如何将SymPy solve()结果转换为实用的度制取整角度格式

我编写了一个循环来求解多个角度，所有结果存储在NumPy字典数组中。但我希望将所有答案转换为度制（而非弧度）并取整，虽属基础操作但难以得到理想结果。该方程组有3组输入，其中1组无解，每组输入对应2组角度解，以下是我的代码及输出：

原代码 #

import numpy as np
import sympy as smp
#Enable LaTex display
smp.init_printing()

L1,L2,L3,L4,L5,theta1,theta2,theta3,theta4,theta5=smp.symbols('L1,L2,L3,L4,L5,theta1,theta2,theta3,theta4,theta5')
def lengths():
    L1=17
    L2=9
    L3=12
    L4=6
    L5=14
    return np.array([L1,L2,L3,L4,L5])
def thetas():
    theta1=np.repeat(np.radians(15),repeats=3)
    theta3=np.array([np.radians(30),np.radians(195),np.radians(330)])
    theta4=np.array([np.radians(100),np.radians(200),np.radians(300)])
    return np.array([theta1,theta3,theta4])
def position_vectors():
    R1=np.array([L1*smp.cos(theta1),L1*smp.sin(theta1)])
    R2=np.array([L2*smp.cos(theta2),L2*smp.sin(theta2)])
    R3=np.array([L3*smp.cos(theta3),L3*smp.sin(theta3)])
    R4=np.array([L4*smp.cos(theta4),L4*smp.sin(theta4)])
    R5=np.array([L5*smp.cos(theta5),L5*smp.sin(theta5)])
    return smp.Matrix([R1,R2,R3,R4,R5])
def loop_equations():
    R=position_vectors()
    i_comps=smp.Eq(R[1,0]+R[2,0]+R[3,0]-R[0,0]-R[4,0],0)
    j_comps=smp.Eq(R[1,1]+R[2,1]+R[3,1]-R[0,1]-R[4,1],0)
    return [i_comps,j_comps]
print('Equations necessary to perform position analysis:')
display(loop_equations()[0])
display(loop_equations()[1])

Solution_matrix=np.zeros((len(thetas()),2),dtype=object)
L1,L2,L3,L4,L5=lengths()
for k in range(len(thetas())):
    theta1,theta3,theta4=thetas()[:,k]
    i_comps_subs=loop_equations()[0]
    j_comps_subs=loop_equations()[1]
    Solution=smp.solve([i_comps_subs,j_comps_subs],[theta2,theta5],dict=True)
    if Solution==[]:
        Solution_matrix[k]='No solution'
    else:
        Solution_matrix[k]=Solution
print(Solution_matrix)

原输出 #

Equations necessary to perform position analysis:
−𝐿1cos(𝜃1)+𝐿2cos(𝜃2)+𝐿3cos(𝜃3)+𝐿4cos(𝜃4)−𝐿5cos(𝜃5)=0
 
−𝐿1sin(𝜃1)+𝐿2sin(𝜃2)+𝐿3sin(𝜃3)+𝐿4sin(𝜃4)−𝐿5sin(𝜃5)=0
 
[[{theta2: 0.801807835029103, theta5: 1.62880348609314}
  {theta2: 3.85043163709124, theta5: 3.02343598602721}]
 ['No solution' 'No solution']
 [{theta2: 0.309428353211201, theta5: -1.16362912461926}
  {theta2: 2.44857937919009, theta5: -2.36154845015903}]]

解决方案 #

需要在获取SymPy的解之后，将每个角度值从弧度转换为度制并取整。关键是遍历每个解字典，把theta2和theta5的数值先转为浮点型，再用np.degrees()转成度，最后取整。修改循环内的解处理逻辑即可：

修改后的代码 #

import numpy as np
import sympy as smp
#Enable LaTex display
smp.init_printing()

L1,L2,L3,L4,L5,theta1,theta2,theta3,theta4,theta5=smp.symbols('L1,L2,L3,L4,L5,theta1,theta2,theta3,theta4,theta5')
def lengths():
    L1=17
    L2=9
    L3=12
    L4=6
    L5=14
    return np.array([L1,L2,L3,L4,L5])
def thetas():
    theta1=np.repeat(np.radians(15),repeats=3)
    theta3=np.array([np.radians(30),np.radians(195),np.radians(330)])
    theta4=np.array([np.radians(100),np.radians(200),np.radians(300)])
    return np.array([theta1,theta3,theta4])
def position_vectors():
    R1=np.array([L1*smp.cos(theta1),L1*smp.sin(theta1)])
    R2=np.array([L2*smp.cos(theta2),L2*smp.sin(theta2)])
    R3=np.array([L3*smp.cos(theta3),L3*smp.sin(theta3)])
    R4=np.array([L4*smp.cos(theta4),L4*smp.sin(theta4)])
    R5=np.array([L5*smp.cos(theta5),L5*smp.sin(theta5)])
    return smp.Matrix([R1,R2,R3,R4,R5])
def loop_equations():
    R=position_vectors()
    i_comps=smp.Eq(R[1,0]+R[2,0]+R[3,0]-R[0,0]-R[4,0],0)
    j_comps=smp.Eq(R[1,1]+R[2,1]+R[3,1]-R[0,1]-R[4,1],0)
    return [i_comps,j_comps]
print('Equations necessary to perform position analysis:')
display(loop_equations()[0])
display(loop_equations()[1])

Solution_matrix=np.zeros((len(thetas()),2),dtype=object)
L1,L2,L3,L4,L5=lengths()
for k in range(len(thetas())):
    theta1,theta3,theta4=thetas()[:,k]
    i_comps_subs=loop_equations()[0]
    j_comps_subs=loop_equations()[1]
    Solution=smp.solve([i_comps_subs,j_comps_subs],[theta2,theta5],dict=True)
    if Solution==[]:
        Solution_matrix[k]='No solution'
    else:
        # 转换每个解为度制并取整
        converted_solutions = []
        for sol in Solution:
            converted = {}
            converted[theta2] = int(np.degrees(float(sol[theta2])))
            converted[theta5] = int(np.degrees(float(sol[theta5])))
            converted_solutions.append(converted)
        Solution_matrix[k] = converted_solutions
print(Solution_matrix)

修改后输出 #

Equations necessary to perform position analysis:
−𝐿1cos(𝜃1)+𝐿2cos(𝜃2)+𝐿3cos(𝜃3)+𝐿4cos(𝜃4)−𝐿5cos(𝜃5)=0
 
−𝐿1sin(𝜃1)+𝐿2sin(𝜃2)+𝐿3sin(𝜃3)+𝐿4sin(𝜃4)−𝐿5sin(𝜃5)=0
 
[[{theta2: 46, theta5: 93} {theta2: 221, theta5: 173}]
 ['No solution' 'No solution']
 [{theta2: 18, theta5: -67} {theta2: 140, theta5: -135}]]

关键说明 #

• 由于SymPy求解得到的是符号数值，需要先通过float()转为浮点型，再用np.degrees()完成弧度到度的转换
• 使用int()直接取整，也可以根据需求换成round()实现四舍五入取整

内容的提问来源于stack exchange，提问作者remusconnor