将交易日列表转换为节假日

Question

我正在使用下面的函数，它使用交易日列表，这似乎会减慢速度。是否可以将该列表转换为节假日以使其更快？

import pandas_market_calendars as mcal
def next_trading_day(start_day, num_trading_days, direction):
    '''returns the next/previous trading day. Business_days determines how many days
    back or into the future, direction determines wether back (-1) or forward (1)'''
    for i in range(0, num_trading_days, direction):
        next_day = start_day +datetime.timedelta(days=direction)    
        while next_day.weekday() in [5,6] or next_day not in set(mcal.get_calendar('NYSE').valid_days(start_date='2016-12-20', end_date='2020-01-10')):
            next_day +=  datetime.timedelta(days=direction)
        start_day = next_day
    return start_day

请注意，我不能使用 Python 必须提供的其他假期功能，因为它们与交易日无关，而是与联邦假期等有关。所以它必须基于 pandas_market_calendars。

Answer 1

解决方案

无需遍历每个日期。请参阅此实现：

逻辑：

1. 在日期范围内搜索有效交易日。使用大于您需要的日期范围 (num_trading_days)。我们另外添加100days 以扩展搜索日期范围。
2. 接下来我们从可用的交易日期中选择第一个num_trading_days。
3. 并选择第一个num_trading_days 的最后一天作为next_day。
4. 最后我们返回一个元组：(next_day, valid_trading_days)。

#!pip install pandas_market_calendars
import pandas_market_calendars as mcal
import numpy as np
import pandas as pd
import datetime

def next_trading_day(start_day=None, num_trading_days=7, direction=1, SAFE_DELTA = 100, as_string=True):
    """Returns the next/previous trading date separated by a certain number of 
    trading days.

    This function returns the next/previous trading day. The parameter num_trading_days 
    determines how many days back or into the future, direction determines whether 
    backward (-1) or forward (1).
    Parameters
    ----------
    start_day: datetime.date or a str object. (default is today's UTC date).
        Date of starting day.
    num_trading_days: int. (default is 7)
    direction: int. (default is 1 (forward))
        Use 1 for forward/future and -1 for backward/past.
    as_string: bool (default is True)
        This controls the data type of the returned value of next_day. 
        Returns the next_day as a string if True, else as a datetime.date object.
    SAFE_DELTA: int. (default is 100)
        SAFE_DELTA = 100 adds 100 additional days to extend the range of dates and 
        then finds which dates are valid trading dates.

    Returns
    -------
    next_day: str or datetime.date object.
    valid_trading_days: pandas.Series object of datetime.date objects.
    """

    if start_day is None:
        start_day = datetime.datetime.utcnow().date()
    start = pd.to_datetime(start_day)
    end = start + np.timedelta64(num_trading_days+SAFE_DELTA, 'D')*direction
    business_days = mcal.get_calendar('NYSE').valid_days(start_date=start, end_date=end)
    valid_trading_days = pd.DataFrame(business_days[:num_trading_days], columns=['DT']).DT.dt.date
    next_day = business_days[num_trading_days].date()
    if as_string:
        next_day = next_day.strftime("%Y-%m-%d")
    # if the series of trading days from start_day to next_day are not required, 
    # then use: 
    # >>> valid_trading_days = None
    # You may also comment out the valid_trading_days variable creation line above. 
    # And change the function-output accordingly.
    return (next_day, valid_trading_days)

测试

start_day='2019-12-18' # (Wednesday)
num_trading_days = 7
next_day, valid_trading_days = next_trading_day(start_day = start_day, 
                                                num_trading_days = num_trading_days, 
                                                direction = 1, 
                                                SAFE_DELTA = 100)
print('next_day: {} (after {} trading days)\n'.format(next_day, num_trading_days))
print('valid_trading_days: \n\n{}'.format(valid_trading_days))

输出：

next_day: 2019-12-30 (after 7 trading days)

valid_trading_days: 

0    2019-12-18
1    2019-12-19
2    2019-12-20
3    2019-12-23
4    2019-12-24
5    2019-12-26
6    2019-12-27
Name: DT, dtype: object