熊猫滚动标准差

Question

是否有其他人rolling.std()在熊猫中遇到新问题？弃用的方法是rolling_std().新方法运行正常但产生一个不随时间序列滚动的常数.

示例代码如下.如果您交易股票,您可能会认识到布林带的公式.我得到的输出rolling.std()每天跟踪股票,显然没有滚动.

这在大熊猫0.19.1.任何帮助,将不胜感激.

import datetime
import pandas as pd
import pandas_datareader.data as web

start = datetime.datetime(2012,1,1)
end = datetime.datetime(2012,12,31)
g = web.DataReader(['AAPL'], 'yahoo', start, end)
stocks = g['Close']
stocks['Date'] = pd.to_datetime(stocks.index)
stocks['AAPL_LO'] = stocks['AAPL'] - stocks['AAPL'].rolling(20).std() * 2
stocks['AAPL_HI'] = stocks['AAPL'] + stocks['AAPL'].rolling(20).std() * 2
stocks.dropna(axis=0, how='any', inplace=True)

Answer 1

import pandas as pd
from pandas_datareader import data as pdr
import numpy as np
import datetime

end = datetime.date.today()
begin=end-pd.DateOffset(365*10)
st=begin.strftime('%Y-%m-%d')
ed=end.strftime('%Y-%m-%d')


data = pdr.get_data_yahoo("AAPL",st,ed)

def bollinger_strat(data, window, no_of_std):
    rolling_mean = data['Close'].rolling(window).mean()
    rolling_std = data['Close'].rolling(window).std()

    df['Bollinger High'] = rolling_mean + (rolling_std * no_of_std)
    df['Bollinger Low'] = rolling_mean - (rolling_std * no_of_std)     

bollinger_strat(data,20,2)