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

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。。。

其他回答

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

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)

使用Canlendar模块

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

如果您将日期作为字符串,那么使用panda的时间戳可能更容易

import pandas as pd
df = pd.Timestamp("2019-04-12")
print(df.dayofweek, df.weekday_name)

输出:

4 Friday

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