python中根据时间获取周数，通过周数获取时间

Posted 2020-11-16 倒旦田头

# 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！