可以使用datetime和timedelta-将天添加到开始日期：

import datetime

# 0 == Monday, 1== Tuesday .... the list is the list of lessons taught on that day
teach = {0: ["A","B"], 1:["C","D"],2:["E"],3:["A","E"], 4:["E","B","E"]}

# get a starting month
while True:
    try:
        month = int(input("Month [1-12]: "))
    except:
        continue

    if 1 <= month <= 12:
        break

# date to start at
start = datetime.datetime(2019,month,1)

# create all days of that month 
m = [[date, teach.get(date.weekday())]    # date  + lessons
     for date in (start + datetime.timedelta(days=n) for n in range(32))  # all days
     if date.month == month]  # only those that fit into the month

print()
for day in m:
    print(day) 

输出：

Month [1-12]: 4
[datetime.datetime(2019, 4, 1, 0, 0), ['A', 'B']]
[datetime.datetime(2019, 4, 2, 0, 0), ['C', 'D']]
[datetime.datetime(2019, 4, 3, 0, 0), ['E']]
[datetime.datetime(2019, 4, 4, 0, 0), ['A', 'E']]
[datetime.datetime(2019, 4, 5, 0, 0), ['E', 'B', 'E']]
[datetime.datetime(2019, 4, 6, 0, 0), None]
[datetime.datetime(2019, 4, 7, 0, 0), None]
[datetime.datetime(2019, 4, 8, 0, 0), ['A', 'B']]
[datetime.datetime(2019, 4, 9, 0, 0), ['C', 'D']]
[datetime.datetime(2019, 4, 10, 0, 0), ['E']]
[datetime.datetime(2019, 4, 11, 0, 0), ['A', 'E']]
[datetime.datetime(2019, 4, 12, 0, 0), ['E', 'B', 'E']]
[datetime.datetime(2019, 4, 13, 0, 0), None]
[datetime.datetime(2019, 4, 14, 0, 0), None]
[datetime.datetime(2019, 4, 15, 0, 0), ['A', 'B']]
[datetime.datetime(2019, 4, 16, 0, 0), ['C', 'D']]
[datetime.datetime(2019, 4, 17, 0, 0), ['E']]
[datetime.datetime(2019, 4, 18, 0, 0), ['A', 'E']]
[datetime.datetime(2019, 4, 19, 0, 0), ['E', 'B', 'E']]
[datetime.datetime(2019, 4, 20, 0, 0), None]
[datetime.datetime(2019, 4, 21, 0, 0), None]
[datetime.datetime(2019, 4, 22, 0, 0), ['A', 'B']]
[datetime.datetime(2019, 4, 23, 0, 0), ['C', 'D']]
[datetime.datetime(2019, 4, 24, 0, 0), ['E']]
[datetime.datetime(2019, 4, 25, 0, 0), ['A', 'E']]
[datetime.datetime(2019, 4, 26, 0, 0), ['E', 'B', 'E']]
[datetime.datetime(2019, 4, 27, 0, 0), None]
[datetime.datetime(2019, 4, 28, 0, 0), None]
[datetime.datetime(2019, 4, 29, 0, 0), ['A', 'B']]
[datetime.datetime(2019, 4, 30, 0, 0), ['C', 'D']]

独行：

• datetime.date.weekday()
• datetime.timedelta
• Why dict.get(key) instead of dict[key]?
• list comprehensions

你必须把学校停课的日子去掉——这只看工作日

Answer 2

^{}正是您所需要的。这个接口有点奇怪，因为源材料和主要用例是icalendar的递归规则，dateutil遵循这个术语，但除此之外，它应该处理您的用例