奇异矩阵 - python
Singular matrix - python
以下代码显示了矩阵的奇异性问题,因为在 Pycharm 中工作,我得到
raise LinAlgError("Singular matrix")
numpy.linalg.linalg.LinAlgError: Singular matrix
我猜问题出在 K 上,但我不明白具体是怎么回事:
from numpy import zeros
from numpy.linalg import linalg
import math
def getA(kappa):
matrix = zeros((n, n), float)
for i in range(n):
for j in range(n):
matrix[i][j] = 2*math.cos((2*math.pi/n)*(abs(j-i))*kappa)
return matrix
def getF(csi, a):
csiInv = linalg.inv(csi)
valueF = csiInv * a * csiInv * a
traceF = valueF.trace()
return 0.5 * traceF
def getG(csi, f, a):
csiInv = linalg.inv(csi)
valueG = (csiInv * a * csiInv) / (2 * f)
return valueG
def getE(g, k):
KInv = linalg.inv(k)
Ktrans = linalg.transpose(k)
KtransInv = linalg.inv(Ktrans)
e = KtransInv * g * KInv
return e
file = open('transformed.txt', 'r')
n = 4
transformed = zeros(n)
for counter, line in enumerate(file):
if counter == n:
break
transformed[counter] = float(line)
CSI = zeros((n, n))
for i in range(n):
for j in range(n):
CSI[i][j] = transformed[abs(i-j)]
A = getA(1)
F = getF(CSI, A)
G = getG(CSI, F, A)
K = zeros((n, n), float)
for j in range(n):
K[0][j] = 0.0001
for i in range(1, n):
for j in range(n):
K[i][j] = ((3.0*70.0*70.0*0.3)/(2.0*300000.0*300000.0))*((j*(i-j))/i)*(1.0+(70.0/300000.0)*j)
E = getE(G, K)
print G
print K
有没有人有任何修复建议?谢谢
将非常"close"奇异的矩阵求逆通常会导致计算问题。一个快速的技巧是在求逆之前向矩阵的对角线添加一个非常小的值。
def getE(g, k):
m = 10^-6
KInv = linalg.inv(k + numpy.eye(k.shape[1])*m)
Ktrans = linalg.transpose(k)
KtransInv = linalg.inv(Ktrans + + numpy.eye(Ktrans.shape[1])*m)
e = KtransInv * g * KInv
return e
我认为这足以完成家庭作业。但是,如果你想真正部署一些计算稳健的东西,你应该研究反转的替代方法。
numerically stable inverse of a 2x2 matrix
以下代码显示了矩阵的奇异性问题,因为在 Pycharm 中工作,我得到
raise LinAlgError("Singular matrix")
numpy.linalg.linalg.LinAlgError: Singular matrix
我猜问题出在 K 上,但我不明白具体是怎么回事:
from numpy import zeros
from numpy.linalg import linalg
import math
def getA(kappa):
matrix = zeros((n, n), float)
for i in range(n):
for j in range(n):
matrix[i][j] = 2*math.cos((2*math.pi/n)*(abs(j-i))*kappa)
return matrix
def getF(csi, a):
csiInv = linalg.inv(csi)
valueF = csiInv * a * csiInv * a
traceF = valueF.trace()
return 0.5 * traceF
def getG(csi, f, a):
csiInv = linalg.inv(csi)
valueG = (csiInv * a * csiInv) / (2 * f)
return valueG
def getE(g, k):
KInv = linalg.inv(k)
Ktrans = linalg.transpose(k)
KtransInv = linalg.inv(Ktrans)
e = KtransInv * g * KInv
return e
file = open('transformed.txt', 'r')
n = 4
transformed = zeros(n)
for counter, line in enumerate(file):
if counter == n:
break
transformed[counter] = float(line)
CSI = zeros((n, n))
for i in range(n):
for j in range(n):
CSI[i][j] = transformed[abs(i-j)]
A = getA(1)
F = getF(CSI, A)
G = getG(CSI, F, A)
K = zeros((n, n), float)
for j in range(n):
K[0][j] = 0.0001
for i in range(1, n):
for j in range(n):
K[i][j] = ((3.0*70.0*70.0*0.3)/(2.0*300000.0*300000.0))*((j*(i-j))/i)*(1.0+(70.0/300000.0)*j)
E = getE(G, K)
print G
print K
有没有人有任何修复建议?谢谢
将非常"close"奇异的矩阵求逆通常会导致计算问题。一个快速的技巧是在求逆之前向矩阵的对角线添加一个非常小的值。
def getE(g, k):
m = 10^-6
KInv = linalg.inv(k + numpy.eye(k.shape[1])*m)
Ktrans = linalg.transpose(k)
KtransInv = linalg.inv(Ktrans + + numpy.eye(Ktrans.shape[1])*m)
e = KtransInv * g * KInv
return e
我认为这足以完成家庭作业。但是,如果你想真正部署一些计算稳健的东西,你应该研究反转的替代方法。
numerically stable inverse of a 2x2 matrix