Python优化移动平均线

Question

我想创建具有滚动功能和scipy优化的移动平均策略，但我的代码不优化滚动。它给出了我作为第一个x0值输入的结果。我在谷歌搜索它有方法来创建所有滚动的可能性，但它需要很多时间。有没有办法有效地优化。这是我的代码，提前感谢;

import pandas as pd
import os
import numpy as np
from datetime import datetime
import scipy.optimize as opt
#read file
data = pd.read_csv(r'C:\Users\Kaan\USDTRY-2018_06_01-2018_09_07.csv', encoding='utf-8', header=None, index_col=0)
data.columns = ['buy','sell',1,2]
data1 = data[['buy','sell']].head(100000)


# Optimization -------------------------------////////////////////------------------------------------

def objective(x):
    x1 = x[0]
    x2 = x[1]
    x3 = x[2]
    x4 = x[3]
    data3 = pd.DataFrame(data=data1)  
    data3['sma1']=data3['buy'].rolling(int(x3)).mean()
    data3['sma2']=data3['buy'].rolling(int(x4)).mean()
    data3['sma1-sma2'] = np.round(data3['sma1']-data3['sma2'],5) 
    data3['pos'] = np.where(data3['sma1-sma2'] >= x1, 1, 0)
    data3['pos'] = np.where((data3['sma1-sma2'] < -x1) ,-1, data3['pos'])
    data3['pos'] = np.where(abs(data3['sma1-sma2']) > x2, 0, data3['pos'])
    data3['return'] = np.round(np.log(data3['buy'] / data3['buy'].shift(1)),5)
    data3['st'] = data3['pos'].shift(1)*data3['return']
    return -1*data3['st'].cumsum().apply(np.exp).tail(1)[0]
def constraints1(x):
    return x[3] * x[2] - 0
def constraints2(x):
    return x[3] - x[2] - 0
b = (0.0,1000000.0)
bonds = (b,b,b,b)
x0=[0.00205062, 0.19746918, 893, 1990]
print(objective(x0))
con1 = {'type':'ineq','fun':constraints1}
con2 = {'type':'ineq','fun':constraints2}
cons = [con1,con2]
sol = opt.minimize(objective, x0, bounds=bonds, constraints=cons)
print(sol)  

Answer 1

另外，在定量策略中使用EWMA是normal，因为它保留了记忆。如果你想要一个超快的EWMA解决方案，那就看看我的其他answer计算快速EWMA

根据numba examples，这应该会显着提高您的代码速度

import numpy as np
from numba import guvectorize

@guvectorize(['void(float64[:], intp[:], float64[:])'], '(n),()->(n)')
def move_mean(a, window_arr, out):
    window_width = window_arr[0]
    asum = 0.0
    count = 0
    for i in range(window_width):
        asum += a[i]
        count += 1
        out[i] = asum / count
    for i in range(window_width, len(a)):
        asum += a[i] - a[i - window_width]
        out[i] = asum / count

另一种选择是用guvectorize取代jit