Python，求解器方法还是当前代码的优化？

Question

我想向矩阵中添加更多的数据来分析和解决问题，但是因为它目前正在执行粗暴操作，所以如果我在分析中添加另一列，则会超出python的极限。有没有解决方法可以找到类似的结果，而不是不得不通过组合来暴力行为？ sample.csv也在下面列出。感谢您的任何建议。

import csv 
import itertools as it 
import numpy as np 

C = 2618.08 
B = 933.15 
A = 932.37 
adjust = 1 


D = csv.reader(open('sample.csv')) 

float_ABC = [] 
OUT = np.zeros((3, 9)) - 100 

for row in D: 
     float_ABC.append([str(x) for x in row]) 

float_ABC = float_ABC.astype(np.float) 

Alpha=float_ABC[:, [0,3,6,9,12,15]] 
Beta=float_ABC[:, [2,5,8,11,14,17]] 
Phi=float_ABC[:, [1,4,7,10,13,16]] 

plines1 = it.product(Alpha[0],Alpha[1],Alpha[2],Alpha[3], 
        Alpha[4],Alpha[5],Alpha[6],Alpha[7], 
        Alpha[8]) 

plines2 = it.product(Beta[0],Beta[1],Beta[2],Beta[3], 
        Beta[4],Beta[5],Beta[6],Beta[7], 
        Beta[8]) 

plines3 = it.product(Phi[0],Phi[1],Phi[2],Phi[3], 
        Phi[4],Phi[5],Phi[6],Phi[7], 
        Phi[8]) 


for count in range(0,6**9): 
    sumA = next(plines1) 
    sumB = next(plines2) 
    sumC = next(plines3) 

    if (sum(sumC)+B)/(sum(sumA)+C) <= (B+adjust)/(C) and \ 
     (sum(sumC)+B)/(sum(sumA)+C) >= (B+adjust-10)/(C) and \ 
     (sum(sumB)+A)/(sum(sumA)+C) > (sum(OUT[2])+A)/(sum(OUT[0])+C): 
     print("#",count,"- new option found!") 
     OUT = np.vstack((sumA,sumC,sumB)) 

和sample.csv：

13.4,-18.81,-24.75,5.82,-8.21,-10.8,0,0,0,3.3,1.56,2.05,-2.1,5.36,7.05,2.6,5.65,7.44 
0,-11.01,-14.49,0,-4.87,-6.41,0,0,0,0.6,2.24,2.95,1,4,5.26,1.7,2.73,3.59 
0,-40.74,-53.6,0,-17.86,-23.5,0,0,0,3.5,6.53,8.59,2.9,9.36,12.31,1.9,2.61,3.44 
1000,-1000,-1000,0,0,0,20.76,21.78,15.66,18.48,23.44,16.96,27.72,26.46,19.92,32.28,29.58,23.08 
1000,-1000,-1000,-2.28,-6.12,-4.16,-2.28,-2.53,-1.73,0,0,0,1.92,-1.85,-1.26,1.08,-1.27,-0.86 
1000,-1000,-1000,0,0,0,6.78,7.38,5.07,6.66,8.93,6.14,8.46,8.41,5.78,9.42,10.37,7.14 
1000,-1000,-1000,0,0,0,28.8,34.28,27.86,37.2,39.64,33.32,45.6,42.76,36.63,54,45.88,40.03 
1000,-1000,-1000,0,-4.95,-3.36,0,0,0,1.8,0.59,0.4,1.2,1.85,1.27,3.72,0.17,0.11 
1000,-1000,-1000,0,0,0,27.6,19.3,13.71,32.76,23.68,17.15,37.8,20.56,14.71,22.56,27.58,21.06 

Answer 1

这个答案是处理这个问题更像codereview然后用算法帮助。

首先，你可以iterate over all three plines1 at the same time using zip

for sumA, sumB, sumC in zip(plines1, plines2, plines3): 
    pass 

但随后拿到步骤的运行计数你是你可以使用enumerate：

for count, (sumA, sumB, sumC) in enumerate(zip(plines1, plines2, plines3)): 
    pass 

我也注意到你重新计算(B+adjust)/(C)和(B+adjust-10)/(C)每次迭代根本没有在循环中改变，所以在循环之前计算一次，而不是每次迭代都会节省一些执行时间：

high_check = (B+adjust)/(C) 
low_check = (B+adjust-10)/(C) 

for count, (sumA, sumB, sumC) in enumerate(zip(plines1, plines2, plines3)): 

    if (low_check <= (sum(sumC)+B)/(sum(sumA)+C) <= high_check 
      and <OTHER CHECK>): 
     ... 

以及计算sum(sumA)（和sumB，sumC）一遍又一遍地是unecessarily昂贵，而且有些混乱，因为sumA代表值的元组，它会更有意义，一旦计算的款项，并采取元组(sumA, sumB, sumC)如称为matrix一个值（第2元组是足够接近）

for count, matrix in enumerate(zip(plines1, plines2, plines3)): 
    sumA, sumB, sumC = map(sum, matrix) 
    if (low_check <= (sumC+B)/(sumA+C) <= high_check 
      and <OTHER CHECK>): 
     ... 
     OUT = np.vstack(matrix) 

同样仅重新计算(sum(OUT[2])+A)/(sum(OUT[0])+C)只有当OUT变化将减少重新计算不变值所需的执行时间：

OUT_check = (sum(OUT[2])+A)/(sum(OUT[0])+C) 

for ... in ...: 

    if ( ... 
      and (sumB+A)/(sumA+C) > OUT_check): 
     ... 
     OUT_check = (sum(OUT[2])+A)/(sum(OUT[0])+C) 

这样的改变代码段是这样的：

plines1 = it.product(*Alpha) #star notation just unpacks all the elements into arguments 
plines2 = it.product(*Beta) 
plines3 = it.product(*Phi) 

high_check = (B+adjust)/(C) 
low_check = (B+adjust-10)/(C) 
OUT_check = (sum(OUT[2])+A)/(sum(OUT[0])+C) 

for count, matrix in enumerate(zip(plines1, plines2, plines3)): 
    sumA, sumB, sumC = map(sum, matrix) 
    if (low_check <= (sumC+B)/(sumA+C) <= high_check 
      and (sumB+A)/(sumA+C) > OUT_check): 
     print("#",count,"- new option found!") 
     OUT = np.vstack(matrix) 
     OUT_check = (sum(OUT[2])+A)/(sum(OUT[0])+C)