如何在 Python 中获取一个月的第三个星期五？

Question

我正在尝试从雅虎获取股票数据。 财务使用 Python 2.7,9. 但我只需要本月第三个星期五的数据，我有一个 function 来获取数据。 但需要一种获取日期的方法：我想要这样的东西：

def get_third_fris(how_many):
    # code and stuff
    return list_of_fris

因此调用get_third_fris(6)将返回当前日期之后的第 3 个星期五的 6 项长列表。 日期需要是 Unix 时间戳。

（我对time或datetime几乎没有经验，所以请解释你的代码在做什么。）

谢谢！

Answer 1

您可以使用calendar模块列出周，然后获取该周的星期五。

import calendar

c = calendar.Calendar(firstweekday=calendar.SUNDAY)

year = 2015; month = 2

monthcal = c.monthdatescalendar(year,month)
third_friday = [day for week in monthcal for day in week if \
                day.weekday() == calendar.FRIDAY and \
                day.month == month][2]

您可以格式化为 Unix 时间戳，但这很重要。 我会向您推荐这个优秀的答案，其中包含基于您的日期是否时区感知的信息。

Answer 2

我们不需要导入除日期时间以外的任何内容。 我们可以假设一周有 7 天，而工作日 0 == 星期一。

import datetime

def third_friday(year, month):
    """Return datetime.date for monthly option expiration given year and
    month
    """
    # The 15th is the lowest third day in the month
    third = datetime.date(year, month, 15)
    # What day of the week is the 15th?
    w = third.weekday()
    # Friday is weekday 4
    if w != 4:
        # Replace just the day (of month)
        third = third.replace(day=(15 + (4 - w) % 7))
    return third

Answer 3

您可以使用标准的 Python 函数来查找本月的第三个星期五：

from datetime import timedelta, date
import calendar

def next_third_friday(d):
    """ Given a third friday find next third friday"""
    d += timedelta(weeks=4)
    return d if d.day >= 15 else d + timedelta(weeks=1)

def third_fridays(d, n):
    """Given a date, calculates n next third fridays"""

    # Find closest friday to 15th of month
    s = date(d.year, d.month, 15)
    result = [s + timedelta(days=(calendar.FRIDAY - s.weekday()) % 7)]

    # This month's third friday passed. Find next.
    if result[0] < d:
        result[0] = next_third_friday(result[0])

    for i in range(n - 1):
        result.append(next_third_friday(result[-1]))

    return result

我们可以应用上面的函数来获取下一个星期五的时间戳：

import time

def timestamp(d):
    return int(time.mktime(d.timetuple()))

fridays = third_fridays(date.today(), 2)

print(fridays)
print(map(timestamp, fridays))

输出：

[datetime.date(2015, 3, 20), datetime.date(2015, 4, 17)]   
[1426802400, 1429218000]

Answer 4

假设你想要一个每第三个星期五的范围，你可以只使用熊猫，示例代码：

import pandas as pd

pd.date_range('2017-12-02','2020-08-31',freq='WOM-3FRI')

Output:
DatetimeIndex(['2017-12-15', '2018-01-19', '2018-02-16', '2018-03-16',
               '2018-04-20', '2018-05-18', '2018-06-15', '2018-07-20',
               '2018-08-17', '2018-09-21', '2018-10-19', '2018-11-16',
               '2018-12-21', '2019-01-18', '2019-02-15', '2019-03-15',
               '2019-04-19', '2019-05-17', '2019-06-21', '2019-07-19',
               '2019-08-16', '2019-09-20', '2019-10-18', '2019-11-15',
               '2019-12-20', '2020-01-17', '2020-02-21', '2020-03-20',
               '2020-04-17', '2020-05-15', '2020-06-19', '2020-07-17',
               '2020-08-21'],
              dtype='datetime64[ns]', freq='WOM-3FRI')

Answer 5

一个更直接的答案怎么样：

import calendar 
c = calendar.Calendar(firstweekday=calendar.SATURDAY)
monthcal = c.monthdatescalendar(my_year, my_month)
monthly_expire_date = monthcal[2][-1]

Answer 6

它易于使用 dateutil 获取下一个星期五

import dateutil.parser as dparse
from datetime import timedelta
next_friday = dparse.parse("Friday")
one_week = timedelta(days=7)
friday_after_next = next_friday + one_week
last_friday = friday_after_next + one_week

