python获取指定日期和转换的整理

一、代码

import datetime
import calendar
from chinese_calendar import is_workday
import pandas as pd


# 判断是否为星期1
def is_monday():
    today = datetime.datetime.now().weekday()
    if today == 0:
        return True


# 判断是否为星期2
def is_tuesday():
    today = datetime.datetime.now().weekday()
    if today == 1:
        return True


# 判断是否为星期3
def is_wednesday():
    today = datetime.datetime.now().weekday()
    if today == 2:
        return True


# 判断是否为星期4
def is_thursday():
    today = datetime.datetime.now().weekday()
    if today == 3:
        return True


# 判断是否为星期5
def is_friday():
    today = datetime.datetime.now().weekday()
    if today == 4:
        return True


# 判断是否为星期6
def is_saturday():
    today = datetime.datetime.now().weekday()
    if today == 5:
        return True


# 判断是否为星期日
def is_sunday():
    today = datetime.datetime.now().weekday()
    if today == 6:
        return True


# 判断上几工作日
def is_word_day(date_num):
    day = datetime.datetime.now() + datetime.timedelta(days=date_num)
    res = is_workday(day)
    if res:
        return day.strftime("%Y-%m-%d")
    else:
        date_num += 1
        is_word_day(date_num)


# 上一自然日
def up_day():
    return (datetime.datetime.now() + datetime.timedelta(days=-1)).strftime("%Y-%m-%d")


# 获取指定前后天
def get_day(date_num):
    day = datetime.datetime.now() + datetime.timedelta(days=date_num)
    return day.day


# excel中5位日期数字的转换
def turn_time(delta_days):
    """
    delta_days:就是那个五位数
    """
    # 上一步得到的就是一个pandas的时间戳形式，然后我们转换成合适的时间格式就好了
    # Timestamp('1899-12-30 00:00:00')
    real_time = pd.to_datetime('1899-12-30') + pd.Timedelta(str(delta_days) + 'D')  # eg:'44701D'
    return real_time.strftime("%Y/%m/%d")


# 获取上个月末
def before_month_last_day():
    today = datetime.datetime.now()
    first = datetime.date(day=1, month=today.month, year=today.year)
    last_day = first - datetime.timedelta(days=1)
    return last_day.strftime("%Y%m%d")


# 获取上上工作日
def get_2up_work(up_date=1, is_work=0):
    '''
    :param up_date:可以用来调整上几个工作日
    :param is_work:
    :return:
    '''
    day = datetime.datetime.now() + datetime.timedelta(days=-up_date)
    res = is_workday(day)
    if res:
        is_work += 1
        up_date += 1
        if is_work == 2:
            return day.strftime("%Y%m%d")
        return get_2up_work(up_date, is_work)
    else:
        up_date += 1
        return get_2up_work(up_date, is_work)


# 获取上个月初
def before_month_fristday():
    today = datetime.datetime.now()
    first = datetime.date(day=1, month=today.month - 1, year=today.year)
    return first.strftime("%Y%m%d")


# 获取上个星期几
def get_up_day(day=1):
    '''
    :param day: 获取上个星期几，1-7
    :return:
    '''
    cal_dict = {1: calendar.MONDAY, 2: calendar.TUESDAY, 3: calendar.WEDNESDAY, 4: calendar.THURSDAY,
                5: calendar.FRIDAY, 6: calendar.SATURDAY, 7: calendar.SUNDAY}
    date = datetime.datetime.now()
    oneday = datetime.timedelta(days=1)
    while date.weekday() != cal_dict.get(day, calendar.MONDAY):
        date -= oneday
    return date.strftime("%Y%m%d")