numpy 我怎样才能把4个矩阵连接成一个大矩阵而不看起来很奇怪呢？

rnmwe5a2 于 2023-03-23 发布在其他

关注(0)|答案(2)|浏览(108)

我在解一个矩阵形式的微分方程，每次我把4个小矩阵连接成一个大矩阵，打印出来的结果太奇怪了，我都不知道对不对。

import numpy as np
from sympy import *
from sympy import init_printing
# from scipy.integrate import odeint
from scipy.integrate import solve_ivp
init_printing()

params = {'m1': 1, 'm2': 2, 'k1': 1, 'k2': 2, 'c1': 0.1, 'c2': 0.1} 
params.update({'N' : 3}) 
params['dimM'] = 4*params['N'] 

##---------------------------------------## M1 et M2 ##--------------------------------------------------##
M1 = np.zeros([2*params['N'],2*params['N']])
print('M1 = \n', M1)

M2 = np.eye(2*params['N'])
print('M2 = \n', M2)

# m1 pour les indices pair
# m2 pour les indices impair

##---------------------------------------## M3 ##--------------------------------------------------------##

v3 = np.zeros(2*params['N'])
v3[::2]= -(params['k1']/params['m2'] - params['k2']/params['m2'])
v3[1::2]= -(params['k1']/params['m1'] - params['k2']/params['m1'])
print("Vecteur v3:", v3, "\n")

v3_b = np.zeros(2*params['N']-1)
v3_b[::2] = params['k2']/params['m1']
v3_b[1::2] = params['k1']/params['m2']
print("Vecteur v3_b:", v3_b, "\n")

M3 = np.diag(v3, 0) + np.diag(v3_b, +1) + np.diag(v3_b, -1)
print('M3 = \n', M3)

##---------------------------------------## M4 ##--------------------------------------------------------##

v4 = np.zeros(2*params['N'])
v4[::2]= -(params['c1']/params['m2'] - params['c2']/params['m2'])
v4[1::2]= -(params['c1']/params['m1'] - params['c2']/params['m1'])
print("Vecteur v4:", v4, "\n")

v4_b = np.zeros(2*params['N']-1)
v4_b[::2] = params['c2']/params['m1']
v4_b[1::2] = params['c1']/params['m2']
print("Vecteur v4_b:", v4_b, "\n")

M4 = np.diag(v4, 0) + np.diag(v4_b, +1) + np.diag(v4_b, -1)
print('M4 = \n', M4)

##---------------------------------------## Matrice complete M ##----------------------------------------##

M = [[M1, M2], [M3,M4]]
print('M = \n', M)

如果你执行这段代码，你就可以看到所有的矩阵。M1，M2，M3，M4都运行得很好。但是M显示了这个奇怪的混乱：

M = 
 [[array([[0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 0., 0., 0.]]), array([[1., 0., 0., 0., 0., 0.],
       [0., 1., 0., 0., 0., 0.],
       [0., 0., 1., 0., 0., 0.],
       [0., 0., 0., 1., 0., 0.],
       [0., 0., 0., 0., 1., 0.],
       [0., 0., 0., 0., 0., 1.]])], [array([[0.5, 2. , 0. , 0. , 0. , 0. ],
       [2. , 1. , 0.5, 0. , 0. , 0. ],
       [0. , 0.5, 0.5, 2. , 0. , 0. ],
       [0. , 0. , 2. , 1. , 0.5, 0. ],
       [0. , 0. , 0. , 0.5, 0.5, 2. ],
       [0. , 0. , 0. , 0. , 2. , 1. ]]), array([[0.  , 0.1 , 0.  , 0.  , 0.  , 0.  ],
       [0.1 , 0.  , 0.05, 0.  , 0.  , 0.  ],
       [0.  , 0.05, 0.  , 0.1 , 0.  , 0.  ],
       [0.  , 0.  , 0.1 , 0.  , 0.05, 0.  ],
       [0.  , 0.  , 0.  , 0.05, 0.  , 0.1 ],
       [0.  , 0.  , 0.  , 0.  , 0.1 , 0.  ]])]]

numpy

来源：https://stackoverflow.com/questions/75787566/how-can-i-concatenate-4-matrices-into-one-big-matrix-without-it-looking-all-weir

2条答案

按热度按时间

ldioqlga1#

经过简短的谷歌搜索，Numpy有一个连接功能。
因此，这个问题的答案很简单，只需使用numpy的concatenate函数，如下所示：

M = np.zeros([4*params['N'],4*params['N']])
Ma = np.concatenate((M1, M2), axis=1)
Mb = np.concatenate((M3, M4), axis=1)
M = np.concatenate((Ma, Mb), axis=0)
print('Ma = \n', Ma)
print('Mb = \n', Mb)
print('M = \n', M)

赞(0）回复(0）举报 2023-03-23

piv4azn72#

您可以使用np.block：

import numpy as np

N = 4

M1 = np.random.randint(0, 10, size=(N, N))
M2 = np.random.randint(10, 100, size=(N, N))
M3 = - np.random.randint(0, 10, size=(N, N))
M4 = - np.random.randint(10, 100, size=(N, N))

M = np.block([[M1, M2], [M3, M4]])

>>> M.shape
(8, 8)
>>> M
M =
 [[  0   0   1   4  95  58  87  41]
 [  5   6   1   4  72  84  35  30]
 [  6   1   0   0  78  73  57  77]
 [  4   0   8   2  69  30  61  92]
 [ -8  -8  -5  -5 -52 -61 -21 -49]
 [ -5  -5  -4  -5 -61 -18 -52 -90]
 [ -5  -4  -5  -3 -98 -67 -75 -11]
 [ -8  -5  -8  -7 -75 -62 -11 -39]]

赞(0）回复(0）举报 2023-03-23

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numpy 我怎样才能把4个矩阵连接成一个大矩阵而不看起来很奇怪呢？

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