python - 根据索引/列名将小矩阵中的值添加到更大的矩阵中
问题描述
我有四个 4x4 矩阵（K1、K2、K3、K4），它们的值将被添加到一个大的 8x8 矩阵 K 中。如下所示，每个矩阵都有索引和列名，我想从基于它们的索引/列名称的小矩阵与较大的矩阵相对应。例如：从 K1 (v1, v1) 等于 1，因此应该在大矩阵中将 1 添加到 (v1, v1) 中。
到目前为止，4 个小矩阵是根据给定数学问题中的一些手动插入的值/变量创建的，因此索引/列名称可以在下一个问题中更改。总会有四个 4x4 矩阵，但它们的索引/列名会因大矩阵中的矩阵而异。是否有可能创建一个带有 if 语句的 for 循环来遍历小矩阵的索引/列名的名称，以便在大矩阵中添加什么？我被困住并感谢我能得到的所有帮助！
这是到目前为止的代码：
import pandas as pd
from sympy.interactive import printing
printing.init_printing(use_latex=True)
from sympy import Eq, solve_linear_system, Matrix
from numpy import linalg
import numpy as np
from IPython.display import display
import sympy as sy
import matplotlib.pyplot as plt
sy.init_printing()
A, E, L, Rx, P, rot3, rot2 = sy.symbols("A, E, L, Rx, P, rot3, rot2", real=True)
r2 = round(2**(1/2), 2)
r3 = round(3**(1/2), 2)
x1_e1 = 0
x2_e1 = 0
y1_e1 = 0
y2_e1 = L
L1 = L
c1 = (x2_e1 - x1_e1) / L1
s1 = (y2_e1 - y1_e1) / L1
column_names1 = ['u1', 'v1', 'u3', 'v3']
row_names1 = column_names1
x1_e2 = (2*L)/r3
x2_e2 = (r3*L)/2
y1_e2 = 0
y2_e2 = L/2
L2 = L/(r3)
c2 = (x2_e2 - x1_e2) / L2
s2 = (y2_e2 - y1_e2) / L2
column_names2 = ['u2', 'v2', 'u4', 'v4']
row_names2 = column_names2
x1_e3 = ((r3*L))/2
x2_e3 = 0
y1_e3 = L/2
y2_e3 = L
L3 = L
c3 = (x2_e3 - x1_e3) / L3
s3 = (y2_e3 - y1_e3) / L3
column_names3 = ['u4', 'v4', 'u3', 'v3']
row_names3 = column_names3
x1_e4 = 0
x2_e4 = (r3*L)/2
y1_e4 = 0
y2_e4 = L/2
L4 = L
c4 = (x2_e4 - x1_e4) / L4
s4 = (y2_e4 - y1_e4) / L4
column_names4 = ['u1', 'v1', 'u4', 'v4']
row_names4 = column_names4
def calc_K() :
A1 = A
E1 = E
AEL1 = round(((A1*E1)/L1)/((A*E)/L), 2)
K_1 = np.zeros((4,4),dtype=np)
K_1[0][0] = round(c1**2, 2)
K_1[0][1] = round(c1*s1, 2)
K_1[0][2] = round(-(c1**2), 2)
K_1[0][3] = round(-(c1*s1), 2)
K_1[1][0] = round(c1*s1, 2)
K_1[1][1] = round(s1**2, 2)
K_1[1][2] = round(-(c1*s1), 2)
K_1[1][3] = round(-(s1**2), 2)
K_1[2][0] = round(-(c1**2), 2)
K_1[2][1] = round(-(c1*s1), 2)
K_1[2][2] = round(c1**2, 2)
K_1[2][3] = round(c1*s1, 2)
K_1[3][0] = round(-(c1*s1), 2)
K_1[3][1] = round(-(s1**2), 2)
K_1[3][2] = round(c1*s1, 2)
K_1[3][3] = round(s1**2, 2)
K__1 = (AEL1 * K_1)
K1 = pd.DataFrame(K__1, columns=column_names1, index=row_names1)
print('')
print('---------- K1 ----------')
print('AE/L *')
print(K1)
print('')
A2 = A
E2 = E
AEL2 = round(((A2*E2)/L2)/((A*E)/L), 2)
K_2 = np.zeros((4,4),dtype=np)
K_2[0][0] = round(c2**2, 2)
K_2[0][1] = round(c2*s2, 2)
K_2[0][2] = round(-(c2**2), 2)
K_2[0][3] = round(-(c2*s2), 2)
K_2[1][0] = round(c2*s2, 2)
K_2[1][1] = round(s2**2, 2)
K_2[1][2] = round(-(c2*s2), 2)
K_2[1][3] = round(-(s2**2), 2)
K_2[2][0] = round(-(c2**2), 2)
K_2[2][1] = round(-(c2*s2), 2)
K_2[2][2] = round(c2**2, 2)
K_2[2][3] = round(c2*s2, 2)
K_2[3][0] = round(-(c2*s2), 2)
K_2[3][1] = round(-(s2**2), 2)
K_2[3][2] = round(c2*s2, 2)
K_2[3][3] = round(s2**2, 2)
K__2 = (AEL2 * K_2)
K2 = pd.DataFrame(K__2, columns=column_names2, index=row_names2)
print('')
