Python如何在numpy中组合两个矩阵
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Python how to combine two matrices in numpy
提问by ilovecp3
new to Python, struggling in numpy, hope someone can help me, thank you!
Python新手，在numpy中挣扎，希望有人能帮帮我，谢谢！
from numpy import *
A = matrix('1.0 2.0; 3.0 4.0')
B = matrix('5.0 6.0')
C = matrix('1.0 2.0; 3.0 4.0; 5.0 6.0')
print "A=",A
print "B=",B
print "C=",C
results:
结果：
A= [[ 1. 2.]
[ 3. 4.]]
B= [[ 5. 6.]]
C= [[ 1. 2.]
[ 3. 4.]
[ 5. 6.]]
Question: how to use A and B to generate C, like in matlab C=[A;B]?
问题：如何使用 A 和 B 生成 C，就像在 matlab 中一样C=[A;B]？
采纳答案by Ashwini Chaudhary
Use numpy.concatenate:
>>> import numpy as np
>>> np.concatenate((A, B))
matrix([[ 1., 2.],
[ 3., 4.],
[ 5., 6.]])
回答by Roman Pekar
You can use numpy.vstack:
您可以使用numpy.vstack：
>>> np.vstack((A,B))
matrix([[ 1., 2.],
[ 3., 4.],
[ 5., 6.]])
回答by yourstruly
If You want to work on existing array C, you could do it inplace:
如果你想在现有的数组 C 上工作，你可以就地做：
>>> from numpy import *
>>> A = matrix('1.0 2.0; 3.0 4.0')
>>> B = matrix('5.0 6.0')
>>> shA=A.shape
>>> shA
(2L, 2L)
>>> shB=B.shape
>>> shB
(1L, 2L)
>>> C = zeros((shA[0]+shB[0],shA[1]))
>>> C
array([[ 0., 0.],
[ 0., 0.],
[ 0., 0.]])
>>> C[:shA[0]]
array([[ 0., 0.],
[ 0., 0.]])
>>> C[:shA[0]]=A
>>> C[shA[0]:shB[0]]=B
>>> C
array([[ 1., 2.],
[ 3., 4.],
[ 0., 0.]])
>>> C[shA[0]:shB[0]+shA[0]]
array([[ 0., 0.]])
>>> C[shA[0]:shB[0]+shA[0]]=B
>>> C
array([[ 1., 2.],
[ 3., 4.],
[ 5., 6.]])
回答by Py Foliage
For advanced combining (you can give it loop if you want to combine lots of matrices):
对于高级组合（如果你想组合大量矩阵，你可以给它循环）：
# Advanced combining
import numpy as np
# Data
A = np.matrix('1 2 3; 4 5 6')
B = np.matrix('7 8')
print('Original Matrices')
print(A)
print(B)
# Getting the size
shA=np.shape(A)
shB=np.shape(B)
rowTot=shA[0]+shB[0]
colTot=shA[1]+shB[1]
rowMax=np.max((shA[0],shB[0]))
colMax=np.max((shA[1],shB[1]))
# Allocate zeros to C
CVert=np.zeros((rowTot,colMax)).astype('int')
CHorz=np.zeros((rowMax,colTot)).astype('int')
CDiag=np.zeros((rowTot,colTot)).astype('int')
# Replace C
CVert[0:shA[0],0:shA[1]]=A
CVert[shA[0]:rowTot,0:shB[1]]=B
print('Vertical Combine')
print(CVert)
CHorz[0:shA[0],0:shA[1]]=A
CHorz[0:shB[0],shA[1]:colTot]=B
print('Horizontal Combine')
print(CHorz)
CDiag[0:shA[0],0:shA[1]]=A
CDiag[shA[0]:rowTot,shA[1]:colTot]=B
print('Diagonal Combine')
print(CDiag)
The result:
结果：
# Result
# Original Matrices
# [[1 2 3]
# [4 5 6]]
# [[7 8]]
# Vertical Combine
# [[1 2 3]
# [4 5 6]
# [7 8 0]]
# Horizontal Combine
# [[1 2 3 7 8]
# [4 5 6 0 0]]
# Diagonal Combine
# [[1 2 3 0 0]
# [4 5 6 0 0]
# [0 0 0 7 8]]
Credit: I edit yourstruly answer and implement what I already have on my code
信用：我编辑了你的答案并实现了我的代码中已有的内容
