Python3.x 获取字符串中的两个工作日之间的差异

您应该使用datetime.weekday()方法来提取感兴趣的天数的星期几.假设您有类似于上面显示的格式的日期，则很容易进行转换，并且只需将日期索引用于您的"允许开始日期"(星期一=0).

然后，你可以设置一个小函数，在你的sorted人列表中寻找下一个开始日期，并计算出你需要等待多少天.

下面的示例做到了这一点，并根据需要"滚动"了周末.

代码:

from datetime import datetime, timedelta
from bisect import bisect_left

start_date = "2022-09-09"

valid_start_dates = [1, 4] # It can be any days in sorted order

start_date_object = datetime.strptime(start_date, "%Y-%m-%d")

d=start_date_object.weekday()
print(f'the numbered day of the week is: {d}')

def days_till_start(day, valid_start_days):
    idx = bisect_left(valid_start_days, day)
    if idx >= len(valid_start_days): # wrap around to next start
        return valid_start_days[0] + 7 - day
    elif day == valid_start_days[idx]:
        return 0
    else:
        return valid_start_days[idx] - day

print(days_till_start(d, valid_start_dates))

start_dates = ['2022-09-05', '2022-09-06', '2022-09-07', '2022-09-08', '2022-09-09', '2022-09-10']

start_wkdys = [datetime.strptime(d, "%Y-%m-%d").weekday() for d in start_dates]

for d in start_wkdys:
    print(f'day index is: {d}')
    print(f'next start date is {days_till_start(d, valid_start_dates)} away')
    print()

输出:

the numbered day of the week is: 4
0

day index is: 0
next start date is 1 away

day index is: 1
next start date is 0 away

day index is: 2
next start date is 2 away

day index is: 3
next start date is 1 away

day index is: 4
next start date is 0 away

day index is: 5
next start date is 3 away