Numpy 如何在Python中求解未知成员相乘的非线性方程组?
我有以下非线性方程组: 这与Cholesky的分解非常相似,但不幸的是它不是 我已经找到了非常熟悉的解决方案:但我不知道如何动态设置系统的所有方程 如何在Numpy、Scipy或Python中的任何其他软件包中求解此系统?以下是一个函数,可根据系统中的描述动态设置方程式:Numpy 如何在Python中求解未知成员相乘的非线性方程组?,numpy,math,scipy,Numpy,Math,Scipy,我有以下非线性方程组: 这与Cholesky的分解非常相似,但不幸的是它不是 我已经找到了非常熟悉的解决方案:但我不知道如何动态设置系统的所有方程 如何在Numpy、Scipy或Python中的任何其他软件包中求解此系统?以下是一个函数,可根据系统中的描述动态设置方程式: import sympy as sp def not_choleskys_decomposition(N): # Initialisation of list of variables. c = []
import sympy as sp
def not_choleskys_decomposition(N):
# Initialisation of list of variables.
c = []
K = []
for n in range(N + 1):
c.append(sp.Symbol("c_{}".format(n), real=True))
K.append(sp.Symbol("K_{}".format(n), real=True))
# Setup your N+1 equations.
equations = []
for n in range(N + 1):
num_c = N + 1 - n
c_terms = [c_i * c_j for c_i, c_j in zip(c[:num_c], c[n:][:num_c])]
lhs = sp.Add(*c_terms)
equations.append(sp.Eq(lhs, K[n]))
return equations, c, K
打印验证可使用将数学直接转换为latex的sympy.latex
函数或sympy.pprint
函数完成
>>> test, _, _ = not_choleskys_decomposition(4)
>>> for t in test:
... sp.pprint(t)
c₀⋅c₀ + c₁⋅c₁ + c₂⋅c₂ + c₃⋅c₃ + c₄⋅c₄ = K₀
c₀⋅c₁ + c₁⋅c₂ + c₂⋅c₃ + c₃⋅c₄ = K₁
c₀⋅c₂ + c₁⋅c₃ + c₂⋅c₄ = K₂
c₀⋅c₃ + c₁⋅c₄ = K₃
c₀⋅c₄ = K₄
可以使用属于每个sympy.Eq
对象的.subs
方法替换变量:
>>> equations, c_sym, K_sym = not_choleskys_decomposition(4)
>>> dict_replace = {c_sym[0]: 1.0}
>>> new_equation = equations[0].subs(dict_replace)
>>> sympy.pprint(new_equation)
c₁⋅c₂ + 1.0⋅c₁ + c₂⋅c₃ + c₃⋅c₄ = K₁
下面是一个示例函数,用于替换c
或K
的列表
def substitute_sym(equations, sym_list, val_list):
# Confirm all dimensions are consistent.
try:
assert len(equations) == len(sym_list) == len(val_list)
except AssertionError:
raise IndexError("Inconsistent dimensions.")
# Replace c symbols with values in equations.
substituted_equations = []
for eq in equations:
_eq = eq.subs(dict(zip(sym_list, val_list)))
substituted_equations.append(_eq)
return substituted_equations
在c
上测试它:
在K
上测试它:
可以使用sympy.solvers.solveset.nonlinsolve(系统,*符号)
求解给定的K
c
如果您选择使用它,我希望这有助于您开始使用SymPy
。我很少使用它的非线性解算器,所以我不能保证任何超过示例的东西。争论也是相关的,解决问题的人脾气暴躁。您会注意到,对于每个索引,我将K
替换为sp.Rational(0.5)
。这样做是为了避免在为某些解算器定义float
类型时抛出错误。祝你好运
编辑:
还请注意,您不需要在Symphy中使用解算器。我很少这样做。我使用Python中的符号数学包进行LaTeX和方程操作
编辑:
使其与scipy.optimize.fsolve一起工作
from sympy.utilities.lambdify import lambdify
from scipy.optimize import fsolve
def lambdify_equations_for_solving_c(equations, c_sym, K_sym, K_val):
f = []
equations_subbed = substitute_sym(equations, K_sym, K_val)
for eq in equations_subbed:
# Note that I reformulate equation into an expression here
# for determining the roots.
f.append(lambdify(c_sym, eq.args[0] - eq.args[1]))
return f
N = 4
equations, c_sym, K_sym = not_choleskys_decomposition(N)
K_vals = [0.1, 0, 0.1, 0, 0]
f = lambdify_equations_for_solving_c(equations, c_sym, K_sym, K_vals)
def fsolve_friendly(p):
return [_f(*p) for _f in f]
c_sol = fsolve(fsolve_friendly, x0=(0, 0.1, -1.1, 0, 0))
fsolve
返回非收敛结果的警告
/home/ggarrett/anaconda3/envs/sigh/lib/python3.7/site-packages/scipy/optimize/minpack.py:162: RuntimeWarning: The iteration is not making good progress, as measured by the
improvement from the last five Jacobian evaluations.
