使用Sympy:TypeError:“>=”在“NoneType”和“int”的实例之间不受支持

Question

因此，我有以下代码：

from sympy import Symbol, solve, nsolve

x1 = Symbol('x1')
x2 = Symbol('x2')
w1 = Symbol('w1')
w2 = Symbol('w2')

eq1 = w1 + w2
eq2 = (w1 * x1) + (w2 * x2)
eq3 = (w1 * x1**2) + (w2 * x2**2)
eq4 = (w1 * x1**3) + (w2 * x2**3)

print(nsolve((eq1, eq2, eq3, eq4), (x1, x2, w1, w2), (2, 0, 2/3, 0)))

关于这个问题：

我的回答是：

 x = findroot(f, x0, J=J, **kwargs)
  File "C:\Users\Sabri\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mpmath\calculus\optimization.py", line 969, in findroot
    for x, error in iterations:
  File "C:\Users\Sabri\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mpmath\calculus\optimization.py", line 660, in __iter__
    s = self.ctx.lu_solve(Jx, fxn)
  File "C:\Users\Sabri\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mpmath\matrices\linalg.py", line 226, in lu_solve
    A, p = ctx.LU_decomp(A)
  File "C:\Users\Sabri\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mpmath\matrices\linalg.py", line 142, in LU_decomp
    ctx.swap_row(A, j, p[j])
  File "C:\Users\Sabri\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mpmath\matrices\matrices.py", line 876, in swap_row
    A[i,k], A[j,k] = A[j,k], A[i,k]
  File "C:\Users\Sabri\AppData\Local\Packages\PythonSoftwareFoundation.Python.3.8_qbz5n2kfra8p0\LocalCache\local-packages\Python38\site-packages\mpmath\matrices\matrices.py", line 490, in __getitem__
    if key[0] >= self.__rows or key[1] >= self.__cols:
TypeError: '>=' not supported between instances of 'NoneType' and 'int'

以下是此处的文档：

https://docs.sympy.org/latest/modules/solvers/solvers.html

目前还不清楚是什么导致了这个问题，除了其中一个变量可能是“无”。在谷歌上，许多类似错误的问题都被明确地显示出来，但这里的情况并非如此。有什么建议吗

编辑：我得到这个答案：

使用此代码：

from scipy.optimize import fsolve


def func(p):
    x1, x2, w1, w2 = p
    return (w1 + w2, (w1 * x1) + (w2 * x2), (w1 * x1**2) + (w2 * x2**2), (w1 * x1**3) + (w2 * x2**3))

x1, x2, w1, w2 = fsolve(func, (2, 0, 2/3, 0))

print(x1, x2, w1, w2)

所以代码应该返回一个结果，但我不确定为什么它对Sympy不起作用。谢谢

Answer 1

Symphy能够解非线性方程组，因此，如果您不想猜测，可以：

>>> from sympy import Rational, nonlinsolve
>>> eq1 = w1 + w2 - 2
>>> eq2 = (w1 * x1) + (w2 * x2) - 0
>>> eq3 = (w1 * x1**2) + (w2 * x2**2) - Rational(2, 3)  # 2/3 gives float 0.66..
>>> eq4 = (w1 * x1**3) + (w2 * x2**3) - 0

>>> nonlinsolve((eq1, eq2, eq3, eq4), (x1, x2, w1, w2))
FiniteSet((-sqrt(3)/3, sqrt(3)/3, 1, 1), (sqrt(3)/3, -sqrt(3)/3, 1, 1))

这就给出了两个解决方案

Answer 2

1. 使用nsolve()时，RHS都是零。将右侧的非零项（2,0,2/3,0）移动到左侧
2. nsolve()的第三个参数是接近解的初始猜测
   有根据的猜测是(x1,x2,w1,w2) = (-1,1,1,1)区间[-1,1]中的(x1,x2,w1,w2) = (-1,1,1,1)权重相等
   我试过其他一些猜测。其中一些造成了同样的错误

输出：

w1 + w2 - 2
w1*x1 + w2*x2
w1*x1**2 + w2*x2**2 - 0.666666666666667
w1*x1**3 + w2*x2**3
Matrix([[-0.577350269189626], [0.577350269189626], [1.00000000000000], [1.00000000000000]])

结果与维基百科中关于table的n=2案例一致

代码：

from sympy import Symbol, solve, nsolve

x1 = Symbol('x1')
x2 = Symbol('x2')
w1 = Symbol('w1')
w2 = Symbol('w2')

eq1 = w1 + w2 - 2
eq2 = (w1 * x1) + (w2 * x2) - 0
eq3 = (w1 * x1**2) + (w2 * x2**2) - 2 / 3
eq4 = (w1 * x1**3) + (w2 * x2**3) - 0

print(eq1, eq2, eq3, eq4, sep='\n')
print(nsolve((eq1, eq2, eq3, eq4), (x1, x2, w1, w2), (-1, 1, 0.5, 0.5)))