如果您希望生成一个包含日期范围（Date）的列，并生成一个指向第一个日期并指定周日（Week Day）的列（我将使用从2008-01-01到2020-02-01的日期）：

import pandas as pd
dr = pd.date_range(start='2008-01-01', end='2020-02-1')
df = pd.DataFrame()
df['Date'] = dr
df['Week Day'] = pd.to_datetime(dr).weekday

输出如下：

工作日从0到6不等，其中0对应于周一，6对应于周日。

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

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)

以下是以DD-MM-YYYY格式输入日期的代码。您可以通过更改“%d-%m-%Y”的顺序以及更改分隔符来更改输入格式。

import datetime
try:
    date = input()
    date_time_obj = datetime.datetime.strptime(date, '%d-%m-%Y')
    print(date_time_obj.strftime('%A'))
except ValueError:
    print("Invalid date.")

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

import datetime
int(datetime.datetime.today().strftime('%w'))+1

这应该会给你一个真实的数字-1=星期天，2=星期一，等等。。。