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如何在Python中获取日期的星期几？

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我想找出以下内容：给定一个日期（ datetime 对象），一周的相应日期是什么 .

例如星期日是第一天，星期一：第二天......依此类推

然后，如果输入是今天的日期 .

示例

>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday()  # what I look for

输出可能 6 （自周五以来）

python datetime weekday

15 回答

  • 10
    回复于
  • 622

    使用 weekday() （docs）：

    >>> import datetime
>>> datetime.datetime.today()
datetime.datetime(2012, 3, 23, 23, 24, 55, 173504)
>>> datetime.datetime.today().weekday()
4

    从文档：

    以星期为单位返回星期几，其中星期一为0，星期日为6 .

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  • 20

    在1700/1/1之后没有导入日期的解决方案

    def weekDay(year, month, day):
    offset = [0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334]
    week   = ['Sunday', 
              'Monday', 
              'Tuesday', 
              'Wednesday', 
              'Thursday',  
              'Friday', 
              'Saturday']
    afterFeb = 1
    if month > 2: afterFeb = 0
    aux = year - 1700 - afterFeb
    # dayOfWeek for 1700/1/1 = 5, Friday
    dayOfWeek  = 5
    # partial sum of days betweem current date and 1700/1/1
    dayOfWeek += (aux + afterFeb) * 365                  
    # leap year correction    
    dayOfWeek += aux / 4 - aux / 100 + (aux + 100) / 400     
    # sum monthly and day offsets
    dayOfWeek += offset[month - 1] + (day - 1)               
    dayOfWeek %= 7
    return dayOfWeek, week[dayOfWeek]

print weekDay(2013, 6, 15) == (6, 'Saturday')
print weekDay(1969, 7, 20) == (0, 'Sunday')
print weekDay(1945, 4, 30) == (1, 'Monday')
print weekDay(1900, 1, 1)  == (1, 'Monday')
print weekDay(1789, 7, 14) == (2, 'Tuesday')
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  • 1

    如果日期是日期时间对象，则这是一种解决方案 .

    import datetime
def dow(date):
    days=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
    dayNumber=date.weekday()
    print days[dayNumber]
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  • 3

    使用Canlendar模块

    import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])
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  • 3

    如果您想要英文日期：

    >>> from datetime import date
>>> datetime.datetime.today().strftime('%A')
'Wednesday'

    阅读更多：https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior

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  • 5

    如果您想要英文日期：

    from datetime import date
import calendar
my_date = date.today()
calendar.day_name[my_date.weekday()]  #'Wednesday'
    回复于
  • 34

    我为codechef question解决了这个问题 .

    import datetime
dt = '21/03/2012'
day, month, year = (int(x) for x in dt.split('/'))    
ans = datetime.date(year, month, day)
print ans.strftime("%A")
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  • 205

    这是我的python3实现 .

    months = {'jan' : 1, 'feb' : 4, 'mar' : 4, 'apr':0, 'may':2, 'jun':5, 'jul':6, 'aug':3, 'sep':6, 'oct':1, 'nov':4, 'dec':6}
dates = {'Sunday':1, 'Monday':2, 'Tuesday':3, 'Wednesday':4, 'Thursday':5, 'Friday':6, 'Saterday':0}
ranges = {'1800-1899':2, '1900-1999':0, '2000-2099':6, '2100-2199':4, '2200-2299':2}

def getValue(val, dic):
    if(len(val)==4):
        for k,v in dic.items():
            x,y=int(k.split('-')[0]),int(k.split('-')[1])
            val = int(val)
            if(val>=x and val<=y):
                return v
    else:
        return dic[val]

def getDate(val):
    return (list(dates.keys())[list(dates.values()).index(val)]) 



def main(myDate):
    dateArray = myDate.split('-')
    # print(dateArray)
    date,month,year = dateArray[2],dateArray[1],dateArray[0]
    # print(date,month,year)

    date = int(date)
    month_v = getValue(month, months)
    year_2 = int(year[2:])
    div = year_2//4
    year_v = getValue(year, ranges)
    sumAll = date+month_v+year_2+div+year_v
    val = (sumAll)%7
    str_date = getDate(val)

    print('{} is a {}.'.format(myDate, str_date))

if __name__ == "__main__":
    testDate = '2018-mar-4'
    main(testDate)
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  • 0

    为了让星期日从1到星期六为7，这是你问题最简单的解决方案：

    datetime.date.today().toordinal()%7 + 1

    他们都是：

    import datetime

today = datetime.date.today()
sunday = today - datetime.timedelta(today.weekday()+1)

for i in range(7):
    tmp_date = sunday + datetime.timedelta(i)
    print tmp_date.toordinal()%7 + 1, '==', tmp_date.strftime('%A')

    输出：

    1 == Sunday
2 == Monday
3 == Tuesday
4 == Wednesday
5 == Thursday
6 == Friday
7 == Saturday
    回复于
  • 9

    datetime库有时会出现strptime（）的错误，所以我切换到dateutil库 . 以下是如何使用它的示例：

    from dateutil import parser
parser.parse('January 11, 2010').strftime("%a")

    您从中获得的输出是 'Mon' . 如果您希望输出为'Monday'，请使用以下命令：

    parser.parse('January 11, 2010').strftime("%A")

    这对我来说非常快 . 我在使用datetime库时遇到了问题，因为我想存储工作日名称而不是工作日编号，而使用datetime库的格式导致了问题 . 如果你没有遇到这个问题，太好了！如果你是，你可以肯定地去做，因为它有一个更简单的语法 . 希望这可以帮助 .

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  • 1

    假设您获得了日，月和年，您可以：

    import datetime
DayL = ['Mon','Tues','Wednes','Thurs','Fri','Satur','Sun']
date = DayL[datetime.date(year,month,day).weekday()] + 'day'
#Set day, month, year to your value
#Now, date is set as an actual day, not a number from 0 to 6.

print(date)
    回复于
  • 6

    如果您有理由避免使用datetime模块，那么此功能将起作用 .

    注意：假设从Julian到公历的变化发生在1582年 . 如果您感兴趣的日历不适用，那么如果年份> 1582则更改行：相应地 .

    def dow(year,month,day):
    """ day of week, Sunday = 1, Saturday = 7
     http://en.wikipedia.org/wiki/Zeller%27s_congruence """
    m, q = month, day
    if m == 1:
        m = 13
        year -= 1
    elif m == 2:
        m = 14
        year -= 1
    K = year % 100    
    J = year // 100
    f = (q + int(13*(m + 1)/5.0) + K + int(K/4.0))
    fg = f + int(J/4.0) - 2 * J
    fj = f + 5 - J
    if year > 1582:
        h = fg % 7
    else:
        h = fj % 7
    if h == 0:
        h = 7
    return h
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  • 0

    如果您不仅仅依赖于 datetime 模块， calendar 可能是更好的选择 . 例如，这将为您提供日期代码：

    calendar.weekday(2017,12,22);

    这会给你一天的感觉：

    days = ["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
days[calendar.weekday(2017,12,22)]

    或者以python的风格，作为一个班轮：

    ["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"][calendar.weekday(2017,12,22)]
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  • 19

    这里是如何将日期列表转换为日期

    import datetime,time
ls={'1/1/2007','1/2/2017'}
dt=datetime.datetime.strptime(ls[1], "%m/%d/%Y")
print(dt)
print(dt.month)
print(dt.year)
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