Heiken Ashi使用pandas python

Question

我正在定义一个函数Heiken Ashi,它是技术分析中流行的图表类型之一.我正在使用Pandas编写一个函数,但没有遇到任何困难.这就是Heiken Ashi [HA]的样子 -

                 Heikin-Ashi Candle Calculations
           HA_Close = (Open + High + Low + Close) / 4
           HA_Open = (previous HA_Open + previous HA_Close) / 2
           HA_Low = minimum of Low, HA_Open, and HA_Close
           HA_High = maximum of High, HA_Open, and HA_Close

               Heikin-Ashi Calculations on First Run
            HA_Close = (Open + High + Low + Close) / 4
                   HA_Open = (Open + Close) / 2
                           HA_Low = Low
                           HA_High = High

在使用for循环和纯python的各种网站上有很多东西,但我认为Pandas也可以做得很好.这是我的进步 -

   def HA(df):

       df['HA_Close']=(df['Open']+ df['High']+ df['Low']+ df['Close'])/4

       ha_o=df['Open']+df['Close']  #Creating a Variable
       #(for 1st row)

       HA_O=df['HA_Open'].shift(1)+df['HA_Close'].shift(1) #Another variable
       #(for subsequent rows)

       df['HA_Open']=[ha_o/2 if df['HA_Open']='nan' else HA_O/2]     
       #(error Part Where am i going wrong?)

       df['HA_High']=df[['HA_Open','HA_Close','High']].max(axis=1)

       df['HA_Low']=df[['HA_Open','HA_Close','Low']].min(axis=1)

       return df

任何人都可以帮我这个吗？`它不起作用....我试过这个 -

  import pandas_datareader.data as web 
  import HA
  import pandas as pd
  start='2016-1-1'
  end='2016-10-30'
  DAX=web.DataReader('^GDAXI','yahoo',start,end)

这是我写的新代码

    def HA(df):
            df['HA_Close']=(df['Open']+ df['High']+ df['Low']+df['Close'])/4
...:        ha_o=df['Open']+df['Close']
...:        df['HA_Open']=0.0
...:        HA_O=df['HA_Open'].shift(1)+df['HA_Close'].shift(1)
...:        df['HA_Open']= np.where( df['HA_Open']==np.nan, ha_o/2, HA_O/2 )
...:        df['HA_High']=df[['HA_Open','HA_Close','High']].max(axis=1)
...:        df['HA_Low']=df[['HA_Open','HA_Close','Low']].min(axis=1)
...:        return df

但HA_Open结果仍然不尽如人意

Answer 1

根据我的测试,这是最快,最准确,最有效的实现:

def HA(df):
    df['HA_Close']=(df['Open']+ df['High']+ df['Low']+df['Close'])/4

    idx = df.index.name
    df.reset_index(inplace=True)

    for i in range(0, len(df)):
        if i == 0:
            df.set_value(i, 'HA_Open', ((df.get_value(i, 'Open') + df.get_value(i, 'Close')) / 2))
        else:
            df.set_value(i, 'HA_Open', ((df.get_value(i - 1, 'HA_Open') + df.get_value(i - 1, 'HA_Close')) / 2))

    if idx:
        df.set_index(idx, inplace=True)

    df['HA_High']=df[['HA_Open','HA_Close','High']].max(axis=1)
    df['HA_Low']=df[['HA_Open','HA_Close','Low']].min(axis=1)
    return df

这是我的测试算法(基本上我使用了这篇文章中提供的算法来对速度结果进行基准测试):

import quandl
import time

df = quandl.get("NSE/NIFTY_50", start_date='1997-01-01')

def test_HA():
    print('HA Test')
    start = time.time()
    HA(df)
    end = time.time()
    print('Time taken by set and get value functions for HA {}'.format(end-start))

    start = time.time()
    df['HA_Close_t']=(df['Open']+ df['High']+ df['Low']+df['Close'])/4

    from collections import namedtuple
    nt = namedtuple('nt', ['Open','Close'])
    previous_row = nt(df.ix[0,'Open'],df.ix[0,'Close'])
    i = 0
    for row in df.itertuples():
        ha_open = (previous_row.Open + previous_row.Close) / 2
        df.ix[i,'HA_Open_t'] = ha_open
        previous_row = nt(ha_open, row.Close)
        i += 1

    df['HA_High_t']=df[['HA_Open_t','HA_Close_t','High']].max(axis=1)
    df['HA_Low_t']=df[['HA_Open_t','HA_Close_t','Low']].min(axis=1)
    end = time.time()
    print('Time taken by ix (iloc, loc) functions for HA {}'.format(end-start))

这是我在i7处理器上得到的输出(请注意结果可能会因处理器速度而异,但我认为结果会相似):

HA Test
Time taken by set and get value functions for HA 0.05005788803100586
Time taken by ix (iloc, loc) functions for HA 0.9360761642456055

我与熊猫的经验表明,类似的功能ix,loc,iloc在比较较慢set_value和get_value功能.此外,使用shift函数计算自身列的值会产生错误的结果.

Answer 2

我对 Python 或 Pandas 不太了解，但经过一些研究，我认为这将是一个很好的解决方案。

请随时添加任何评论。我非常感激。

我使用了namedtuples和itertuples（如果循环遍历DataFrame，这似乎是最快的）。

我希望它有帮助！

def HA(df):
    df['HA_Close']=(df['Open']+ df['High']+ df['Low']+df['Close'])/4

    nt = namedtuple('nt', ['Open','Close'])
    previous_row = nt(df.ix[0,'Open'],df.ix[0,'Close'])
    i = 0
    for row in df.itertuples():
        ha_open = (previous_row.Open + previous_row.Close) / 2
        df.ix[i,'HA_Open'] = ha_open
        previous_row = nt(ha_open, row.Close)
        i += 1

    df['HA_High']=df[['HA_Open','HA_Close','High']].max(axis=1)
    df['HA_Low']=df[['HA_Open','HA_Close','Low']].min(axis=1)
    return df