我是Python的新手,正在嘗試實現Gauss-Newton方法,特別是在Wikipedia頁面上的示例(Gauss–Newton algorithm,第3個示例).以下是我到目前為止所做的：

import scipy

import numpy as np

import math

import scipy.misc

from matplotlib import pyplot as plt, cm, colors

S = [0.038,0.194,.425,.626,1.253,2.500,3.740]

rate = [0.050,0.127,0.094,0.2122,0.2729,0.2665,0.3317]

iterations = 5

rows = 7

cols = 2

B = np.matrix([[.9],[.2]]) # original guess for B

Jf = np.zeros((rows,cols)) # Jacobian matrix from r

r = np.zeros((rows,1)) #r equations

def model(Vmax, Km, Sval):

return ((vmax * Sval) / (Km + Sval))

def partialDerB1(B2,xi):

return round(-(xi/(B2+xi)),10)

def partialDerB2(B1,B2,xi):

return round(((B1*xi)/((B2+xi)*(B2+xi))),10)

def residual(x,y,B1,B2):

return (y - ((B1*x)/(B2+x)))

for i in range(0,iterations):

sumOfResid=0

#calculate Jr and r for this iteration.

for j in range(0,rows):

r[j,0] = residual(S[j],rate[j],B[0],B[1])

sumOfResid = sumOfResid + (r[j,0] * r[j,0])

Jf[j,0] = partialDerB1(B[1],S[j])

Jf[j,1] = partialDerB2(B[0],B[1],S[j])

Jft = np.transpose(Jf)

B = B + np.dot((np.dot(Jft,Jf)**-1),(np.dot(Jft,r)))

print B

殘差的平方和在每次迭代時都增加而不是趨向于0,并且我得到的B向量增加.

我在了解問題所在時遇到了麻煩,我們將不勝感激.

解決方法:

您在Beta更新的代碼中出了錯：應該是

B = B - np.dot(np.dot( inv(np.dot(Jft, Jf)), Jft), r)

而不是矩陣上的**-1來計算逆矩陣

import scipy

import numpy as np

from numpy.linalg import inv

import math

import scipy.misc

#from matplotlib import pyplot as plt, cm, colors

S = [0.038,0.194,.425,.626,1.253,2.500,3.740]

rate = [0.050,0.127,0.094,0.2122,0.2729,0.2665,0.3317]

iterations = 5

rows = 7

cols = 2

B = np.matrix([[.9],[.2]]) # original guess for B

print(B)

Jf = np.zeros((rows,cols)) # Jacobian matrix from r

r = np.zeros((rows,1)) #r equations

def model(Vmax, Km, Sval):

return ((Vmax * Sval) / (Km + Sval))

def partialDerB1(B2,xi):

return round(-(xi/(B2+xi)),10)

def partialDerB2(B1,B2,xi):

return round(((B1*xi)/((B2+xi)*(B2+xi))),10)

def residual(x,y,B1,B2):

return (y - ((B1*x)/(B2+x)))

#

for _ in xrange(iterations):

sumOfResid=0

#calculate Jr and r for this iteration.

for j in xrange(rows):

r[j,0] = residual(S[j],rate[j],B[0],B[1])

sumOfResid += (r[j,0] * r[j,0])

Jf[j,0] = partialDerB1(B[1],S[j])

Jf[j,1] = partialDerB2(B[0],B[1],S[j])

Jft = Jf.T

B -= np.dot(np.dot( inv(np.dot(Jft,Jf)),Jft),r)

print B

標簽：scipy,python,numpy

來源： https://codeday.me/bug/20191120/2044228.html

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