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[英]Sympy: How to compute Lie derivative of matrix with respect to a vector field
[英]How to minimize sympy partial derivative expression after matrix/vector multiplication?
我正在使用 sympy 計算矩陣/向量乘法的偏導數(參見下面的代碼),但結果表達式很復雜,我想知道是否有辦法簡化它。
from sympy import symbols, MatrixSymbol, diff, Matrix, symarray, expand, factor, simplify
import numpy as np
Ca = Matrix(symarray('Ca', (2, 4)))
Cb = Matrix(symarray('Cb', (2, 4)))
Cc = Matrix(symarray('Cc', (2, 4)))
qi = Matrix(symarray('qi', (4, 1)))
qj = Matrix(symarray('qj', (4, 1)))
R90 = Matrix(symarray('R90', (2, 2)))
u = (Ca*qi - Cb*qj)
v = (Cc*qj - Cb*qj)
u_tilde = R90*u
# Equation
constr_eq = (u_tilde).T*v
# Partial derivatives
u_diff = u.diff(qi)
constr_eq_diff_wrt_qi = constr_eq.diff(qi)
constr_eq_diff_wrt_qj = constr_eq.diff(qj)
output
print('u_diff : ', u_diff)
是
[[[[Ca_0_0], [Ca_1_0]]], [[[Ca_0_1], [Ca_1_1]]], [[[Ca_0_2], [Ca_1_2]]], [[[Ca_0_3], [Ca_1_3]]]]
我想只是
Ca
因此 output 的
'constr_eq_diff_wrt_qi'
和
constr_eq_diff_wrt_qj
是不可讀的。
謝謝你。 伊沃
基於@asmeurer 幫助這里是一個工作示例:
from sympy import MatrixSymbol, diff
# Initialize simbolic matrices
Ca = MatrixSymbol('Ca', 2, 4)
Cb = MatrixSymbol('Cb', 2, 4)
Cc = MatrixSymbol('Cc', 2, 4)
qi = MatrixSymbol('qi', 4, 1)
qj = MatrixSymbol('qj', 4, 1)
R90 = MatrixSymbol('R90', 2, 2)
# Initialize vectors
u = (Ca*qi - Cb*qj)
v = (Cc*qj - Cb*qj)
u_tilde = R90*u
# Constraint Equation
constr_eq = (u_tilde).T*v
# Partial derivatives
generalized_coordinates = [qi, qj]
# Get contraint equations partial derivatives
for _ in generalized_coordinates:
print('jacobian wrt ' + str(_) + ' : ', constr_eq.diff(_))
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