下面是解决此问题的简单代码片段

import datetime

intDay = datetime.date(year=2000, month=12, day=1).weekday()
days = ["Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday"]
print(days[intDay])

输出应为：

Friday

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)]

使用Canlendar模块

import calendar
a=calendar.weekday(year,month,day)
days=["MONDAY","TUESDAY","WEDNESDAY","THURSDAY","FRIDAY","SATURDAY","SUNDAY"]
print(days[a])

import numpy as np

def date(df):
    df['weekday'] = df['date'].dt.day_name()

    conditions = [(df['weekday'] == 'Sunday'),
              (df['weekday'] == 'Monday'),
              (df['weekday'] == 'Tuesday'),
              (df['weekday'] == 'Wednesday'),
              (df['weekday'] == 'Thursday'),
              (df['weekday'] == 'Friday'),
              (df['weekday'] == 'Saturday')]

    choices = [0, 1, 2, 3, 4, 5, 6]

    df['week'] = np.select(conditions, choices)

    return df

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

使用以下代码：

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