这利用了这样一个事实，即星期五之间总是有一周的时间......尽管我不确定这是否能回答您的问题，但它至少应该为您提供一个良好的起点

Answer 7

使用dateutil.relativedelta ：

from dateutil.relativedelta import relativedelta, FR # $ pip install python-dateutil

def third_friday_dateutil(now):
    """the 3rd Friday of the month, not the 3rd Friday after today."""
    now = now.replace(day=1) # 1st day of the month
    now += relativedelta(weeks=2, weekday=FR)
    return now

或者使用dateutil.rrule ：

from datetime import date, timedelta
from dateutil.rrule import rrule, MONTHLY, FR

def third_friday_rrule(now):
    return rrule(MONTHLY, count=1, byweekday=FR, bysetpos=3, dtstart=now.replace(day=1))[0]

def get_third_fris_rrule(how_many):
    return list(rrule(MONTHLY, count=how_many, byweekday=FR, bysetpos=3, dtstart=date.today()+timedelta(1)))

这是一个蛮力解决方案（快 15 倍）：

#!/usr/bin/env python
import calendar
from datetime import date, timedelta
from itertools import islice

DAY = timedelta(1)
WEEK = 7*DAY

def fridays(now):
    while True:
        if now.weekday() == calendar.FRIDAY:
            while True:
                yield now
                now += WEEK
        now += DAY

def next_month(now):
    """Return the first date that is in the next month."""
    return (now.replace(day=15) + 20*DAY).replace(day=1)

def third_friday_brute_force(now):
    """the 3rd Friday of the month, not the 3rd Friday after today."""
    return next(islice(fridays(now.replace(day=1)), 2, 3))

def get_third_fris(how_many):
    result = []
    now = date.today()
    while len(result) < how_many:
        fr = third_friday_brute_force(now)
        if fr > now: # use only the 3rd Friday after today
            result.append(fr)
        now = next_month(now)
    return result

print(get_third_fris(6))

输出

[datetime.date(2015, 3, 20),
 datetime.date(2015, 4, 17),
 datetime.date(2015, 5, 15),
 datetime.date(2015, 6, 19),
 datetime.date(2015, 7, 17),
 datetime.date(2015, 8, 21)]

请参阅在 Python 中将 datetime.date 转换为 UTC 时间戳

这是与其他解决方案和测试的比较（针对所有可能的 400 年模式） 。

Answer 8

我概括了我的答案，以便任何人都可以在一个月的任何Nth工作日使用它并使用最少的默认库。 我的用途是查找当年的 DST（夏令时）日期（3 月的第 2 个星期日和 11 月的第 1 个星期日）。

# Libraries:
from datetime import datetime

# Function:
def get_nth_day_of_month(year, month, Nth, weekday):
   # Process is to find out what weekday the 1st of the month is
   # And then go straight to the desired date by calculating it
   first_of_month_weekday = datetime(year, month, 1).weekday()
   day_desired = 7 * (Nth-1) + (weekday - first_of_month_weekday)
   if day_desired < 1 : day_desired += 7 #correction for some 1st-weekday situations
   return datetime(year, month, day_desired)

# Config: 
year = 2022
month = 3 #DST starts in March
weekday = 6 #sunday
Nth = 2  #2nd sunday

dst_start = get_nth_day_of_month(year, month, Nth, weekday)

就我而言，这会产生今年夏令时的开始：

In [2]: dst_start
Out [2]: datetime.datetime(2022, 3, 13, 0, 0)

然后在 2022 年夏令时结束时：

month = 11
Nth = 1

dst_end = get_nth_day_of_month(year, month, Nth, weekday)

结果是：

In[4]: dst_end
Out[4]: datetime.datetime(2022, 11, 5, 0, 0)

因此，在 2022 年，DST 从 2022-03-13 运行到 2022-11-05。

标准：天数从星期一 = 0 到星期日 = 6

Answer 9

没有外部库的纯 python。
返回预期的月份日期。
注意：基于@autonopy 的回答，但有效。

from datetime import datetime
def get_nth_day_of_month(year, month, Nth, weekday):
    first_of_month_weekday = datetime(year, month, 1).weekday()
    # Find weekday offset from beginning of month
    day_offset = (weekday - first_of_month_weekday) + 1
    if day_offset < 1:
        day_offset += 7  # correction for some 1st-weekday situations
    # Add N weeks
    return 7 * (Nth - 1) + day_offset