print('---------- K2 ----------')
print('AE/L *')
print(K2)
print('')
A3 = A
E3 = E
AEL3 = round(((A3*E3)/L3)/((A*E)/L), 2)
K_3 = np.zeros((4,4),dtype=np)
K_3[0][0] = round(c3**2, 2)
K_3[0][1] = round(c3*s3, 2)
K_3[0][2] = round(-(c3**2), 2)
K_3[0][3] = round(-(c3*s3), 2)
K_3[1][0] = round(c3*s3, 2)
K_3[1][1] = round(s3**2, 2)
K_3[1][2] = round(-(c3*s3), 2)
K_3[1][3] = round(-(s3**2), 2)
K_3[2][0] = round(-(c3**2), 2)
K_3[2][1] = round(-(c3*s3), 2)
K_3[2][2] = round(c3**2, 2)
K_3[2][3] = round(c3*s3, 2)
K_3[3][0] = round(-(c3*s3), 2)
K_3[3][1] = round(-(s3**2), 2)
K_3[3][2] = round(c3*s3, 2)
K_3[3][3] = round(s3**2, 2)
K__3 = (AEL3 * K_3)
K3 = pd.DataFrame(K__3, columns=column_names3, index=row_names3)
print('')
print('---------- K3 ----------')
print('AE/L *')
print(K3)
print('')
A4 = A
E4 = E
AEL4 = round(((A4*E4)/L4)/((A*E)/L), 2)
K_4 = np.zeros((4,4),dtype=np)
K_4[0][0] = round(c4**2, 2)
K_4[0][1] = round(c4*s4, 2)
K_4[0][2] = round(-(c4**2), 2)
K_4[0][3] = round(-(c4*s4), 2)
K_4[1][0] = round(c4*s4, 2)
K_4[1][1] = round(s4**2, 2)
K_4[1][2] = round(-(c4*s4), 2)
K_4[1][3] = round(-(s4**2), 2)
K_4[2][0] = round(-(c4**2), 2)
K_4[2][1] = round(-(c4*s4), 2)
K_4[2][2] = round(c4**2, 2)
K_4[2][3] = round(c4*s4, 2)
K_4[3][0] = round(-(c4*s4), 2)
K_4[3][1] = round(-(s4**2), 2)
K_4[3][2] = round(c4*s4, 2)
K_4[3][3] = round(s4**2, 2)
K__4 = (AEL4 * K_4)
K4 = pd.DataFrame(K__4, columns=column_names4, index=row_names4)
print('')
print('---------- K4 ----------')
print('AE/L *')
print(K4)
print('')
column_names5 = ['u1', 'v1', 'u2', 'v2', 'u3', 'v3', 'u4', 'v4']
row_names5 = column_names5
KK__1 = np.zeros((8,8),dtype=np)
KK_1 = pd.DataFrame(KK__1, columns=column_names5, index=row_names5)
print(KK_1)
这是现在的输出：
---------- K1 ----------
AE/L *
u1 v1 u3 v3
u1 0 0 0 0
v1 0 1 0 -1
u3 0 0 0 0
v3 0 -1 0 1
---------- K2 ----------
AE/L *
u2 v2 u4 v4
u2 0.432 -0.761 -0.432 0.761
v2 -0.761 1.30 0.761 -1.30
u4 -0.432 0.761 0.432 -0.761
v4 0.761 -1.30 -0.761 1.30
---------- K3 ----------
AE/L *
u4 v4 u3 v3
u4 0.75 -0.43 -0.75 0.43
v4 -0.43 0.25 0.43 -0.25
u3 -0.75 0.43 0.75 -0.43
v3 0.43 -0.25 -0.43 0.25
---------- K4 ----------
AE/L *
u1 v1 u4 v4
u1 0.75 0.43 -0.75 -0.43
v1 0.43 0.25 -0.43 -0.25
u4 -0.75 -0.43 0.75 0.43
v4 -0.43 -0.25 0.43 0.25
---------- K ----------
u1 v1 u2 v2 u3 v3 u4 v4
u1 0 0 0 0 0 0 0 0
v1 0 0 0 0 0 0 0 0
u2 0 0 0 0 0 0 0 0
v2 0 0 0 0 0 0 0 0
u3 0 0 0 0 0 0 0 0
v3 0 0 0 0 0 0 0 0
u4 0 0 0 0 0 0 0 0
v4 0 0 0 0 0 0 0 0
期望的输出：
------------------------ K ------------------------
u1 v1 u2 v2 u3 v3 u4 v4
u1 0.75 0.43 0 0 0 0 -0.75 -0.43
v1 0.43 1.25 0 0 0 -1 -0.43 -0.25
u2 0 0 0.25 -0.43 0 0 -0.25 0.43
v2 0 0 -0.43 0.75 0 0 0.43 -0.75
u3 0 0 0 0 0 0 -0.75 0.43
v3 0 -1 0 0 0 -1 0.43 -0.25
u4 -0.75 -0.43 -0.25 0.43 0.25 0.43 1.75 -0.43
v4 -0.43 -0.25 0.43 -0.75 0.43 -0.25 -0.43 1.25
解决方案
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