股票数据缺口填补

Question

我在测试一种在填补缺口上交易的策略。

鉴于AAPL在2022-10-04年创造了一个向上的缺口，在2022-10-07年填补了这一缺口(而且将有多个缺口)，那么如何确定最近的向上缺口是否以节奏式的方式填补？

代码：

import pandas_datareader as pdr

df = pdr.data.DataReader('AAPL', 'yahoo', start='2022-07-28', end='2022-09-01')
df['upward_gap'] = df['Low'] > df['High'].shift(1)  # identify upward gap
df['upward_gap_no'] = df['upward_gap'].cumsum()

当前产出：

                  High         Low  ...  upward_gap  upward_gap_no
Date                                ...                           
2022-08-08  167.809998  164.199997  ...       False              0
2022-08-09  165.820007  163.250000  ...       False              0
2022-08-10  169.339996  166.899994  ...        True              1
2022-08-11  170.990005  168.190002  ...       False              1
2022-08-12  172.169998  169.399994  ...       False              1
2022-08-15  173.389999  171.350006  ...       False              1
2022-08-16  173.710007  171.660004  ...       False              1
2022-08-17  176.149994  172.570007  ...       False              1
2022-08-18  174.899994  173.119995  ...       False              1
2022-08-19  173.740005  171.309998  ...       False              1
2022-08-22  169.860001  167.139999  ...       False              1
2022-08-23  168.710007  166.649994  ...       False              1
2022-08-24  168.110001  166.250000  ...       False              1
2022-08-25  170.139999  168.350006  ...        True              2
2022-08-26  171.050003  163.559998  ...       False              2
2022-08-29  162.899994  159.820007  ...       False              2
2022-08-30  162.559998  157.720001  ...       False              2
2022-08-31  160.580002  157.139999  ...       False              2

预期产出(对任何更好的代表性开放)：

                  High         Low  ...  upward_gap  upward_gap_no
Date                                ...                           
2022-08-08  167.809998  164.199997  ...       False              0
2022-08-09  165.820007  163.250000  ...       False              0
2022-08-10  169.339996  166.899994  ...        True              1  - 1st upward gap
2022-08-11  170.990005  168.190002  ...       False              1
2022-08-12  172.169998  169.399994  ...       False              1
2022-08-15  173.389999  171.350006  ...       False              1
2022-08-16  173.710007  171.660004  ...       False              1
2022-08-17  176.149994  172.570007  ...       False              1
2022-08-18  174.899994  173.119995  ...       False              1
2022-08-19  173.740005  171.309998  ...       False              1
2022-08-22  169.860001  167.139999  ...       False              1
2022-08-23  168.710007  166.649994  ...       False              1
2022-08-24  168.110001  166.250000  ...       False              1
2022-08-25  170.139999  168.350006  ...        True              2  - 2nd upward gap
2022-08-26  171.050003  163.559998  ...       False              0  - Both 1st & 2nd gap filled
2022-08-29  162.899994  159.820007  ...       False              0
2022-08-30  162.559998  157.720001  ...       False              0
2022-08-31  160.580002  157.139999  ...       False              0

Answer 1

好吧--虽然这并不是你所期望的，但我想我有些东西能满足你的要求。但是，我还没有对其他层次的差距做过任何事情，但是，我已经考虑到了基本的差距(或者您所称的upward gap 1)。

<> check_gap值()用于标识包含high_check号的行。

• (E 110黄色E 211)是d12值，用于标识空白何时填补。H 213H 114D 15编号(E 116绿色E 217/code>)一直保持d18，直到d19值下降到d20编号以下。H 221/code>代码H 122>代码<代码<<>代码>
• >代码>
• >代码><<225>代码>但是由于它们在另一个upward_gap.

中而被忽略。

import pandas_datareader as pdr

df = pdr.data.DataReader('AAPL', 'yahoo', start='2022-07-28', end='2022-09-01')
df['check_gap'] = (df['Low'] > df['High'].shift(1)).shift(-1).fillna(False)  # identify upward gap

start_checking = False
previous = False
skip_one = False
high_check = 0
upward_gap = []
high_check_list = []

for (cg, low, high) in zip(df['check_gap'], df['Low'], df['High']):
    if cg and not start_checking:
        start_checking = True
        high_check = high
        skip_one = True
    
    if start_checking and not skip_one:
        if low < high_check:
            start_checking = False
            high_check = 0
    elif skip_one:
        skip_one = False
        
    if start_checking and cg and not previous:
        high_check_list.append(0)
        upward_gap.append(False)
    else:
        upward_gap.append(start_checking)
        high_check_list.append(high_check)
        
    previous = start_checking
    
    
df['high_check'] = high_check_list
df['upward_gap'] = upward_gap

df

输出：

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..。

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