我想了解以下内容:给定一个日期(datetime对象),一周中对应的日期是什么?
例如,星期天是第一天,星期一是第二天。。等等
然后如果输入的内容类似于今天的日期。
实例
>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday() # what I look for
产量可能是6(因为现在是星期五)
我想了解以下内容:给定一个日期(datetime对象),一周中对应的日期是什么?
例如,星期天是第一天,星期一是第二天。。等等
然后如果输入的内容类似于今天的日期。
实例
>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday() # what I look for
产量可能是6(因为现在是星期五)
当前回答
使用Canlendar模块
import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])
其他回答
我为CodeChef问题解决了这个问题。
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"))
在MATLAB中,高斯方法
day_name={'Sun','Mon','Tue','Wed','Thu','Fri','Sat'}
month_offset=[0 3 3 6 1 4 6 2 5 0 3 5]; % common year
% input date
y1=2022
m1=11
d1=22
% is y1 leap
if mod(y1,4)==0 && mod(y1,100)==0 && mod(y1,400)==0
month_offset=[0 3 4 0 2 5 0 3 6 1 4 6]; % offset for leap year
end
% Gregorian calendar
weekday_gregor=rem( d1+month_offset(m1) + 5*rem(y1-1,4) + 4*rem(y1-1,100) + 6*rem(y1-1,400),7)
day_name{weekday_gregor+1}
0:星期日1:星期一。。6:星期六
1700/1/1之后的日期不需要导入的解决方案
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')
一个简单、直接但尚未提及的选项:
import datetime
...
givenDateObj = datetime.date(2017, 10, 20)
weekday = givenDateObj.isocalendar()[2] # 5
weeknumber = givenDateObj.isocalendar()[1] # 42
要让星期天1点到星期六7点,这是解决问题的最简单方法:
datetime.date.today().toordinal()%7 + 1
所有这些:
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