我想了解以下内容:给定一个日期(datetime对象),一周中对应的日期是什么?

例如,星期天是第一天,星期一是第二天。。等等

然后如果输入的内容类似于今天的日期。

实例

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

产量可能是6(因为现在是星期五)


当前回答

假设您有timeStamp:字符串变量YYYY-MM-DD HH:MM:SS

步骤1:使用blow代码将其转换为dateTime函数。。。

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

步骤2:现在您可以提取所有必需的功能,如下所示,这将为每个字段创建新的列-小时、月、星期、年、日期

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)

其他回答

一个简单、直接但尚未提及的选项:

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

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)]

假设你得到了日期、月份和年份,你可以做到:

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)

当星期一为0,星期日为6时,使用date.weekday()

or

date.isoweekday(),当周一是1,周日是7

使用weekday():

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

根据文档:

以整数形式返回星期几,其中星期一为0,星期日为6。