我试图以最有效的方式解决带状稀疏矩阵的逆,以便我可以将其合并到我的实时系统中。我正在生成代表卷积运算的稀疏带状矩阵。目前,我正在使用spsolve
图书馆的scipy.sparse.linalg
。我发现使用solve_banded
库中的scipy.linalg
有更好的方法。然而,solve_banded
需要(l,u)
,这是非零的上下对角线和ab
的数量,(l + u + 1, M)
阵列像带状矩阵。我不知道如何转换我的代码,以便我可以使用solve_banded
。对此方面的任何帮助都非常感谢。
import numpy as np
from scipy import linalg
import math
import time
from scipy.sparse import spdiags
from scipy.sparse.linalg import spsolve
def ABC(deg, fc, N):
r"""Generate sparse-banded matrices
"""
omc = 2*math.pi*fc
t = ((1-math.cos(omc))/(1+math.cos(omc)))**deg
p = 1
for k in np.arange(deg):
p = np.convolve(p,np.array([-1,1]),'full')
P = spdiags(np.kron(p,np.ones((N,1))).T, np.arange(deg+1), N-deg, N)
B = P.T.dot(P)
q = np.sqrt(t)
for k in np.arange(deg):
q = np.convolve(q,np.array([1,1]),'full')
Q = spdiags(np.kron(q,np.ones((N,1))).T, np.arange(deg+1), N-deg, N)
C = Q.T.dot(Q)
A = B + C
return A,B,C
if __name__ == '__main__':
mu = 0.1
deg = 3
wc = 0.1
for i in np.arange(1,7,1):
# some dense random vector
x = np.random.rand(10**i,1)
# generate sparse banded matrices
A,_,C = ABC(deg, wc, 10**i)
# another banded matrix
G = mu*A.dot(A.T) + C.dot(C.T)
# SCIPY SPSOLVE
st = time.time()
y = spsolve(G,x)
et = time.time()
print("SCIPY SPSOLVE: N = ", 10**i, "Time taken: ", et-st)
结果
SCIPY SPSOLVE: N = 10 Time taken: 0.0
SCIPY SPSOLVE: N = 100 Time taken: 0.0
SCIPY SPSOLVE: N = 1000 Time taken: 0.015689611434936523
SCIPY SPSOLVE: N = 10000 Time taken: 0.020943641662597656
SCIPY SPSOLVE: N = 100000 Time taken: 0.16722917556762695
SCIPY SPSOLVE: N = 1000000 Time taken: 1.7254831790924072
使用solveh_banded
库中的scipy
解决了这个问题。当矩阵是对称的和正定带状矩阵时,非常快速的矩阵求逆技术用于极大的稀疏带状矩阵。
from scipy.linalg import solveh_banded
def sp_inv(A, x):
A = A.toarray()
N = np.shape(A)[0]
D = np.count_nonzero(A[0,:])
ab = np.zeros((D,N))
for i in np.arange(1,D):
ab[i,:] = np.concatenate((np.diag(A,k=i),np.zeros(i,)),axis=None)
ab[0,:] = np.diag(A,k=0)
y = solveh_banded(ab,x,lower=True)
return y