如何在给定日期的情况下获得星期几？

使用 weekday()：

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

从 documentation：

以整数形式返回星期几，其中星期一为 0，星期日为 6。

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如果你想用英文写日期：

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

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如果你想用英文写日期：

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

阅读更多：https://docs.python.org/3/library/datetime.html#strftime-strptime-behavior

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当星期一为 0 且星期日为 6 时使用 date.weekday()

或者

date.isoweekday() 当星期一是 1 和星期日是 7 时

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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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1700/1/1 之后不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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如果您将日期作为字符串，则使用 pandas 的 Timestamp 可能会更容易

import pandas as pd
df = pd.Timestamp("2019-04-12")
print(df.dayofweek, df.weekday_name)

输出：

4 Friday

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这是一个简单的代码片段来解决这个问题

import datetime

intDay = datetime.date(year=2000, month=12, day=1).weekday()
days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
print(days[intDay])

输出应该是：

Friday

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如果日期是日期时间对象，这是一个解决方案。

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

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datetime 库有时会出现 strptime() 错误，所以我切换到 dateutil 库。这是一个如何使用它的示例：

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“星期一”，请使用以下命令：

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

这对我来说很快就奏效了。我在使用 datetime 库时遇到问题，因为我想存储工作日名称而不是工作日编号，并且使用 datetime 库的格式会导致问题。如果您对此没有任何问题，那就太好了！如果你是，你绝对可以这样做，因为它也有更简单的语法。希望这可以帮助。

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假设您有时间戳：字符串变量，YYYY-MM-DD HH:MM:SS

第 1 步：使用打击代码将其转换为 dateTime 函数...

df['timeStamp'] = pd.to_datetime(df['timeStamp'])

第 2 步：现在您可以提取所有必需的功能，如下所示，这将为每个文件-小时、月、星期几、年、日期创建新列

df['Hour'] = df['timeStamp'].apply(lambda time: time.hour)
df['Month'] = df['timeStamp'].apply(lambda time: time.month)
df['Day of Week'] = df['timeStamp'].apply(lambda time: time.dayofweek)
df['Year'] = df['timeStamp'].apply(lambda t: t.year)
df['Date'] = df['timeStamp'].apply(lambda t: t.day)

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假设给定日期、月份和年份，您可以执行以下操作：

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)

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如果您有理由避免使用 datetime 模块，那么此功能将起作用。

注意：假设从儒略历到公历的变化发生在 1582 年。如果您感兴趣的日历不是这样，那么如果年份 > 1582：相应地更改行。

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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这不需要星期几的评论。我推荐这个代码~！

import datetime


DAY_OF_WEEK = {
    "MONDAY": 0,
    "TUESDAY": 1,
    "WEDNESDAY": 2,
    "THURSDAY": 3,
    "FRIDAY": 4,
    "SATURDAY": 5,
    "SUNDAY": 6
}

def string_to_date(dt, format='%Y%m%d'):
    return datetime.datetime.strptime(dt, format)

def date_to_string(date, format='%Y%m%d'):
    return datetime.datetime.strftime(date, format)

def day_of_week(dt):
    return string_to_date(dt).weekday()


dt = '20210101'
if day_of_week(dt) == DAY_OF_WEEK['SUNDAY']:
    None

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如果您不完全依赖 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);

这会给你一天本身：

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

或者以 python 的风格，作为一个衬里：

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

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import datetime
int(datetime.datetime.today().strftime('%w'))+1

这应该给你你的真实日期 - 1 = 星期日，2 = 星期一，等等......

