Function 使用 df.apply 或 Pandas 中的類似內容修改行值

Question

概述

我正在研究 dataframe ，其中df["Pivots"]會在 1 和 -1 之間交替，只要之前由鋸齒形指標識別出高點或低點。

我正在嘗試在 dataframe 上使用 Pandas 實現以下操作，並在df["Pivots"]先前（錯誤地）被分配一個值1時修改相關行，這標志着高，但另一行實際上具有更高的High值。

請參閱下面的屏幕截圖，了解數據的直觀表示和所需的 output。

偽代碼

• 如果當前行在df["Pivots"]中有 -1

• rows_between = index < 當前行和索引 > df["Pivots"]中的最后一個 pivot 值，這將是 1

• 如果df.High in rows_between > df["Pivot Price"]在當前行中，則 actual_high 為df[High].max() in rows_between。

• 從當前行的 df["Pivots" df["Pivots"]和 df["Pivot Price" df["Pivot Price"]中刪除 1 並將其添加到實際高的行中的df["Pivots"]和df["Pivot Price"]

例子

在此示例中， df.High行中的2023-10-08為實際高，並且高於第2023-09-24行中的df["Pivot Price"] 。

這是原裝 dataframe。

這是所需的 output ：

實際的 dataframe 將包含許多行，這只是一個最小的、可重現的示例。

代碼

    df.to_dict()
    
   {'Open': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): 1.0427,
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): 1.0586,
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): 1.0314,
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): 1.0669,
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): 1.0058,
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): 0.9966},
 'High': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): 1.0621,
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): 1.0609,
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): 1.0714,
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): 1.0679,
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): 1.0198,
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): 0.9966},
 'Low': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): 1.0383,
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): 1.0297,
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): 1.0285,
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): 1.004,
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): 0.9941,
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): 0.938},
 'Close': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): 1.0577,
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): 1.0297,
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): 1.0666,
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): 1.0053,
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): 0.9988,
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): 0.9528},
 'Pivots': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): 1,
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): 0,
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): 0,
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): 0,
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): 0,
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): -1},
 'Pivot Price': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): 1.0621,
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): nan,
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): nan,
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): nan,
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): nan,
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): 0.938},
 'Date': {Timestamp('2023-09-24 00:00:00', freq='W-SUN'): Timestamp('2023-09-24 00:00:00'),
  Timestamp('2023-10-01 00:00:00', freq='W-SUN'): Timestamp('2023-10-01 00:00:00'),
  Timestamp('2023-10-08 00:00:00', freq='W-SUN'): Timestamp('2023-10-08 00:00:00'),
  Timestamp('2023-10-15 00:00:00', freq='W-SUN'): Timestamp('2023-10-15 00:00:00'),
  Timestamp('2023-10-22 00:00:00', freq='W-SUN'): Timestamp('2023-10-22 00:00:00'),
  Timestamp('2023-10-29 00:00:00', freq='W-SUN'): Timestamp('2023-10-29 00:00:00')}}

作為參考， 這是生成這些樞軸的代碼。

Answer 1

我想不出使用.apply()的簡短解決方案，但使用一些輔助功能，您可以使用以下代碼解決問題：

import numpy as np

def get_highs_idx(df):
    return df[df['Pivots'] == 1].index.tolist()

def get_lows_idx(df):
    return df[df['Pivots'] == -1].index.tolist()

def get_previous_high_idx(df, low_idx):
    highs_idx = get_highs_idx(df)
    for high_idx in reversed(highs_idx):
        if high_idx < low_idx:
            return high_idx
    return None

def reset_pivot(df, old_high_idx, new_high_idx):
    df.loc[old_high_idx, 'Pivots'] = 0
    df.loc[old_high_idx, 'Pivot Price'] = np.nan
    df.loc[new_high_idx, 'Pivots'] = 1
    df.loc[new_high_idx, 'Pivot Price'] = df.loc[new_high_idx, 'High']
            
def correct_highs(df):
    lows_idx = get_lows_idx(df)
    for low_idx in lows_idx:
        high_idx = get_previous_high_idx(df, low_idx)
        if high_idx is not None:
            new_high_idx = df.loc[high_idx:low_idx, 'High'].idxmax()
            if high_idx != new_high_idx:
                reset_pivot(df, high_idx, new_high_idx)
            
correct_highs(df)    

代碼可能會減少一點，但我認為這種方式更清晰。

編輯

根據您的評論，我也在代碼下方添加了糾正低點的代碼。

def get_previous_low_idx(df, high_idx):
    lows_idx= get_lows_idx(df)
    for low_idx in reversed(lows_idx):
        if low_idx < high_idx:
            return low_idx
    return None

def reset_low_pivot(df, old_low_idx, new_low_idx):
    df.loc[old_low_idx, 'Pivots'] = 0
    df.loc[old_low_idx, 'Pivot Price'] = np.nan
    df.loc[new_low_idx, 'Pivots'] = -1
    df.loc[new_low_idx, 'Pivot Price'] = df.loc[new_low_idx, 'Low']
            
def correct_lows(df):
    highs_idx = get_highs_idx(df)
    for high_idx in highs_idx:
        low_idx = get_previous_low_idx(df, high_idx)
        if low_idx is not None:
            new_low_idx = df.loc[low_idx:high_idx, 'Low'].idxmin()
            if low_idx != new_low_idx:
                reset_low_pivot(df, low_idx, new_low_idx)


            
correct_lows(df)  

我不想影響原始答案，但您可能希望將reset_pivot重命名為reset_high_pivot以保持一致性。

還可以添加高級 function：

def correct_pivots(df):
    correct_highs(df)
    correct_lows(df)