Question

我有以下日期类型：

daytypes = {
    'Holidays_1': (
        date(2017, 4, 20),
        date(2017, 7, 10)
    ),
    'Holidays_2': (
        date(2017, 2, 5),
        date(2017, 5, 12),
        date(2017, 12, 14)
    )
}

我想知道新的日期是否是任何假期前一天的劳动节（周一至周五），而且它本身不是假期。

例如，我有以下三个datetime变量：

from datetime import datetime
dt1 = datetime.strptime("2017-02-04 11:12:20.0", "%Y-%m-%d %H:%M:%S.%f")
dt2 = datetime.strptime("2017-05-11 20:00:00.0", "%Y-%m-%d %H:%M:%S.%f")
dt3 = datetime.strptime("2017-02-06 12:00:00.0", "%Y-%m-%d %H:%M:%S.%f")

只有dt2对应于指定的规则。

我该怎么做？

Answer 1

试试这个

from datetime import datetime as date
from datetime import timedelta, datetime

daytypes = {
    'Holidays_1': (
        date(2017, 4, 20),
        date(2017, 7, 10)
    ),
    'Holidays_2': (
        date(2017, 2, 5),
        date(2017, 5, 12),
        date(2017, 12, 14)
    )
}

holidays = []
for d in daytypes:
    holidays.extend(daytypes[d])

dt2 = datetime.strptime("2017-05-11 20:00:00.0", "%Y-%m-%d %H:%M:%S.%f")

for h in holidays:
    if h-dt2 < timedelta(1) and h-dt2 > timedelta(0):
        return True

Answer 2

您可以使用weekday()功能获取日期对象的星期几。该函数返回一个从0（星期一）到6（星期日）的整数。

例如：

>>> from datetime import datetime, date
>>> dt1 = datetime.strptime("2017-02-04 11:12:20.0", "%Y-%m-%d %H:%M:%S.%f")
>>> holiday1 = date(2017, 02, 05)
>>> # Is the holiday one day ahead of the date?
>>> print date(dt1.year, dt1.month, dt1.day + 1) == holiday1 
True
>>> # Is the date a weekday?
>>> print dt1.weekday() in range(0,4) # Days from 0 to 4 are working days
False

文档在这里：https://docs.python.org/2/library/datetime.html#datetime.date.weekday

编辑：您可以使用for循环迭代每个假期：

for holiday_tuple in daytypes.keys():
    for holiday in holiday_tuple:
        # Do whatever you need here
        print date(dtx.year, dtx.month, dtx.day + 1) == holiday

其中dtx是任何声明的日期时间对象。您还可以使用additionalnal for循环来迭代日期时间对象列表。

Answer 3

这是一个快速完成此任务的功能。该功能将使用您的假期字典，并将根据这些假期评估输入日期，并将返回输入日期是周末，假日的1天内还是假日的某一天内的工作日。如果需要，您可以更改函数以返回更多或更少的细节，但这应该符合您描述的需求。

def check_holidays(holidays, input_date):
    if input_date.weekday() >= 5:
        return 'Is a weekend date'
    for holiday, holiday_dates in holidays.items():
        holiday_range = []
        for h_date in holiday_dates:
            holiday_range.append(h_date)
            holiday_range.append(h_date - timedelta(days=1))
        if input_date.date() in holiday_range:
            return 'Date is within 1 day of a holiday'
    return 'Date is a weekday and not within 1 day of a holiday'

Answer 4

我这样解决了：

holidays = []
for d in daytypes:
    holidays.extend(daytypes[d])

if (dt.isoweekday() >= 1 and dt.isoweekday() <= 5 and date(dt.year, dt.month, dt.day + 1) in holidays):
    print "Date is within 1 day of a holiday."
else:
    print "Date is a weekday and not within 1 day of a holiday."

如何将规则应用于日期时间变量？

4 个答案: