我想了解以下内容:给定一个日期(datetime对象),一周中对应的日期是什么?

例如,星期天是第一天,星期一是第二天。。等等

然后如果输入的内容类似于今天的日期。

实例

>>> today = datetime.datetime(2017, 10, 20)
>>> today.get_weekday()  # what I look for

产量可能是6(因为现在是星期五)


当前回答

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

其他回答

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

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)

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]

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

使用以下代码:

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