如何在python numpy中创建随机正交矩阵

Question

有没有一种方法可以调用在python中创建一个随机的标准正交矩阵？可能使用numpy？或者有没有办法使用多个numpy方法创建一个正交矩阵？谢谢.

Answer 1

版本0.18的scipy有scipy.stats.ortho_group和scipy.stats.special_ortho_group.添加它的拉取请求是https://github.com/scipy/scipy/pull/5622

例如,

In [24]: from scipy.stats import ortho_group  # Requires version 0.18 of scipy

In [25]: m = ortho_group.rvs(dim=3)

In [26]: m
Out[26]: 
array([[-0.23939017,  0.58743526, -0.77305379],
       [ 0.81921268, -0.30515101, -0.48556508],
       [-0.52113619, -0.74953498, -0.40818426]])

In [27]: np.set_printoptions(suppress=True)

In [28]: m.dot(m.T)
Out[28]: 
array([[ 1.,  0., -0.],
       [ 0.,  1.,  0.],
       [-0.,  0.,  1.]])

Answer 2

您可以通过使用均值和方差的高斯随机变量的元素执行矩阵的因式分解来获得随机n x n正交矩阵Q(均匀分布在n x n正交矩阵的流形上).这是一个例子:QRn x n01

import numpy as np
from scipy.linalg import qr

n = 3
H = np.random.randn(n, n)
Q, R = qr(H)

print (Q.dot(Q.T))

[[  1.00000000e+00  -2.77555756e-17   2.49800181e-16]
 [ -2.77555756e-17   1.00000000e+00  -1.38777878e-17]
 [  2.49800181e-16  -1.38777878e-17   1.00000000e+00]]

Answer 3

这是rvs从https://github.com/scipy/scipy/pull/5622/files中提取的方法,只需要很小的改动 - 足以作为一个独立的numpy函数运行.

import numpy as np    

def rvs(dim=3):
     random_state = np.random
     H = np.eye(dim)
     D = np.ones((dim,))
     for n in range(1, dim):
         x = random_state.normal(size=(dim-n+1,))
         D[n-1] = np.sign(x[0])
         x[0] -= D[n-1]*np.sqrt((x*x).sum())
         # Householder transformation
         Hx = (np.eye(dim-n+1) - 2.*np.outer(x, x)/(x*x).sum())
         mat = np.eye(dim)
         mat[n-1:, n-1:] = Hx
         H = np.dot(H, mat)
         # Fix the last sign such that the determinant is 1
     D[-1] = (-1)**(1-(dim % 2))*D.prod()
     # Equivalent to np.dot(np.diag(D), H) but faster, apparently
     H = (D*H.T).T
     return H

它符合Warren的测试,/sf/answers/2689860071/

Answer 4

from scipy.stats import special_ortho_group
num_dim=3
x = special_ortho_group.rvs(num_dim)

文档

Answer 5

创建任意形状 ( n x m) 正交矩阵的简单方法：

import numpy as np

n, m = 3, 5

H = np.random.rand(n, m)
u, s, vh = np.linalg.svd(H, full_matrices=False)
mat = u @ vh

print(mat @ mat.T) # -> eye(n)

请注意，如果n > m，它将获得mat.T @ mat = eye(m)。