python中根据时间获取周数,通过周数获取时间
作者:互联网
# 时间
## 时间和周数
import time import datetime # 获取今天是第几周 print(time.strftime('%W')) # 获取当前是周几(0-6,0代表周一) today=datetime.datetime.now().weekday() # 获取指定日期属于当年的第几周 week=datetime.datetime.strptime('20190825','%Y%m%d').strftime('%W')
## 获取下周的时间范围
import datetime,calendar,pandas as pd def get_N_day(): # 获取当前日期 today1 = datetime.date.today() today2 = datetime.date.today # 指定时间周期 oneday = datetime.timedelta(days=1) oneweek = datetime.timedelta(days=7) # 获取周一和周日的判断条件(周一为0,周日为6) m1 = calendar.MONDAY m2 = calendar.SUNDAY # 获取下周周一的时间 if today1.weekday() != m1: while today1.weekday() != m1: today1 += oneday else: today1+=oneweek # 获取下周周日的时间 if today2.weekday() != m2: today2 += oneweek while today2.weekday() != m2: today2 += oneday else: today2 += oneweek # 转换时间形式 nextMonday = today1.strftime('%Y%m%d') nextSunday = today2.strftime('%Y%m%d') # 获取下周七天的时间列表 date_list = [d.strftime("%Y-%m-%d") for d in pd.date_range(nextMonday, nextSunday, freq="D")] # 转换呈现形式 list_date=[] for i in date_list: one=i.split('-') time=['月','日'] time.insert(0,one[1]) time.insert(2,one[2]) time1=''.join(time) list_date.append(time1) return list_date print(get_N_day())
easy!
标签:today2,today1,python,周数,datetime,获取,time,date 来源: https://www.cnblogs.com/daodantiantou/p/11609985.html