测试：

>>> # first Monday of Nov 2021
>>> get_nth_day_of_month(2021, 11, 1, 0)
1

>>> # first Monday of January 2022
>>> get_nth_day_of_month(2022, 1, 1, 0)
3

>>> # first Monday of May 2022
>>> get_nth_day_of_month(2022, 5, 1, 0)
2

>>> # Mother's day 2022
>>> get_nth_day_of_month(2022, 5, 2, 0)
9

Answer 10

我概括了@pourhaus 的答案以找到任何一个月的第 n 天：

def nth_day_of_month(month, year, day_of_week, n):
    first_possible_day = {1: 1, 2: 8, 3: 15, 4: 22, 5: 29}[n]
    d = datetime.date(year, month, first_possible_day)
    w = d.weekday()
    if w != day_of_week:
        d = d.replace(day=(first_possible_day + (day_of_week - w) % 7))
    return d

Answer 11

假设您使用熊猫：

def exp_friday(df):
mask = np.where((df.index.day > 14) & 
                (df.index.day < 22) & 
                (df.index.dayofweek == 4), True, False)
return df[mask]

Answer 12

这是一个通用函数，以列表形式为您提供特定周的所有日期。

def frecuencia_daymng(self, start_day, year, month, dayofweek):
    """dayofweek starts on MONDAY in 0 index"""
    c = calendar.Calendar(firstweekday=start_day)
    monthcal = c.monthdatescalendar(year, month)
    ldates = []
    for tdate in monthcal:
        if tdate[dayofweek].month == month:
            ldates.append(tdate[dayofweek])
    return ldates

假设您想要 2020 年的所有星期一 10。

frecuencia_daymng(calendar.MONDAY, 2020, 10, 0)

这将为您提供输出。

[datetime.date(2020, 10, 5),
 datetime.date(2020, 10, 12),
 datetime.date(2020, 10, 19),
 datetime.date(2020, 10, 26)]

所以现在你有这个月的第一个，第二个......等星期一。

Answer 13

我的建议是从每月的第一天开始，然后找到最近的星期五。

4 表示为来自 datetime.weekday() 方法的 Friday。

因此，我们然后从 4(Friday) 中减去该月第一天的工作日 如果结果为负，则找到的最接近的星期五是上个月，因此我们添加 7 天，否则我们已经有了第一个星期五。

然后结果就像再添加 14 天以获得第三个星期五一样简单，然后将代表第三个星期五的 timedelta 添加到该月的第一天。

from datetime import datetime, timedelta

def get_third_friday(year, month):
    first_day_of_month = datetime(year, month, 1)
    closest_friday = 4 - first_day_of_month.weekday()
    if closest_friday < 0:
        first_friday = closest_friday + 7
    else:
        first_friday = closest_friday

    third_friday = first_friday + 14
    return first_day_of_month + timedelta(third_friday)

Answer 14

这是一个已经有人想出来的解决方案： relativedelta模块是Python dateutil package ( pip install python-dateutil ) 的扩展。

import datetime
from dateutil import relativedelta


def third_fridays(n):
    first_of_this_month = datetime.date.today().replace(day=1)
    return (
        first_of_this_month
        + relativedelta.relativedelta(weekday=relativedelta.FR(3), months=i)
        for i in range(n)
    )

这里的关键部分当然是weekday=relativedelta.FR(3) ，它准确说明了需要什么：该月的第三个星期五。 这是weekday参数文档的相关部分，

工作日：

relativedelta 模块中可用的工作日实例之一（MO、TU 等）。 这些实例可能会收到一个参数 N，指定第 N 个工作日，它可以是正数或负数（如 MO(+1) 或 MO(-2)）。

（对于 Python 的新手， return (...)是一个生成器表达式，您可以将其视为要迭代的内容，例如， for friday in third_fridays(18): print(friday) ）

Answer 15

from dateutil.relativedelta import *
from datetime import *

def find_mth_friday(your_date,m):
    mth_friday = your_date + relativedelta(day=1, weekday=FR(m)) #sets day=1 in your_date and adds m fridays to it.
    mth_friday_timestamp = int(mth_friday.strftime("%s")) #converting datetime to unix timestamp
    return mth_friday_timestamp

def get_third_fris(n):
    output_timestamps = []
    today = datetime.now() #gets current system date
    for i in range(1,n+1): #value of i varies from 1 to 6 if n=6
        next_month = today + relativedelta(months=+i) #adds i months to current system date
        third_friday = find_mth_friday(next_month,3) #finds third friday of the month using 'find_mth_friday()', the function we defined
        output_timestamps.append(third_friday)
    return output_timestamps

print(get_third_fris(6)) #let's try invoking our function with n=6 dates

这就是你想要的吧？