如何在Python中获取日期的星期几？

Question

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.

例

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

输出可能6(从星期五开始)

Answer 1

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.

Answer 2

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'

Answer 3

如果您想要英文日期:

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

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

Answer 4

使用date.weekday()或date.isoweekday().

Answer 5

我为了一个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")

Answer 6

在1700/1/1之后输入日期的解决方案

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')

Answer 7

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

Answer 8

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]

Answer 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")

你从中得到的输出是'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.我在使用日期时间库时遇到问题,因为我想存储工作日名称而不是工作日编号,而使用日期时间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,那么你可以无限期地使用它,因为它也有更简单的语法.希望这可以帮助.

Answer 10

假设你有timeStamp: String variable, YYYY-MM-DD HH:MM:SS

第 1 步：使用吹码将其转换为 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特征，如下所示，这将为每个小时、月份、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)

Answer 11

这不需要一周中的某一天的评论。
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

Answer 12

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)

Answer 13

如果您有理由避免使用 datetime 模块，那么此功能将起作用。

注意：假定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

Answer 14

如果您不仅仅依赖该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);

这将为您带来这一天：

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

或者以 python 的风格，作为一个衬垫：

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

Answer 15

If you have dates as a string, it might be easier to do it using pandas' Timestamp

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

Output:

4 Friday