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要将星期日作为 1 到星期六作为 7，这是您问题的最简单解决方案：

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

他们全部：

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

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我们可以借助 Pandas：

import pandas as pd

正如上面问题中提到的我们有：

datetime(2017, 10, 20)

如果在 jupyter notebook 中执行这一行，我们会得到如下输出：

datetime.datetime(2017, 10, 20, 0, 0)

使用 weekday() 和 weekday_name：

如果您想要整数格式的工作日，请使用：

pd.to_datetime(datetime(2017, 10, 20)).weekday()

输出将是：

4

如果您希望它作为星期天、星期一、星期五等日期的名称，您可以使用：

pd.to_datetime(datetime(2017, 10, 20)).weekday_name

输出将是：

'Friday'

如果在 Pandas 数据框中有一个日期列，那么：

现在假设您有一个具有如下日期列的 pandas 数据框： pdExampleDataFrame['Dates'].head(5)

0   2010-04-01
1   2010-04-02
2   2010-04-03
3   2010-04-04
4   2010-04-05
Name: Dates, dtype: datetime64[ns]

现在，如果我们想知道星期一、星期二等工作日的名称，我们可以使用 .weekday_name，如下所示：

pdExampleDataFrame.head(5)['Dates'].dt.weekday_name

输出将是：

0    Thursday
1      Friday
2    Saturday
3      Sunday
4      Monday
Name: Dates, dtype: object

如果我们想要这个 Dates 列中的工作日整数，那么我们可以使用：

pdExampleDataFrame.head(5)['Dates'].apply(lambda x: x.weekday())

输出将如下所示：

0    3
1    4
2    5
3    6
4    0
Name: Dates, dtype: int64

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import datetime
import calendar

day, month, year = map(int, input().split())
my_date = datetime.date(year, month, day)
print(calendar.day_name[my_date.weekday()])

输出样本

08 05 2015
Friday

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如果您想生成一个包含日期范围 (Date) 的列，并生成一个转到第一个列并指定星期几 (Week Day)，请执行以下操作（我将使用从 { 3} 到 2020-02-01）：

import pandas as pd
dr = pd.date_range(start='2008-01-01', end='2020-02-1')
df = pd.DataFrame()
df['Date'] = dr
df['Week Day'] = pd.to_datetime(dr).weekday

输出如下：

https://i.stack.imgur.com/OyHnF.png

Week Day 从 0 到 6 不等，其中 0 对应于星期一，6 对应于星期日。

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以下是将小端字符串日期列表转换为 datetime 的方法：

import datetime, time
ls = ['31/1/2007', '14/2/2017']
for d in ls:    
    dt = datetime.datetime.strptime(d, "%d/%m/%Y")
    print(dt)
    print(dt.strftime("%A"))

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一个简单、直接且仍未提及的选项：

import datetime
...
givenDateObj = datetime.date(2017, 10, 20)
weekday      = givenDateObj.isocalendar()[2] # 5
weeknumber   = givenDateObj.isocalendar()[1] # 42

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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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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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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

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如果你是中国用户，你可以使用这个包：https://github.com/LKI/chinese-calendar

import datetime

# 判断 2018年4月30号 是不是节假日
from chinese_calendar import is_holiday, is_workday
april_last = datetime.date(2018, 4, 30)
assert is_workday(april_last) is False
assert is_holiday(april_last) is True

# 或者在判断的同时，获取节日名
import chinese_calendar as calendar  # 也可以这样 import
on_holiday, holiday_name = calendar.get_holiday_detail(april_last)
assert on_holiday is True
assert holiday_name == calendar.Holiday.labour_day.value

# 还能判断法定节假日是不是调休
import chinese_calendar
assert chinese_calendar.is_in_lieu(datetime.date(2006, 2, 1)) is False
assert chinese_calendar.is_in_lieu(datetime.date(2006, 2, 2)) is True

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这是一个新鲜的方法。周日是 0。

from datetime import datetime
today = datetime(year=2022, month=6, day=17)
print(today.toordinal()%7)  # 5
yesterday = datetime(year=1, month=1, day=1)
print(today.toordinal()%7)  # 1

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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.")

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使用此代码：

import pandas as pd
from datetime import datetime
print(pd.DatetimeIndex(df['give_date']).day)

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