Python scipy.linalg.expm与手工编码的差异
我试图实现矩阵指数函数，如Python scipy.linalg.expm与手工编码的差异,python,numpy,scipy,Python,Numpy,Scipy,我试图实现矩阵指数函数，如scipy.linalg.expm中所示。我从中获得灵感。我就这样写了下面的内容 from scipy.linalg import expm from scipy.linalg import eigh from scipy.linalg import inv from math import exp as math_exp from numpy import array, zeros from numpy.random import random_sample from
scipy.linalg.expmfrom scipy.linalg import expm
from scipy.linalg import eigh
from scipy.linalg import inv
from math import exp as math_exp
from numpy import array, zeros
from numpy.random import random_sample
from numpy.testing import assert_allclose
def diag2sqr(x):
'''Makes an square matrix from a diagonal one.
Takes a 1d matrix. Determines its data type.
Finds out the shape of the 1d matrix.
Makes an empty square matrix with both
dimensions equal to largest (nonzero) dimension of
the 1d matrix. It then fills the elements of the
1d matrix into diagonal slots of the empty
square one.
Parameters
----------
x : ndarray
ndarray of be coverted to a square ndarray
Returns
-------
xsqr : ndarray
ndarray with diagonals sameas those of x
all other elements are zero
dtype same as that of x
'''
x_flat = x.ravel()
xsqr = zeros((x_flat.shape[0], x_flat.shape[0]), dtype=x.dtype)
# Making the empty matrix
for i in range(x_flat.shape[0]):
xsqr[i, i] = x_flat[i]
# filling up the ith element
print('xsqr', xsqr)
return xsqr
def kaityo_expm(x, ):
'''Exponentiates an ndarray (kaityo).
Exponentiates a ndarray in the most naive way.
Parameters
----------
x : ndarray
The ndarray to be exponentiated
Returns
-------
kexpm : ndarray
x after exponentiating
'''
rx, ux = eigh(x)
# Find eigenvalues and eigenvectors
# eigenvectors composed to form a unitary
ux_inv = inv(ux)
# Inverse of the unitary
# tx = diag([array([math_exp(i) for i in rx]).ravel()])
# tx = array([math_exp(i) for i in rx])
tx = diag2sqr(array([math_exp(i) for i in rx]))
# Constructing the diagonal matrix
kexpm1 = tx@ux_inv
kexpm = ux@kexpm1
return kexpm
scipy.linalg.expmx = random_sample((10, 10))
assert_allclose(expm(x), kaityo_expm(x))
AssertionError:
Not equal to tolerance rtol=1e-07, atol=0
Mismatch: 100%
Max absolute difference: 7.04655733
Max relative difference: 0.59875635
x: array([[18.032424, 16.224408, 12.432163, 16.614248, 12.85653 , 13.705387,
15.096966, 10.577946, 18.399573, 17.938062],
[16.352809, 17.525898, 12.79079 , 16.295562, 13.512996, 14.407979,...
y: array([[18.649103, 13.157682, 11.264763, 16.099163, 15.2293 , 17.854499,
11.691586, 13.412066, 15.023189, 15.598455],
[13.157682, 13.612502, 9.628261, 12.659313, 13.559437, 13.382417,..
- 他们的不同意见可以接受吗
- 我的实现错了吗
- 如果我的实现是错误的，我如何修复它
- 如果我的实现是正确的，那么何时使用
是安全的scipy.linalg.expm
>>> import numpy as np
>>> import scipy.linalg as ln
>>> A = [[2/3, -4/3, 2],
[5/6, 4/3, -2],
[5/6, -2/3, 0]]
>>> A = np.matrix(A)
>>> print(A)
[[ 0.66666667 -1.33333333 2. ]
[ 0.83333333 1.33333333 -2. ]
[ 0.83333333 -0.66666667 0. ]]
>>> eigvalue, eigvectors = np.linalg.eig(A)
>>> print("eigvalue: ", eigvalue)
>>> print("eigvectors:")
>>> print(eigvectors)
eigvalue: [ 1. -1. 2.]
eigvectors:
[[ 0.81649658 0.27216553 0.87287156]
[ 0.40824829 -0.68041382 -0.21821789]
[ 0.40824829 -0.68041382 0.43643578]]
>>> e_Lambda = np.eye(np.size(A, 0))*(np.exp(eigvalue))
>>> print(e_Lambda)
[[2.71828183 0. 0. ]
[0. 0.36787944 0. ]
[0. 0. 7.3890561 ]]
>>> e_A = eigvectors*e_Lambda*eigvectors.I
>>> print(e_A)
[[ 2.3265481 -6.22769903 7.01116649]
[ 0.97933433 4.27520659 -3.51559341]
[ 0.97933433 -3.11384951 3.87346269]]
>>> e_A2 = ln.expm(A)
>>> print(e_A2)
[[ 2.3265481 -6.22769903 7.01116649]
[ 0.97933433 4.27520659 -3.51559341]
[ 0.97933433 -3.11384951 3.87346269]]
>>> np.testing.assert_allclose(e_A, e_A2)
>>> print(e_A - e_A2)
[[-1.77635684e-15 1.77635684e-15 -8.88178420e-16]
[ 4.44089210e-16 -1.77635684e-15 8.88178420e-16]
[-2.22044605e-16 0.00000000e+00 4.44089210e-16]]
scipy.linalg.expm我用笔记本做了一个测试 如果有人时间不够，这里是Jupyter笔记本版本。两件事：1）
eighinv