在numpy mesgrid上评估sympy lambdify的结果 [英] Evaluating the result of sympy lambdify on a numpy mesgrid

问题描述

我想在一个numpy mgrid上评估sympy.lambdify的输出.我尝试了以下方法:

I would like to evaluate the output of sympy.lambdify on a numpy mgrid. I tried the following:

import sympy as sp
import numpy as np


theta, v = sp.symbols("theta v")
coeff = (-sp.sin(theta/2)*sp.sin(2*v) + sp.sin(v)*sp.cos(theta/2) + 3)
kb = sp.Matrix([[coeff*sp.cos(theta),
                 coeff*sp.sin(theta),
                 sp.sin(theta/2)*sp.sin(v) + sp.sin(2*v)*sp.cos(theta/2)]])
f = sp.lambdify((theta, v), kb, modules='numpy')
f(*np.mgrid[0:2*np.pi:101j, 0:2*np.pi:101j])

但是我得到一个错误，说矩阵必须是二维的.

but I get an error saying the matrix must be 2-dimensional.

推荐答案

我找到了解决方案.

import sympy as sp
import numpy as np


theta, v = sp.symbols("theta v")
coeff = (-sp.sin(theta/2)*sp.sin(2*v) + sp.sin(v)*sp.cos(theta/2) + 3)
kb = sp.Matrix([[coeff*sp.cos(theta),
                 coeff*sp.sin(theta),
                 sp.sin(theta/2)*sp.sin(v) +
                 sp.sin(2*v)*sp.cos(theta/2)]])

f = sp.lambdify((theta, v), kb, [{'ImmutableMatrix': np.array}, "numpy"])
x, y = np.mgrid[0:2*np.pi:101j, 0:2*np.pi:101j]
g = f(x, y)
x, y, z = g[0]

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