python pandas的相对强度指数

Question

我是pandas新手。计算熊猫的RSI指标中相对强度部分的最佳方法是什么？到目前为止，我得到以下信息：

from pylab import *
import pandas as pd
import numpy as np



def Datapull(Stock):
    try:
        df = (pd.io.data.DataReader(Stock,'yahoo',start='01/01/2010'))
        return df
        print 'Retrieved', Stock
        time.sleep(5)
    except Exception, e:
        print 'Main Loop', str(e)


def RSIfun(price, n=14):
    delta = price['Close'].diff()
    #-----------
    dUp=
    dDown=

    RolUp=pd.rolling_mean(dUp, n)
    RolDown=pd.rolling_mean(dDown, n).abs()

    RS = RolUp / RolDown
    rsi= 100.0 - (100.0 / (1.0 + RS))
    return rsi

Stock='AAPL'
df=Datapull(Stock)
RSIfun(df)

到目前为止，我做得对吗？我在方程式的差异部分遇到麻烦，您在其中分开了向上和向下的计算

Answer 1

dUp= delta[delta > 0]
dDown= delta[delta < 0]

您还需要类似：

RolUp = RolUp.reindex_like(delta, method='ffill')
RolDown = RolDown.reindex_like(delta, method='ffill')

否则RS = RolUp / RolDown不会做你想要的

编辑： 似乎这是RS计算的一种更准确的方法：

# dUp= delta[delta > 0]
# dDown= delta[delta < 0]

# dUp = dUp.reindex_like(delta, fill_value=0)
# dDown = dDown.reindex_like(delta, fill_value=0)

dUp, dDown = delta.copy(), delta.copy()
dUp[dUp < 0] = 0
dDown[dDown > 0] = 0

RolUp = pd.rolling_mean(dUp, n)
RolDown = pd.rolling_mean(dDown, n).abs()

RS = RolUp / RolDown