用代碼實(shí)例解析python瀑布線(xiàn)指標(biāo)

# -*- coding: utf-8 -*-
"""
Created on Tue May 23 08:57:02 2017
@author: yunjinqi 
 
E-mail:yunjinqi@qq.com 
 
Differentiate yourself in the world from anyone else.
"""
import pandas as pd
import numpy as np
import datetime
import time
#獲取數(shù)據(jù)
df=pd.read_csv('C:/Users/HXWD/Desktop/000001.csv',encoding='gbk')
df.columns=['date','code','name','close','high','low','open','preclose',
'change','change_per','volume','amt']
df=df[['date','open','high','low','close','volume','amt']]
df.head()
def get_EMA(df,N):
 for i in range(len(df)):
 if i==0:
  df.ix[i,'ema']=df.ix[i,'close']
 if i>0:
  df.ix[i,'ema']=(2*df.ix[i-1,'close']+(N-1)*df.ix[i,'close'])/(N+1)
 
 return df['ema']
def get_PBX(df):
 df['a1']=(get_EMA(df,4)+df['close'].rolling(8).mean()+
  df['close'].rolling(16).mean())/3
 df['a2']=(get_EMA(df,6)+df['close'].rolling(12).mean()+
  df['close'].rolling(24).mean())/3
 df['a3']=(get_EMA(df,9)+df['close'].rolling(18).mean()+
  df['close'].rolling(36).mean())/3
 df['a4']=(get_EMA(df,13)+df['close'].rolling(26).mean()+
  df['close'].rolling(52).mean())/3
 df['a5']=(get_EMA(df,18)+df['close'].rolling(36).mean()+
  df['close'].rolling(72).mean())/3
 df['a6']=(get_EMA(df,24)+df['close'].rolling(48).mean()+
  df['close'].rolling(96).mean())/3
 return df
get_PBX(df)
df.tail() 
 
'''
瀑布線(xiàn)
原理：均線(xiàn)類(lèi)指標(biāo)
用法：
　　瀑布線(xiàn)是趨勢(shì)性指標(biāo)，因其在運(yùn)行的過(guò)程中，形態(tài)與瀑布極其相似，故得名為瀑布線(xiàn)。
 瀑布線(xiàn)在低位粘合，短線(xiàn)瀑布線(xiàn)向上穿越長(zhǎng)線(xiàn)瀑布線(xiàn)并向上發(fā)散，買(mǎi)入。
 瀑布線(xiàn)在高位粘合，短線(xiàn)瀑布線(xiàn)向下穿越長(zhǎng)線(xiàn)瀑布線(xiàn)并向下發(fā)散，賣(mài)出。
 當(dāng)長(zhǎng)中短期瀑布線(xiàn)依次由下向上排列形成多頭排列時(shí)，可持股。
 當(dāng)長(zhǎng)中短期瀑布線(xiàn)依次由上向下排列形成空頭排列時(shí)，應(yīng)空倉(cāng)。 
'''

 >>> -123<0
 True
 >>> True and -1
 -1
 >>> -1 or 1
 -1

>>> 123<0
False
>>> False and -1
False
>>> False or 1
1

class Solution(object):
 def reverse(self, x):
 """
 :type x: int
 :rtype: int
 """
 a = 0
 sign = x<0 and -1 or 1
 x = abs(x)
 while(x != 0):
  a = a*10 + x%10
  x //=10
 return a*sign if a<2**31 else 0