使用python解决非线性超定系统

Question

我正试图找到一个很好的方法来解决python非线性超定系统。我在这里查看优化工具http://docs.scipy.org/doc/scipy/reference/optimize.nonlin.html，但我无法弄清楚如何使用它们。我到目前为止是

#overdetermined nonlinear system that I'll be using 
''' 
a = cos(x)*cos(y)       
b = cos(x)*sin(y)       
c = -sin(y)         
d = sin(z)*sin(y)*sin(x) + cos(z)*cos(y)  
e = cos(x)*sin(z)       
f = cos(z)*sin(x)*cos(z) + sin(z)*sin(x)  
g = cos(z)*sin(x)*sin(y) - sin(z)*cos(y)  
h = cos(x)*cos(z) 
a-h will be random int values in the range 0-10 inclusive 
''' 
import math 
from random import randint 
import scipy.optimize 

def system(p): 
    x, y, z = p 
    return(math.cos(x)*math.cos(y)-randint(0,10), 
      math.cos(x)*math.sin(y)-randint(0,10), 
      -math.sin(y)-randint(0,10), 
      math.sin(z)*math.sin(y)*math.sin(x)+math.cos(z)*math.cos(y)-randint(0,10), 
      math.cos(x)*math.sin(z)-randint(0,10), 
      math.cos(z)*math.sin(x)*math.cos(z)+math.sin(z)*math.sin(x)-randint(0,10), 
      math.cos(z)*math.sin(x)*math.sin(y)-math.sin(z)*math.cos(y)-randint(0,10), 
      math.cos(x)*math.cos(z)-randint(0,10)) 

x = scipy.optimize.broyden1(system, [1,1,1], f_tol=1e-14) 

你能帮我一下吗？

Answer 1

如果我理解你的话，你想找到方程f(x) = b的非线性系统的近似解，其中b是包含随机值b=[a,...,h]的向量。

为了做到这一点，您首先需要从system函数中删除随机值，否则在每次迭代中求解器都会尝试求解不同的方程组。此外，我认为基本Broyden方法仅适用于具有与方程式一样多的未知数的系统。或者，您可以使用scipy.optimize.leastsq。一个可能的解决方案如下所示：

# I am using numpy because it's more convenient for the generation of 
# random numbers. 
import numpy as np 
from numpy.random import randint 
import scipy.optimize 

# Removed random right-hand side values and changed nomenclature a bit. 
def f(x): 
    x1, x2, x3 = x 
    return np.asarray((math.cos(x1)*math.cos(x2), 
         math.cos(x1)*math.sin(x2), 
         -math.sin(x2), 
         math.sin(x3)*math.sin(x2)*math.sin(x1)+math.cos(x3)*math.cos(x2), 
         math.cos(x1)*math.sin(x3), 
         math.cos(x3)*math.sin(x1)*math.cos(x3)+math.sin(x3)*math.sin(x1), 
         math.cos(x3)*math.sin(x1)*math.sin(x2)-math.sin(x3)*math.cos(x2), 
         math.cos(x1)*math.cos(x3))) 

# The second parameter is used to set the solution vector using the args 
# argument of leastsq. 
def system(x,b): 
    return (f(x)-b) 

b = randint(0, 10, size=8) 
x = scipy.optimize.leastsq(system, np.asarray((1,1,1)), args=b)[0] 

我希望这对您有所帮助。但是请注意，找到解决方案的可能性非常小，特别是在间隔[0,10]中生成随机整数，而f的范围限制为[-2,2]时