问题描述
1st 我需要得到两个最长线长度的两个方程
1st i need to get two equation of two longest line length
- 我将长度与 eq 放在列表中，如下所示 [( length 1 , eq 1 ) ,.....]
- 反向排序列表
- 求两条最长直线的两个方程
运行以下代码时
from sympy import *
import math
class shape():
def __init__(self,T):
self.T=T
self.Ax = self.T[0][0]
self.Ay = self.T[0][1]
self.Bx = self.T[1][0]
self.By = self.T[1][1]
self.Cx = self.T[2][0]
self.Cy = self.T[2][1]
#Triangle:
self.C_dashx = 0.5 * (self.Ax + self.Bx)
self.C_dashy = 0.5 * (self.Ay + self.By)
self.B_dashx = 0.5 * (self.Ax + self.Cx)
self.B_dashy = 0.5 * (self.Ay + self.Cy)
self.A_dashx = 0.5 * (self.Bx + self.Cx)
self.A_dashy = 0.5 * (self.By + self.Cy)
if (self.Ax - self.A_dashx) == 0:
self.m_AA =0
else:
self.m_AA = (self.Ay - self.A_dashy) / (self.Ax - self.A_dashx)
if (self.Bx - self.B_dashx) == 0:
self.m_BB =0
else:
self.m_BB = (self.By - self.B_dashy) / (self.Bx - self.B_dashx)
if (self.Bx - self.B_dashx) == 0:
self.m_CC =0
else:
self.m_CC = (self.Cy - self.C_dashy) / (self.Bx - self.B_dashx)
def triangle_intersection(self):
self.x , self.y = symbols('x y')
self.eq1=Eq(self.m_AA*(self.x-self.Ax)+self.Ay-self.y,0)
self.eq2=Eq(self.m_BB*(self.x-self.Bx)+self.By-self.y,0)
self.solve_equation=solve([self.eq1,self.eq2],[self.x,self.y])
self.Zx=(self.solve_equation[self.x]).factor()
self.Zy=(self.solve_equation[self.y]).factor()
print(self.Zx,self.Zy)
def square_intersection(self):
self.Dx = self.T[3][0]
self.Dy = self.T[3][0]
# Line Solpe:
if (self.Bx - self.Ax) == 0:
self.m_AB = 0
else:
self.m_AB = (self.By - self.Ay) / (self.Bx - self.Ax)
if (self.Dx - self.Ax) == 0:
self.m_AD = 0
else:
self.m_AD = (self.Dy - self.Ay) / (self.Dx - self.Ax)
if (self.Cx - self.Bx) == 0:
self.m_BC = 0
else:
self.m_BC = (self.Cy - self.By) / (self.Cx - self.Bx)
if (self.Dx - self.Bx) == 0:
self.m_BD = 0
else:
self.m_BD = (self.Dy - self.By) / (self.Dx - self.Bx)
if (self.Dx - self.Cx) == 0:
self.m_CD = 0
else:
self.m_CD = (self.Dy - self.Cy) / (self.Dx - self.Cx)
if (self.Cx - self.Ax) == 0:
self.m_AC = 0
else:
self.m_AC = (self.Cy - self.Ay) / (self.Cx - self.Ax)
# Equation:
#### WHAT IS PROBLEM HERE###########
self.z, self.t = symbols('z t',positive=True)
self.eq_AB = Eq(self.m_AB * (self.z - self.Ax) + self.Ay - self.t,0)
self.eq_AC = Eq(self.m_AC * (self.z - self.Ax) + self.Ay - self.t,0)
self.eq_AD = Eq(self.m_AD * (self.z - self.Ax) + self.Ay - self.t,0)
self.eq_BC = Eq(self.m_BC * (self.z - self.Bx) + self.By - self.t,0)
self.eq_BD = Eq(self.m_BD * (self.z - self.Bx) + self.By - self.t,0)
self.eq_CD = Eq(self.m_CD * (self.z - self.Cx) + self.Cy - self.t,0)
self.length_AB = math.sqrt((self.Bx - self.Ax)**2 + (self.By - self.Ay)**2)
self.length_AC = math.sqrt((self.Cx - self.Ax)**2 + (self.Cy - self.Ay)**2)
self.length_AD = math.sqrt((self.Dx - self.Ax)**2 + (self.Dy - self.Ay)**2)
self.length_BC = math.sqrt((self.Cx - self.Bx)**2 + (self.Cy - self.By)**2)
self.length_BD = math.sqrt((self.Dx - self.Bx)**2 + (self.Dy - self.By)**2)
self.length_CD = math.sqrt((self.Dx - self.Cx)**2 + (self.Dy - self.Cy)**2)
#### WHAT IS PROBLEM HERE###########
self.lenghts = [(self.length_AB, self.eq_AB),(self.length_AC, self.eq_AC),(self.length_AD, self.eq_AD),(self.length_BC, self.eq_BC)
, (self.length_BD, self.eq_BD), (self.length_CD, self.eq_CD)]
self.newlenghts = sorted(self.lenghts,reverse=True)
self.max1=self.newlenghts[0][1]
self.max2=self.newlenghts[1][1]
self.solve_equation_sq = solve([self.max1, self.max2], [self.z, self.t])
print(self.solve_equation_sq)
self.Rx = (self.solve_equation_sq[self.z]).factor()
self.Ry = (self.solve_equation_sq[self.t]).factor()
print(self.Rx, self.Ry)
test=[(0,0),(4,0),(4,4),(0,4)]
a1=shape(test)
a1.triangle_intersection()
a1.square_intersection()
我收到以下错误
2.66666666666667 1.33333333333333{z: t}回溯(最近一次调用最后一次):文件C:/Users/xxxxx/.PyCharmCE2019.3/config/scratches/programme logic function.py"，第127行，
2.66666666666667 1.33333333333333{z: t}Traceback (most recent call last): File "C:/Users/xxxxx/.PyCharmCE2019.3/config/scratches/programme logic function.py", line 127, in
a1.square_intersection()
文件C:/Users/xxxxx/.PyCharmCE2019.3/config/scratches/programme logic function.py"，第119行，在square_intersection中
File "C:/Users/xxxxx/.PyCharmCE2019.3/config/scratches/programme logic function.py", line 119, in square_intersection
self.Ry = (self.solve_equation_sq[self.t]).factor()
键错误:t
推荐答案
如果你打印出你想解的方程，你会发现它们是相同的，所以返回的唯一解是 {z:t} 不是 {z:smthng, t:smthngelse}.因此，当您请求 t 的值时，它会告诉您解决方案字典中没有 t.现在你必须回去看看你是否正确计算了要求解的方程.
If you print out the equations you are trying to solve you will see that they are the same so the only solution returned is {z:t} not {z:smthng, t:smthngelse}. So when you request the value of t it tells you that there is no t in the solution dictionary. Now you have to go back and see that you properly computed that equations that were to be solved.
还要注意，SymPy 具有 Polygon/Triangle 对象，并且可能已经有一种方法可以回答您所提出的问题.
Note, too, that SymPy has the Polygon/Triangle objects and might already have a method to answer the question that you are asking.
这篇关于求解两个方程时出现 KeyError的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！