我想要找到以下内容:
给定一个日期(datetime
对象),对应的星期几是什么?
例如,星期天是第一天,星期一是第二天……以此类推。
如果输入是今天的日期,该怎么办。
示例
>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday() # what I look for
输出可能为6
(因为今天是星期五)
我们可以借助Pandas:
import pandas as pd
如上所述,我们在问题中有:
datetime(2017, 10, 20)
datetime.datetime(2017, 10, 20, 0, 0)
使用weekday()和weekday_name:
如果您想要以整数格式获取工作日,则可以使用以下方法:
pd.to_datetime(datetime(2017, 10, 20)).weekday()
4
如果你想要像星期日、星期一、星期五等一样将其作为日期名称,你可以使用以下代码:
pd.to_datetime(datetime(2017, 10, 20)).weekday_name
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几的名称,比如星期一、星期二等,我们可以使用如下的.weekday_name
:
pdExampleDataFrame.head(5)['Dates'].dt.weekday_name
0 Thursday
1 Friday
2 Saturday
3 Sunday
4 Monday
Name: Dates, dtype: object
如果我们想要从这个日期列中获取整数形式的星期几,可以使用以下方法:
pdExampleDataFrame.head(5)['Dates'].apply(lambda x: x.weekday())
0 3
1 4
2 5
3 6
4 0
Name: Dates, dtype: int64
如果您想生成一个包含日期范围(Date
)的列,并生成一个列,用于给第一个日期分配星期几(Week Day
),请按照以下步骤进行操作(我将使用从2008-01-01
到2020-02-01
的日期范围):
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:
Week Day
的取值范围为 0 到 6,其中 0 表示周一,6 表示周日。
如果你是中文用户,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
import datetime
...
givenDateObj = datetime.date(2017, 10, 20)
weekday = givenDateObj.isocalendar()[2] # 5
weeknumber = givenDateObj.isocalendar()[1] # 42
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)
import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])
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
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.")
from datetime import timedelta
import datetime as dt
today = dt.date.today()
monday = today - timedelta(days=today.weekday())
friday = monday + timedelta(days=4)
import pandas as pd
from datetime import datetime
print(pd.DatetimeIndex(df['give_date']).day)