warnings.warn(msg, RuntimeWarning)
这是一个完整的答案,说明了如何将方程作为python可调用函数进行求解。解算程序本身就是另一个挑战。对此我不确定。这些方程是一个多项式系统,所以我想知道关于5次或更高的一般多项式在根方面不可解的问题是否会起作用。也许可以证明,在这个特殊的例子中,所有N都有一个解,但我想我建议在寻找符号解之前先研究这个问题。为我不平衡的回答道歉。我更专注于帮助OP创建方程组。我在我的答案中添加了更多内容,使之能够成为一个函数,并与fsolve一起使用。fsolve的结果并不乐观,但这是我在编程方面能提供的最全面的帮助——数学是另一个。嗯,这是一个有趣的问题。你需要符号解还是数值解?(你解决这个问题的目的是什么?)?
>>> from sympy.solvers.solveset import nonlinsolve
>>> test_solve, c_sym, K_sym = not_choleskys_decomposition(1)
>>> test_replaced = substitute_sym(test_solve, K_sym, [sp.Rational(0.5)] * 2)
>>> solved = nonlinsolve(test_replaced, c_sym)
>>> sp.pprint(solved)
⎧⎛ ⎛ 2⎞ ⎞ ⎛ ⎛
⎪⎜ ⎜ ⎛ √6 √2⋅ⅈ⎞ ⎟ ⎛ √6 √2⋅ⅈ⎞ √6 √2⋅ⅈ⎟ ⎜ ⎜ ⎛ √6 √2⋅ⅈ
⎨⎜-⎜-1 + 2⋅⎜- ── - ────⎟ ⎟⋅⎜- ── - ────⎟, - ── - ────⎟, ⎜-⎜-1 + 2⋅⎜- ── + ────
⎪⎝ ⎝ ⎝ 4 4 ⎠ ⎠ ⎝ 4 4 ⎠ 4 4 ⎠ ⎝ ⎝ ⎝ 4 4
⎩
2⎞ ⎞ ⎛ ⎛ 2⎞
⎞ ⎟ ⎛ √6 √2⋅ⅈ⎞ √6 √2⋅ⅈ⎟ ⎜ ⎜ ⎛√6 √2⋅ⅈ⎞ ⎟ ⎛√6 √2⋅ⅈ⎞ √6 √2⋅
⎟ ⎟⋅⎜- ── + ────⎟, - ── + ────⎟, ⎜-⎜-1 + 2⋅⎜── - ────⎟ ⎟⋅⎜── - ────⎟, ── - ───
⎠ ⎠ ⎝ 4 4 ⎠ 4 4 ⎠ ⎝ ⎝ ⎝4 4 ⎠ ⎠ ⎝4 4 ⎠ 4 4
⎞ ⎛ ⎛ 2⎞ ⎞⎫
ⅈ⎟ ⎜ ⎜ ⎛√6 √2⋅ⅈ⎞ ⎟ ⎛√6 √2⋅ⅈ⎞ √6 √2⋅ⅈ⎟⎪
─⎟, ⎜-⎜-1 + 2⋅⎜── + ────⎟ ⎟⋅⎜── + ────⎟, ── + ────⎟⎬
⎠ ⎝ ⎝ ⎝4 4 ⎠ ⎠ ⎝4 4 ⎠ 4 4 ⎠⎪
⎭
from sympy.utilities.lambdify import lambdify
from scipy.optimize import fsolve
def lambdify_equations_for_solving_c(equations, c_sym, K_sym, K_val):
f = []
equations_subbed = substitute_sym(equations, K_sym, K_val)
for eq in equations_subbed:
# Note that I reformulate equation into an expression here
# for determining the roots.
f.append(lambdify(c_sym, eq.args[0] - eq.args[1]))
return f
N = 4
equations, c_sym, K_sym = not_choleskys_decomposition(N)
K_vals = [0.1, 0, 0.1, 0, 0]
f = lambdify_equations_for_solving_c(equations, c_sym, K_sym, K_vals)
def fsolve_friendly(p):
return [_f(*p) for _f in f]
c_sol = fsolve(fsolve_friendly, x0=(0, 0.1, -1.1, 0, 0))
/home/ggarrett/anaconda3/envs/sigh/lib/python3.7/site-packages/scipy/optimize/minpack.py:162: RuntimeWarning: The iteration is not making good progress, as measured by the
improvement from the last five Jacobian evaluations.
warnings.warn(msg, RuntimeWarning)