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)
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:
返回星期几的数字,其中星期一为0,星期日为6。
from datetime import date
import calendar
my_date = date.today()
calendar.day_name[my_date.weekday()] #'Wednesday'
my_date.strftime('%A')
更为有效。 - Nathan Tew>>> from datetime import datetime
>>> datetime.now().strftime('%A')
'Wednesday'
>>> datetime.now().strftime('%a')
'Wed'
我为 CodeChef 的一个问题找到了解决方法。
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"))
一个在1700/1/1之后的日期无需导入任何模块即可解决的方法
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')
aux
是从year -1700
推导出来的,所以我们需要加上100使其成为400的倍数。例如:2000 - 1700
= 300,所以+ 100
给我们400。不确定为什么在那个时候要使用aux
来确定闰年,而不是直接使用year
本身。 - not2savvyoffset
是什么? - Shyam3089如果你有日期字符串,使用Pandas的时间戳可能更容易
import pandas as pd
df = pd.Timestamp("2019-04-12")
print(df.dayofweek, df.weekday_name)
输出:
4 Friday
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
如果日期是一个日期时间对象,这是一个解决方案。
import datetime
def dow(date):
days=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
dayNumber=date.weekday()
print days[dayNumber]
假设你有一个 timeStamp: 字符串变量,格式为YYYY-MM-DD HH:MM:SS
步骤1: 使用以下代码将其转换为dateTime函数...
df['timeStamp'] = pd.to_datetime(df['timeStamp'])
步骤2:现在您可以按照以下方式提取所有所需的特征,这将为每个字段创建新列- 小时、月份、星期几、年份、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)
int(datetime.datetime.today().strftime('%w'))
- mrooney.strftime('%A')
[链接]获取星期几名称。 - rborodinov