Heiken Ashi 使用 Pandas python
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/40613480/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Heiken Ashi Using pandas python
提问by Wacky Tacky
I was defining a function Heiken Ashi which is one of the popular chart type in Technical Analysis.
I was writing a function on it using Pandas but finding little difficulty.
This is how Heiken Ashi [HA] looks like-
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 在它上面写一个函数，但发现没什么困难。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
There is a lot of stuff available on various websites using for loop and pure python but i think Pandas can also do job well. This is my progress-
使用 for 循环和纯 python 的各种网站上有很多可用的东西，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
Can Anyone Help me with this please?` It doesnt work.... I tried on this-
任何人都可以帮我吗？`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)
This is the New Code i wrote
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
But still the HA_Open result was not satisfactory
但是HA_Open的结果还是不尽如人意
回答by arkochhar
Here is the fastest, accurate and efficient implementation as per my tests:
根据我的测试，这是最快、准确和有效的实现：
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
Here is my test algorithm (essentially I used the algorithm provided in this post to benchmark the speed results):
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))
Here is the output I got on my i7 processor (please note the results may vary depending on your processor speed but I assume that the results will be similar):
这是我在 i7 处理器上得到的输出（请注意，结果可能会因您的处理器速度而异，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
My experience with Pandas shows that functions like ix, loc, ilocare slower in comparison to set_valueand get_valuefunctions. Moreover computing value for a column on itself using shiftfunction gives erroneous results.
我对 Pandas 的经验表明，与和函数相比ix，loc、、 等函数iloc要慢一些。此外，使用函数计算列本身的值会给出错误的结果。set_valueget_valueshift
回答by Jason Melo
I'm not that knowledgeable regarding Python, or Pandas, but after some research, this is what I could figure would be a good solution.
我对 Python 或 Pandas 不太了解，但经过一些研究，我认为这是一个很好的解决方案。
Please, feel free to add any comments. I very much appreciate.
请随时添加任何评论。我非常感谢。
I used namedtuples and itertuples(seem to be the fastest, if looping through a DataFrame).
我使用了namedtuples和itertuples（如果循环遍历数据帧，似乎是最快的）。
I hope it helps!
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
回答by Derek R.
def heikenashi(df):
df['HA_Close'] = (df['Open'] + df['High'] + df['Low'] + df['Close']) / 4
df['HA_Open'] = (df['Open'].shift(1) + df['Open'].shift(1)) / 2
df.iloc[0, df.columns.get_loc("HA_Open")] = (df.iloc[0]['Open'] + df.iloc[0]['Close'])/2
df['HA_High'] = df[['High', 'Low', 'HA_Open', 'HA_Close']].max(axis=1)
df['HA_Low'] = df[['High', 'Low', 'HA_Open', 'HA_Close']].min(axis=1)
df = df.drop(['Open', 'High', 'Low', 'Close'], axis=1) # remove old columns
df = df.rename(columns={"HA_Open": "Open", "HA_High": "High", "HA_Low": "Low", "HA_Close": "Close", "Volume": "Volume"})
df = df[['Open', 'High', 'Low', 'Close', 'Volume']] # reorder columns
return df
回答by Pythonic
Will be faster with numpy.
使用 numpy 会更快。
def HEIKIN(O, H, L, C, oldO, oldC):
HA_Close = (O + H + L + C)/4
HA_Open = (oldO + oldC)/2
elements = numpy.array([H, L, HA_Open, HA_Close])
HA_High = elements.max(0)
HA_Low = elements.min(0)
out = numpy.array([HA_Close, HA_Open, HA_High, HA_Low])
return out
回答by Mott The Tuple
Unfortunately, set_value(), and get_value() are deprecated. Building off arkochhar's answer, I was able to get a 75% speed increase by using the following list comprehension method with my own OHLC data (7000 rows of data). It is faster than using at and iat as well.
不幸的是，不推荐使用 set_value() 和 get_value()。根据 arkochhar 的答案，通过使用以下列表理解方法和我自己的 OHLC 数据（7000 行数据），我能够将速度提高 75%。它也比使用 at 和 iat 更快。
def HA( dataframe ):
df = dataframe.copy()
df['HA_Close']=(df.Open + df.High + df.Low + df.Close)/4
df.reset_index(inplace=True)
ha_open = [ (df.Open[0] + df.Close[0]) / 2 ]
[ ha_open.append((ha_open[i] + df.HA_Close.values[i]) / 2) \
for i in range(0, len(df)-1) ]
df['HA_Open'] = ha_open
df.set_index('index', 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
回答by Joseph Da
I adjusted the code to make it work with Python 3.7
我调整了代码以使其适用于 Python 3.7
def HA(df):
df_HA = df
df_HA['Close']=(df['Open']+ df['High']+ df['Low']+df['Close'])/4
#idx = df_HA.index.name
#df_HA.reset_index(inplace=True)
for i in range(0, len(df)):
if i == 0:
df_HA['Open'][i]= ( (df['Open'][i] + df['Close'][i] )/ 2)
else:
df_HA['Open'][i] = ( (df['Open'][i-1] + df['Close'][i-1] )/ 2)
#if idx:
#df_HA.set_index(idx, inplace=True)
df_HA['High']=df[['Open','Close','High']].max(axis=1)
df_HA['Low']=df[['Open','Close','Low']].min(axis=1)
return df_HA
