在numpy mesgrid上評估sympy lambdify的結果

Question

我想在numpy的mgrid上評估sympy.lambdify的輸出。我試過以下內容：

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]) 

但我得到一個錯誤，說矩陣必須是二維的。

Answer 1

我找到了解決方案。

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]