Question

我遇到了一个问题，我已经花了很多时间而找不到最佳解决方案，情况是：

1。我创建了一个运行良好的循环，然后必须退出循环，因为我已经达到了所需的结果。

2。我必须再次调用该循环才能对第一个循环的结果进行其他操作，但是我需要在上一个循环中创建的数组'index_val'，问题是第二部分要正常工作，必须重复代码（与我现在评论的相同）。

import numpy as np

data = [[  7,     8,     9,    1,    2,     3],
        [ 10,    11,    12,    4,    5,    6],
        [  1,     2,     3,    4,    5,     6]]

index = []
Vector_master = np.zeros((12,1))  
Matrix_master = np.zeros((12,12))  

for i in range(len(data)):
    Vector_n = np.zeros((12,1)) 
    Matrix_n = np.zeros((12,12)) 

    index.append([data[i][0], data[i][1], data[i][2],
                    data[i][3], data[i][4], data[i][5]])   
    index_val = np.array([index[i]])
    index_val.shape = (index_val.size,)
    index_val -= 1
    print(index_val)

    w = data[i][0]*data[i][1]
    L = data[i][3]
    s_elemental = np.array([0,  w/2, w**2,  0,  w/2,  -w**2])    

    Vector_n[index_val, 0] = s_elemental
    for row in range(12):
        Vector_master[row]+=Vector_n[row] 

    matrix = np.array([[ L,      0,     0,   L,     0,      0],
                       [  0,   L**3,  L**2,   0,  L**3,   L**2],
                       [  0,   L**2,     L,   0,  L**2,      L],
                       [ -L,      0,     0,   L,     0,      0],
                       [  0,   L**3,  L**2,   0,  L**3,   L**2],
                       [  0,   L**2,     L,   0,  L**2,     L]])

    Matrix_n[np.ix_(index_val, index_val)] = matrix 
    for row in range(12):
        for col in range(12):
            Matrix_master[row][col] +=  Matrix_n[row][col] 

k11 = Matrix_master[ :6 , :6] 
Vector_cut = Vector_master[ :6]

vector_a = np.dot(k11, Vector_cut)  
Vector_b = np.zeros((6,1))   
vector_c = np.append(vector_a, Vector_b)  

print('\n')
for i in range(len(data)):
#    index.append([data[i][0], data[i][1], data[i][2],
#                    data[i][3], data[i][4], data[i][5]])   
#    index_val = np.array([index[i]])
#    index_val.shape = (index_val.size,)
#    index_val -= 1
    print('\n',index_val)
    vector_result = vector_c[index_val]
    print(vector_result)

此刻，我在屏幕上打印的内容如下：

 [0 1 2 3 4 5]
[      0. -241439.  -62687.       0. -428416. -107104.]

 [0 1 2 3 4 5]
[      0. -241439.  -62687.       0. -428416. -107104.]

 [0 1 2 3 4 5]
[      0. -241439.  -62687.       0. -428416. -107104.]

您应打印以下内容：

 [6 7 8 0 1 2]
[0. 0. 0. 0. -241439. -62687.]

 [9 10 11 3 4 5]
[0. 0. 0. 0. -428416. -107104.]

 [0 1 2 3 4 5]
[0. -241439. -62687. 0. -428416. -107104.]

我真的很感谢您的帮助，请问我的英语，这不是我的母语，

Answer 1

您可以只创建一个index_vals列表，该列表将在运行第一个循环时保存所有index_val。

在第二个循环中，您可以访问它。

import numpy as np

data = [[  7,     8,     9,    1,    2,     3],
        [ 10,    11,    12,    4,    5,    6],
        [  1,     2,     3,    4,    5,     6]]

index = []
Vector_master = np.zeros((12,1))  
Matrix_master = np.zeros((12,12))  

index_vals = [] # new list
for i in range(len(data)):
    Vector_n = np.zeros((12,1)) 
    Matrix_n = np.zeros((12,12)) 

    index.append([data[i][0], data[i][1], data[i][2],
                    data[i][3], data[i][4], data[i][5]])   
    index_val = np.array([index[i]])
    index_val.shape = (index_val.size,)
    index_val -= 1
    print(index_val)

    # appending
    index_vals.append(index_val)

    w = data[i][0]*data[i][1]
    L = data[i][3]
    s_elemental = np.array([0,  w/2, w**2,  0,  w/2,  -w**2])    

    Vector_n[index_val, 0] = s_elemental
    for row in range(12):
        Vector_master[row]+=Vector_n[row] 

    matrix = np.array([[ L,      0,     0,   L,     0,      0],
                       [  0,   L**3,  L**2,   0,  L**3,   L**2],
                       [  0,   L**2,     L,   0,  L**2,      L],
                       [ -L,      0,     0,   L,     0,      0],
                       [  0,   L**3,  L**2,   0,  L**3,   L**2],
                       [  0,   L**2,     L,   0,  L**2,     L]])

    Matrix_n[np.ix_(index_val, index_val)] = matrix 
    for row in range(12):
        for col in range(12):
            Matrix_master[row][col] +=  Matrix_n[row][col] 

k11 = Matrix_master[ :6 , :6] 
Vector_cut = Vector_master[ :6]

vector_a = np.dot(k11, Vector_cut)  
Vector_b = np.zeros((6,1))   
vector_c = np.append(vector_a, Vector_b)  

for i in range(len(data)):
    print('\n',index_vals[i])
    vector_result = vector_c[ index_vals[i] ]
    print(vector_result)

[6 7 8 0 1 2]
[ 9 10 11  3  4  5]
[0 1 2 3 4 5]

 [6 7 8 0 1 2]
[      0.       0.       0.       0. -241439.  -62687.]

 [ 9 10 11  3  4  5]
[      0.       0.       0.       0. -428416. -107104.]

 [0 1 2 3 4 5]
[      0. -241439.  -62687.       0. -428416. -107104.]

避免循环后重复代码的优雅方法

1 个答案: