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。
>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday() # what I look for
输出可能是6
(自周五以来)
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。
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Julian到公历的变化发生在1582年。如果您感兴趣的日历不适用,那么如果年份> 1582,则更改行:相应地。
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
如果您不仅仅依赖于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);
这会给你一天的感觉:
days = ["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
days[calendar.weekday(2017,12,22)]
或者以python的风格,作为一个班轮:
["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"][calendar.weekday(2017,12,22)]
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()])
08 05 2015
Friday
假设你有timeStamp:字符串变量,YYYY-MM-DD HH:MM:SS
第1步:使用blow代码将其转换为dateTime函数...
df['timeStamp'] = pd.to_datetime(df['timeStamp'])
第2步:现在您可以提取以下所有必需的功能,这将为每个日期,月份,星期几,年份,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)
要将星期日作为1到星期六7,这是您问题的最简单的解决方案:
datetime.date.today().toordinal()%7 + 1
他们都是:
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
这是如何将日期列表转换为日期
import datetime,time
ls={'1/1/2007','1/2/2017'}
dt=datetime.datetime.strptime(ls[1], "%m/%d/%Y")
print(dt)
print(dt.month)
print(dt.year)
我们可以接受熊猫的帮助:
import pandas as pd
如上所述问题我们有:
datetime(2017, 10, 20)
如果在jupyter笔记本中执行此行,我们有一个这样的输出:
datetime.datetime(2017, 10, 20, 0, 0)
使用weekday()和weekday_name:
如果你想要整数格式的工作日,那么使用:
pd.to_datetime(datetime(2017, 10, 20)).weekday()
输出将是:
4
如果你想把它当作星期日,星期一,星期五等一天的名字,你可以使用:
pd.to_datetime(datetime(2017, 10, 20)).weekday_name
输出将是:
'Friday'
如果在Pandas数据帧中有一个日期列,那么:
现在假设你有一个像这样的日期列的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,如果我们想知道工作日的名称,如周一,周二,..等,我们可以使用.weekday_name
如下:
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
如果我们想要从这个Dates列中得到工作日的整数,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
如果您将日期作为字符串,则使用pandas的Timestamp可能更容易
import pandas as pd
df = pd.Timestamp("2019-04-12")
print(df.dayofweek, df.weekday_name)
输出:
4 Friday
使用日历模块
import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])
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)
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'
import datetime
int(datetime.datetime.today().strftime('%w'))+1
这应该给你你的真实日数 - 1 =星期日,2 =星期一等...
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
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")
1700/1/1之后的日期没有导入的解决方案
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')
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]
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")
你从中获得的输出是'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。
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)