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
以下是以DD-MM-YYYY格式输入日期的代码。您可以通过更改“%d-%m-%Y”的顺序以及更改分隔符来更改输入格式。
import datetime
try:
date = input()
date_time_obj = datetime.datetime.strptime(date, '%d-%m-%Y')
print(date_time_obj.strftime('%A'))
except ValueError:
print("Invalid date.")
其他回答
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)
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)
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
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)]
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()])
输出示例
08 05 2015
Friday