Python，求解器方法还是优化当前代码？

Question

我正在尝试向矩阵中添加更多数据以分析和求解，但就目前而言，它正在执行粗略操作，如果我将另一列添加到分析。是否有可用的求解器方法可以找到类似的结果，而不必通过组合进行暴力破解？ 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

这个答案更像是codereview然后帮助算法处理问题。

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

但要获得运行步数，您可以使用 enumerate:

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

我还注意到您在每次迭代时都重新计算了 (B+adjust)/(C) 和 (B+adjust-10)/(C)，而这两个迭代在循环中根本没有改变，因此在循环之前计算一次而不是每次迭代肯定会节省您的执行时间时间：

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> ):
        ...

以及一遍又一遍地计算 sum(sumA)（对于 sumB、sumC）是不必要的代价，并且由于 sumA 表示值的元组而有些混乱，计算更有意义求和一次并将元组 (sumA, sumB, sumC) 作为一个名为 matrix 的值（2d 元组足够接近）

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)

类似地，仅当 OUT 更改时才重新计算 (sum(OUT[2])+A)/(sum(OUT[0])+C) 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)

所以更改后的代码部分将如下所示：

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)

Python，求解器方法还是优化当前代码？

Python, solver method or optimization of current code?

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