请写一个python示例，已知五自由度的机械臂求其运动学逆解，其改进dh参数分别为a0=0，a1=35，a2=503，a3=383，a4=427。alpha0=0，alpha1=-pi/2，alpha2=0，alpha3=0，alpha4=0。d1=493，d2=0，d3=0，d4=0，d5=0

好的，以下是一个简单的Python示例代码来计算五自由度机械臂的运动学逆解。我们使用 SymPy 库来进行符号计算。

from sympy import symbols, cos, sin, pi, simplify
from sympy.matrices import Matrix

# 定义符号变量
q1, q2, q3, q4, q5 = symbols('q1:6')
d1, d2, d3, d4, d5 = symbols('d1:6')
a0, a1, a2, a3, a4 = symbols('a0:5')
alpha0, alpha1, alpha2, alpha3, alpha4 = symbols('alpha0:5')

# DH 参数
s = {alpha0: 0,   a0: 0,    d1: 493,  q1: q1,
     alpha1: -pi/2, a1: 35,   d2: 0,    q2: q2-pi/2,
     alpha2: 0,   a2: 503,  d3: 0,    q3: q3,
     alpha3: 0,   a3: 383,  d4: 0,    q4: q4,
     alpha4: 0,   a4: 427,  d5: 0,    q5: 0}

# 处理转换矩阵
def transform(alpha, a, d, q):
    T = Matrix([[            cos(q),           -sin(q),           0,             a],
                [ sin(q)*cos(alpha), cos(q)*cos(alpha), -sin(alpha), -sin(alpha)*d],
                [ sin(q)*sin(alpha), cos(q)*sin(alpha),  cos(alpha),  cos(alpha)*d],
                [                 0,                 0,           0,             1]])
    return T

# 运动学正解
T0_1 = transform(alpha0, a0, d1, q1).subs(s)
T1_2 = transform(alpha1, a1, d2, q2).subs(s)
T2_3 = transform(alpha2, a2, d3, q3).subs(s)
T3_4 = transform(alpha3, a3, d4, q4).subs(s)
T4_5 = transform(alpha4, a4, d5, q5).subs(s)

T0_5 = simplify(T0_1 * T1_2 * T2_3 * T3_4 * T4_5)

# 计算逆解
# 末端执行器的位置和姿态
px = T0_5[0, 3]
py = T0_5[1, 3]
pz = T0_5[2, 3]
nx = T0_5[0, 2]
ny = T0_5[1, 2]
nz = T0_5[2, 2]
ox = T0_5[0, 0]
oy = T0_5[1, 0]
oz = T0_5[2, 0]
ax = ny*oz - nz*oy
ay = nz*ox - nx*oz
az = nx*oy - ny*ox

# 逆解
q1_sol = atan2(py, px)
q3_sol = acos((px**2 + py**2 + pz**2 - a2**2 - a3**2 - a4**2) / (2*a3*a4))
q2_sol = atan2(pz-a1, sqrt(px**2 + py**2)) - atan2(a4*sin(q3_sol), a3+a4*cos(q3_sol))
q4_sol = atan2(nz*sin(q1_sol) - nz*cos(q1_sol)*cos(q2_sol) + ny*cos(q1_sol)*sin(q2_sol) - nx*sin(q1_sol)*sin(q2_sol), -oz*sin(q1_sol) + oz*cos(q1_sol)*cos(q2_sol) - oy*cos(q1_sol)*sin(q2_sol) + ox*sin(q1_sol)*sin(q2_sol))
q5_sol = atan2(-ax*sin(q1_sol) + ax*cos(q1_sol)*cos(q2_sol) - ay*cos(q1_sol)*sin(q2_sol) + az*sin(q1_sol)*sin(q2_sol), ax*cos(q1_sol)*sin(q2_sol) - ay*sin(q1_sol)*sin(q2_sol) - az*cos(q1_sol)*cos(q2_sol))

# 打印结果
print("q1:", q1_sol.evalf())
print("q2:", q2_sol.evalf())
print("q3:", q3_sol.evalf())
print("q4:", q4_sol.evalf())
print("q5:", q5_sol.evalf())

请注意，这只是一个简单的示例代码，可能需要根据你的具体情况进行调整。