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
以下是如何将小端字符串日期列表转换为datetime:
import datetime, time
ls = ['31/1/2007', '14/2/2017']
for d in ls:
dt = datetime.datetime.strptime(d, "%d/%m/%Y")
print(dt)
print(dt.strftime("%A"))
其他回答
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)
这不需要每周的评论。我推荐这个代码~!
import datetime
DAY_OF_WEEK = {
"MONDAY": 0,
"TUESDAY": 1,
"WEDNESDAY": 2,
"THURSDAY": 3,
"FRIDAY": 4,
"SATURDAY": 5,
"SUNDAY": 6
}
def string_to_date(dt, format='%Y%m%d'):
return datetime.datetime.strptime(dt, format)
def date_to_string(date, format='%Y%m%d'):
return datetime.datetime.strftime(date, format)
def day_of_week(dt):
return string_to_date(dt).weekday()
dt = '20210101'
if day_of_week(dt) == DAY_OF_WEEK['SUNDAY']:
None
1700/1/1之后的日期不需要导入的解决方案
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')
import datetime
int(datetime.datetime.today().strftime('%w'))+1
这应该会给你一个真实的数字-1=星期天,2=星期一,等等。。。
如果您有理由避免使用datetime模块,那么此函数将起作用。
注:从儒略历到公历的变化被认为发生在1582年。如果您感兴趣的日历并非如此,那么如果年份>1582,则相应地更改行。
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