编辑:我找到了一个有效的解决方案,但我仍然希望能够对此处发生的事情做出更多解释:
from scipy import optimize
from sympy import lambdify, DeferredVector
v = DeferredVector('v')
f_expr = (v[0] ** 2 + v[1] ** 2)
f = lambdify(v, f_expr, 'numpy')
zero = optimize.root(f, x0=[0, 0], method='krylov')
zero
原始问题:
下面我们有由M
和f1(x1, x2)
组成的矩阵f2(x1, x2)
。我希望在x1
时知道x2
和M = [f1, f2] = [0, 0]
的值。
以下代码正在运行,减去根查找行,这些行已被注释掉。
import numpy as np
import sympy as sp
import matplotlib.pyplot as plt
from scipy import optimize
from sympy import init_printing, symbols, lambdify, Matrix
from sympy import pi, exp, cos, sin
x1, x2 = symbols('x1 x2')
# Expressions
f1_expr = sin(4 * pi * x1 * x2) - 2 * x2 - x1
f2_expr = ((4 * pi - 1) / (4 * pi)) * (exp(2 * x1) - exp(1)) + 4 * exp(1) * (x2 ** 2) - 2 * exp(1) * x1
# Expressions -> NumPy function
f1 = lambdify((x1, x2), f1_expr, 'numpy')
f2 = lambdify((x1, x2), f2_expr, 'numpy')
# Matrix and it's Jacobian
M_expr = Matrix([f1_expr, f2_expr])
M_jacob_expr = M_expr.jacobian([x1, x2])
# Matrix -> NumPy function
M = lambdify((x1, x2), M_expr, [{'ImmutableMatrix': np.array}, "numpy"])
M_jacob = lambdify((x1, x2), M_jacob_expr, [{'ImmutableMatrix': np.array}, "numpy"])
# Data points
x1pts = np.arange(-2, 3, 0.01)
x2pts = np.arange(-3, 3, 0.01)
xx1pts, xx2pts = np.meshgrid(x1pts, x2pts)
# Solve matrix for two heat maps
z1, z2 = M(xx1pts, xx2pts)
z1 = z1.reshape(z1.shape[1], z1.shape[2])
z2 = z2.reshape(z2.shape[1], z2.shape[2])
# All of these commented lines throw errors.
# Find roots with SymPy
#zero1 = sp.mpmath.findroot(f1_expr, x0=(-0.3, 0.05))
#zeros = sp.mpmath.findroot(M_expr, x0=(-0.3, 0.05))
# Can I use NumPy somehow?
#zero2 = optimize.newton_krylov(f2, (-0.3, 0.05))
#zeros = optimize.newton_krylov(M, (-0.3, 0.05))
################
# Plotting below
fig, (ax1, ax2) = plt.subplots(nrows=1, ncols=2)
im1 = ax1.contourf(x1pts, x2pts, z1)
im2 = ax2.contourf(x1pts, x2pts, z2)
ax1.set_xlabel('x1')
ax1.set_ylabel('x2')
ax1.set_title('f1(x1, x2)')
ax2.set_xlabel('x1')
ax2.set_ylabel('x2')
ax2.set_title('f1(x1, x2)')
fig.colorbar(im1)
plt.tight_layout()
plt.show()
plt.close(fig)
答案 0 :(得分:1)
由于以下几个原因,这是打破的:
scipy.optimize
经常坚持输入函数只包含一个参数。包装器可以避免这种情况,如此处所示。
SciPy包中的一些线性代数函数表明输入参数与输入函数的返回值匹配。
以下代码正在运行:
import numpy as np
import pandas as pd
import sympy as sp
import matplotlib.pyplot as plt
from scipy import optimize
from sympy import init_printing, symbols, lambdify, Matrix, latex
from sympy import pi, exp, log, sqrt, sin, cos, tan, sinh, cosh, tanh
from sympy.abc import a, b, c, x, y, z, r, w
x1, x2 = symbols('x1 x2')
# Expressions
f1_expr = sin(4 * pi * x1 * x2) - 2 * x2 - x1
f2_expr = ((4 * pi - 1) / (4 * pi)) * (exp(2 * x1) - exp(1)) + 4 * exp(1) * (x2 ** 2) - 2 * exp(1) * x1
f_expr = [f1_expr, f2_expr]
# Expressions -> NumPy function
f1 = lambdify((x1, x2), f1_expr, 'numpy')
f2 = lambdify((x1, x2), f2_expr, 'numpy')
f = np.array([f1, f2])
def _f(args):
return [f1(args[0], args[1]), f2(args[0], args[1])]
# Matrix and it's Jacobian
M_expr = Matrix([f1_expr, f2_expr])
M_jacob_expr = M_expr.jacobian([x1, x2])
# Matrix -> NumPy function
M = lambdify((x1, x2), M_expr, [{'ImmutableMatrix': np.array}, "numpy"])
M_jacob = lambdify((x1, x2), M_jacob_expr, [{'ImmutableMatrix': np.array}, "numpy"])
# Data points
x1pts = np.arange(-2, 3, 0.01)
x2pts = np.arange(-2, 3, 0.01)
xx1pts, xx2pts = np.meshgrid(x1pts, x2pts)
# Solved over ranges for plots
z1, z2 = M(xx1pts, xx2pts)
z1 = z1.reshape(z1.shape[1], z1.shape[2])
z2 = z2.reshape(z2.shape[1], z2.shape[2])
# Find roots
results = optimize.root(_f, x0=[-0.3, 0.05], method='Krylov')
zeros = results.get('x')
# First figure
fig, ax = plt.subplots(1)
im = ax.contourf(x1pts, x2pts, z1)
ax.scatter(zeros[0], zeros[1], linewidth=5, color='k')
ax.set_xlabel('x1')
ax.set_ylabel('x2')
ax.set_title('f1(x1, x2)')
plt.colorbar(im)
plt.tight_layout()
plt.set_cmap('seismic')
fig.savefig('img30_1.png')
plt.show()
plt.close(fig)
# Second figure
fig, ax = plt.subplots(1)
im = ax.contourf(x1pts, x2pts, z2)
ax.scatter(zeros[0], zeros[1], linewidth=5, color='white')
ax.set_xlabel('x1')
ax.set_ylabel('x2')
ax.set_title('f2(x1, x2)')
plt.colorbar(im)
plt.tight_layout()
plt.set_cmap('seismic')
fig.savefig('img30_2.png')
plt.show()
plt.close(fig)