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。

如果日期是datetime对象，这是一个解决方案。

import datetime
def dow(date):
    days=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"]
    dayNumber=date.weekday()
    print days[dayNumber]

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)

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

如果您有理由避免使用datetime模块，那么此函数将起作用。

注：从儒略历到公历的变化被认为发生在1582年。如果您感兴趣的日历并非如此，那么如果年份>1582，则相应地更改行。

def dow(year,month,day):
    """ day of week, Sunday = 1, Saturday = 7
     http://en.wikipedia.org/wiki/Zeller%27s_congruence """
    m, q = month, day
    if m == 1:
        m = 13
        year -= 1
    elif m == 2:
        m = 14
        year -= 1
    K = year % 100    
    J = year // 100
    f = (q + int(13*(m + 1)/5.0) + K + int(K/4.0))
    fg = f + int(J/4.0) - 2 * J
    fj = f + 5 - J
    if year > 1582:
        h = fg % 7
    else:
        h = fj % 7
    if h == 0:
        h = 7
    return h