如果你想用英语约会：

from datetime import date
import calendar
my_date = date.today()
calendar.day_name[my_date.weekday()]  #'Wednesday'

当星期一为0，星期日为6时，使用date.weekday（）

or

date.isoweekday（），当周一是1，周日是7

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

使用weekday（）：

>>> import datetime
>>> datetime.datetime.today()
datetime.datetime(2012, 3, 23, 23, 24, 55, 173504)
>>> datetime.datetime.today().weekday()
4

根据文档：

以整数形式返回星期几，其中星期一为0，星期日为6。

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)]

要让星期天1点到星期六7点，这是解决问题的最简单方法：

datetime.date.today().toordinal()%7 + 1

所有这些：

import datetime

today = datetime.date.today()
sunday = today - datetime.timedelta(today.weekday()+1)

for i in range(7):
    tmp_date = sunday + datetime.timedelta(i)
    print tmp_date.toordinal()%7 + 1, '==', tmp_date.strftime('%A')

输出：

1 == Sunday
2 == Monday
3 == Tuesday
4 == Wednesday
5 == Thursday
6 == Friday
7 == Saturday