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)
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
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)]
其他回答
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)
我为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"))
使用Canlendar模块
import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])
如果你想用英语约会:
from datetime import datetime
datetime.today().strftime('%A')
'Wednesday'
阅读更多信息:https://docs.python.org/3/library/datetime.html#strftime-结构时间行为
如果你想用英语约会:
from datetime import date
import calendar
my_date = date.today()
calendar.day_name[my_date.weekday()] #'Wednesday'