使用 Sympy 导出方程的多元牛顿法

Question

将 Sympy 派生方程转换为与 scipy.optimize.newton 兼容的多变量问题

我提到了。我发现使用 lambdify 将使符号方程与其他包兼容，例如 numpy 和 scipy.

这是我之前 post 的延续：

import numpy as np
import sympy as sym
import scipy.optimize
from sympy import symbols, diff, exp, log, power
from sympy.utilities.lambdify import lambdify

data = [3, 33, 146, 227, 342, 351, 353, 444, 556, 571, 709, 759, 836, 860, 968, 1056, 1726, 1846, 1872, 1986, 2311, 2366, 2608, 2676, 3098, 3278, 3288, 4434, 5034, 5049, 5085, 5089, 5089, 5097, 5324, 5389,5565, 5623, 6080, 6380, 6477, 6740, 7192, 7447, 7644, 7837, 7843, 7922, 8738, 10089, 10237, 10258, 10491, 10625, 10982, 11175, 11411, 11442, 11811, 12559, 12559, 12791, 13121, 13486, 14708, 15251, 15261, 15277, 15806, 16185, 16229, 16358, 17168, 17458, 17758, 18287, 18568, 18728, 19556, 20567, 21012, 21308, 23063, 24127, 25910, 26770, 27753, 28460, 28493, 29361, 30085, 32408, 35338, 36799, 37642, 37654, 37915, 39715, 40580, 42015, 42045, 42188, 42296, 42296, 45406, 46653, 47596, 48296, 49171, 49416, 50145, 52042, 52489, 52875, 53321, 53443, 54433, 55381, 56463, 56485, 56560, 57042, 62551, 62651, 62661, 63732, 64103, 64893, 71043, 74364, 75409, 76057, 81542, 82702, 84566, 88682]
n = len(data)
tn = data[n-1]


b, c = sym.symbols('b c', real=True)

f = -(-n +sum([sym.log(b*c*(num**(c-1))*sym.exp(-b*(num**c))) for num in data]))

bh = lambdify((b,c),diff(f,b),"numpy")
ch = lambdify((b,c),diff(f,c),"numpy")

sol = scipy.optimize.newton([bh,ch],(0.00404,1.0))
print(sol)

我似乎无法使用牛顿法。感谢任何信息或资源。

Answer 1

根据scipy.optimize.newton your x0 should be a scalar and not an array or tuple (which is what you are passing to scipy.optimize.newton() in your code). Fortunately, scipy.optimize.fsolve也可以找到零位并接受数组为x0.

据我所知，使用 sympy 让生活变得更艰难，而您可以简单地使用 numpy，这会加快您的代码速度（您可以避免 for 循环） .以下示例向您展示了如何使用它（希望对您有所帮助）：

import warnings 
import numpy as np
import scipy.optimize as opt
# filter warnings
warnings.filterwarnings("ignore")

data = np.array([3, 33, 146, 227, 342, 351, 353, 444, 556, 571, 709, 759, 836, 860, 968,
        1056, 1726, 1846, 1872, 1986, 2311, 2366, 2608, 2676, 3098, 3278, 3288,
        4434, 5034, 5049, 5085, 5089, 5089, 5097, 5324, 5389,5565, 5623, 6080, 
        6380, 6477, 6740, 7192, 7447, 7644, 7837, 7843, 7922, 8738, 10089,
        10237, 10258, 10491, 10625, 10982, 11175, 11411, 11442, 11811, 12559, 
        12559, 12791, 13121, 13486, 14708, 15251, 15261, 15277, 15806, 16185,
        16229, 16358, 17168, 17458, 17758, 18287, 18568, 18728, 19556, 20567,
        21012, 21308, 23063, 24127, 25910, 26770, 27753, 28460, 28493, 29361,
        30085, 32408, 35338, 36799, 37642, 37654, 37915, 39715, 40580, 42015,
        42045, 42188, 42296, 42296, 45406, 46653, 47596, 48296, 49171, 49416,
        50145, 2042,  52489, 52875, 53321, 53443, 54433, 55381, 56463, 56485,
        56560, 57042, 62551, 62651, 62661, 63732, 64103, 64893, 71043, 74364,
        75409, 76057, 81542, 82702, 84566, 88682])

n    = len(data)
tn   = data[n-1]

f        = lambda b, c: n - np.sum(np.log(b * c * data**(c-1) * np.exp(-b * data**c)))
obj_func = lambda x   : np.array([f(x[0], x[1]) - x[0], f(x[0], x[1]) - x[1]])


x0  = np.array([0.00404, 1.0])
sol = opt.fsolve(obj_func, x0 = x0)
print('b, c = ', sol)

输出：

b, c =  [0.01737803 0.4300348 ]

Answer 2

我们可以使用sympy中的'DeferredVector'来定义符号，而不是定义符号。而不是

b, c = sym.symbols('b c', real=True)
f = -(-n +sum([sym.log(b*c*(num**(c-1))*sym.exp(-b*(num**c))) for num in data]))
bh = lambdify((b,c),diff(f,b),"numpy")
ch = lambdify((b,c),diff(f,c),"numpy"

执行以下操作： x = DeferredVector('x') f = -(-n +sum([sym.log(x[0]x[1](num**(c-1)) for num in data]) )

现在使用 diff w.r.t x[0] 和 x[1] 和运行 scipy.optimize.newton()。

使用 Sympy 导出方程的多元牛顿法

Multivariate Newton's method for equations derived using Sympy

python

scipy

sympy

newtons-method

scipy-optimize