Question

我正在尝试求解非线性方程组。问题是解决方案很复杂，根据Octave / Matlab的描述，虚部很小。我正在尝试将其移至python，但是不幸的是，我不确定如何优雅地处理它。

在Octave中，我可以直接使用fsolve，然后将解决方案通过“实数”函数传递，以获取数字的实数部分。事实是，它很容易解决而不会返回任何错误

不幸的是，在python中使用numpy尝试求解方程式时会返回错误。这是用Python编写的方程式：

import numpy as np
from scipy.optimize import fsolve
import scipy.io as spio

params = dict()
params['cbeta'] = 0.96
params['cdelta'] = 0.1
params['calpha'] = 0.33
params['cgamma'] = 1.2
params['clambda']= 1.0
params['csigma'] = 0.8
params['etau'] = 0.0

def steady_s(vars0):
    # unpacking paramters
    cbeta = params['cbeta']
    cdelta = params['cdelta']
    calpha = params['calpha']
    cgamma = params['cgamma']
    clambda= params['clambda']
    csigma = params['csigma']

    # guesses for initial values
    c = vars0[0]
    y = vars0[1]
    k = vars0[2]
    g = vars0[3]
    r = vars0[4]

    # == functions to minimize to find steady states == #
    f = np.empty((5,))

    # HH Euler
    f[0] = (1.0/c)*cbeta*(r + 1.0 - cdelta) - (1.0+g)/c

    # Goods market clearing
    f[1] = y - c - k*(1.0 + g) + k*(1.0-cdelta)

    # Capital Market clearing
    f[2] = r - (k)**(calpha-1.0)*calpha**2.0

    # production function for final good
    f[3] = y - k**calpha

    # growth rate
    pi = (calpha - 1.0) * k**calpha #small pi, this isnt actual profits
    f[4] = g - (cgamma - 1.0) * clambda * (csigma*clambda*pi)**(csigma/(1.0-csigma))

    return f

# == Initial Guesses == #
vars0 = np.ones((5,))

# == Solving for Steady State == #

xss = fsolve(steady_s, vars0)

在Octave中实现相同的功能可提供以下解决方案：

 Columns 1 through 3:

   0.7851388 + 0.0000000i   0.8520544 + 0.0000000i   0.6155938 + 0.0000000i

 Columns 4 and 5:

   0.0087008 - 0.0000000i   0.1507300 - 0.0000000i

我将此解决方案通过Octave中的“真实”函数传递给我想要的结果。

尤其是，python甚至很难一次求解方程式。特别是，如果我尝试在定义了所有参数的函数外部运行f [4]，它将返回一个nan值。

任何帮助将不胜感激！

对任何我错过/严重格式化的内容表示歉意。

Answer 1

确实，scipy与复数作斗争。但是，一个名为mpmath的项目可以解决您的问题。此处：http://mpmath.org/。它曾经随sympy（sympy.org）一起提供。您可以找到文档here：此解决方案对我有用：

from mpmath import findroot
import numpy as np
import scipy.io as spio

params = dict()
params['cbeta'] = 0.96
params['cdelta'] = 0.1
params['calpha'] = 0.33
params['cgamma'] = 1.2
params['clambda']= 1.0
params['csigma'] = 0.8
params['etau'] = 0.0

def steady_s(c,y,k,g,r):
    # unpacking paramters
    cbeta = params['cbeta']
    cdelta = params['cdelta']
    calpha = params['calpha']
    cgamma = params['cgamma']
    clambda= params['clambda']
    csigma = params['csigma']

    # guesses for initial values
    #c = vars0[0]
    #y = vars0[1]
    #k = vars0[2]
    #g = vars0[3]
    #r = vars0[4]

    # == functions to minimize to find steady states == #
    f = [0,0,0,0,0]

    # HH Euler
    f[0] = (1.0/c)*cbeta*(r + 1.0 - cdelta) - (1.0+g)/c

    # Goods market clearing
    f[1] = y - c - k*(1.0 + g) + k*(1.0-cdelta)

    # Capital Market clearing
    f[2] = r - (k)**(calpha-1.0)*calpha**2.0

    # production function for final good
    f[3] = y - k**calpha

    # growth rate
    pi = (calpha - 1.0) * k**calpha #small pi, this isnt actual profits
    f[4] = g - (cgamma - 1.0) * clambda * (csigma*clambda*pi)**(csigma/(1.0-csigma))

    return f

# == Initial Guesses == #
vars0 = list(np.ones((5,)))

# == Solving for Steady State == #
xss = findroot(steady_s, vars0)

非线性方程组的优化系统-处理复数

1 个答案: