这是一个新的方法。星期天是0。

from datetime import datetime
today = datetime(year=2022, month=6, day=17)
print(today.toordinal()%7)  # 5
yesterday = datetime(year=1, month=1, day=1)
print(today.toordinal()%7)  # 1

如果您是中国用户，您可以使用此软件包：https://github.com/LKI/chinese-calendar

import datetime

# 判断 2018年4月30号 是不是节假日
from chinese_calendar import is_holiday, is_workday
april_last = datetime.date(2018, 4, 30)
assert is_workday(april_last) is False
assert is_holiday(april_last) is True

# 或者在判断的同时，获取节日名
import chinese_calendar as calendar  # 也可以这样 import
on_holiday, holiday_name = calendar.get_holiday_detail(april_last)
assert on_holiday is True
assert holiday_name == calendar.Holiday.labour_day.value

# 还能判断法定节假日是不是调休
import chinese_calendar
assert chinese_calendar.is_in_lieu(datetime.date(2006, 2, 1)) is False
assert chinese_calendar.is_in_lieu(datetime.date(2006, 2, 2)) is True

使用以下代码：

import pandas as pd
from datetime import datetime
print(pd.DatetimeIndex(df['give_date']).day)

在MATLAB中，高斯方法

day_name={'Sun','Mon','Tue','Wed','Thu','Fri','Sat'}
month_offset=[0 3 3 6 1 4 6 2 5 0 3 5];  % common year

% input date
y1=2022
m1=11
d1=22

% is y1 leap
if mod(y1,4)==0 && mod(y1,100)==0 && mod(y1,400)==0
    month_offset=[0 3 4 0 2 5 0 3 6 1 4 6];  % offset for leap year
end

% Gregorian calendar
weekday_gregor=rem( d1+month_offset(m1) + 5*rem(y1-1,4) +  4*rem(y1-1,100) + 6*rem(y1-1,400),7)

day_name{weekday_gregor+1}

0:星期日1:星期一。。6:星期六

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

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)