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')

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')

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

以下是以DD-MM-YYYY格式输入日期的代码。您可以通过更改“%d-%m-%Y”的顺序以及更改分隔符来更改输入格式。

import datetime
try:
    date = input()
    date_time_obj = datetime.datetime.strptime(date, '%d-%m-%Y')
    print(date_time_obj.strftime('%A'))
except ValueError:
    print("Invalid date.")

如果您不完全依赖datetime模块，日历可能是更好的选择。例如，这将为您提供日期代码：

calendar.weekday(2017,12,22);

这将给你一天：

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

或者以python的风格，作为一行：

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

import numpy as np

def date(df):
    df['weekday'] = df['date'].dt.day_name()

    conditions = [(df['weekday'] == 'Sunday'),
              (df['weekday'] == 'Monday'),
              (df['weekday'] == 'Tuesday'),
              (df['weekday'] == 'Wednesday'),
              (df['weekday'] == 'Thursday'),
              (df['weekday'] == 'Friday'),
              (df['weekday'] == 'Saturday')]

    choices = [0, 1, 2, 3, 4, 5, 6]

    df['week'] = np.select(conditions, choices)

    return df