问题描述
我想知道如何使用方程ay + bxy + cx + dy + e = x 绘制椭圆轨道吗?
I wonder how can I draw elliptical orbit by using the equation ay + bxy + cx + dy + e = x ?
我首先确定了a，b，c，d，e常数，现在我假设通过给定x值，我将获得y，这将给我我想要的图形，但我不能通过使用matplotlib来完成.
I have first determined the a,b,c,d,e constants and now I assume by giving x values I will obtain y and this will give me the graph I want but I couldn't do it by using matplotlib.
如果您能帮助我，我将不胜感激！
I would really appreciate, if you could help me!
我在这里添加了代码.
from numpy import linalg
from numpy import linspace
import numpy as np
from numpy import meshgrid
import random
import matplotlib.pyplot as plt
from scipy import optimize
x = [1.02, 0.95, 0.87, 0.77, 0.67, 0.56, 0.44, 0.30, 0.16, 0.01]
y = [0.39, 0.32, 0.27, 0.22, 0.18, 0.15, 0.13, 0.12, 0.12, 0.15]
my_list = [] #It is the main list.
b = [0] * len(x) # That is the list that contains the results that are given as x^2 from the equation.
def fxn(): # That is the function that solves the given equation to find each parameter.
global my_list
global b
for z in range(len(x)):
w = [0] * 5
w[0] = y[z] ** 2
w[1] = x[z] * y[z]
w[2] = x[z]
w[3] = y[z]
w[4] = 1
my_list.append(w)
b[z] = x[z] ** 2
t = linalg.lstsq(my_list, b)[0]
print 'List of list representation is', my_list
print 'x^2, the result of the given equation is', b
print '\nThe list that contains the parameters is', t
fxn()
t = linalg.lstsq(my_list, b)[0]
print '\nThe constant a is', t[0]
print 'The constant b is', t[1]
print 'The constant c is', t[2]
print 'The constant d is', t[3]
print 'The constant e is', t[4]
这是常量值:
a = -4.10267300566
b = 1.10642410023
c = 0.39735696603
d = 3.05101004127
e = -0.370426134994
推荐答案
可以解决y作为x的函数的问题
The problem can be solved for y as a function of x
要注意的是，每个有效x都有y的2个值，并且在椭圆跨度x范围之外没有(或虚构的)y解
The catch is that there are 2 values of y for every valid x, and no (or imaginary) y solutions outside the range of x the ellipse spans
下面是3.5代码，sympy 1.0应该可以，但可以打印，列表组合可能无法向后兼容2.x
below is 3.5 code, sympy 1.0 should be fine but print, list comps may not be backwards compatable to 2.x
from numpy import linalg
from numpy import linspace
import numpy as np
from numpy import meshgrid
import random
import matplotlib.pyplot as plt
from scipy import optimize
from sympy import *
xs = [1.02, 0.95, 0.87, 0.77, 0.67, 0.56, 0.44, 0.30, 0.16, 0.01]
ys = [0.39, 0.32, 0.27, 0.22, 0.18, 0.15, 0.13, 0.12, 0.12, 0.15]
b = [i ** 2 for i in xs] # That is the list that contains the results that are given as x^2 from the equation.
def fxn(x, y): # That is the function that solves the given equation to find each parameter.
my_list = [] #It is the main list.
for z in range(len(x)):
w = [0] * 5
w[0] = y[z] ** 2
w[1] = x[z] * y[z]
w[2] = x[z]
w[3] = y[z]
w[4] = 1
my_list.append(w)
return my_list
t = linalg.lstsq(fxn(xs, ys), b)
def ysolv(coeffs):
x,y,a,b,c,d,e = symbols('x y a b c d e')
ellipse = a*y**2 + b*x*y + c*x + d*y + e - x**2
y_sols = solve(ellipse, y)
print(*y_sols, sep='\n')
num_coefs = [(a, f) for a, f in (zip([a,b,c,d,e], coeffs))]
y_solsf0 = y_sols[0].subs(num_coefs)
y_solsf1 = y_sols[1].subs(num_coefs)
f0 = lambdify([x], y_solsf0)
f1 = lambdify([x], y_solsf1)
return f0, f1
f0, f1 = ysolv(t[0])
y0 = [f0(x) for x in xs]
y1 = [f1(x) for x in xs]
plt.scatter(xs, ys)
plt.scatter(xs, y0, s=100, color = 'red', marker='+')
plt.scatter(xs, y1, s=100, color = 'green', marker='+')
plt.show()
在Spyder中运行上述命令时:
when the above is ran in Spyder:
runfile('C:/Users/john/mypy/mySE_answers/ellipse.py', wdir='C:/Users/john/mypy/mySE_answers')
(-b*x - d + sqrt(-4*a*c*x - 4*a*e + 4*a*x**2 + b**2*x**2 + 2*b*d*x + d**2))/(2*a)
-(b*x + d + sqrt(-4*a*c*x - 4*a*e + 4*a*x**2 + b**2*x**2 + 2*b*d*x + d**2))/(2*a)
为y值生成的函数并非在所有地方都有效:
The generated functions for the y values aren't valid everywhere:
f0(0.1), f1(0.1)
Out[5]: (0.12952825130864626, 0.6411040771593166)
f0(2)
Traceback (most recent call last):
File "<ipython-input-6-9ce260237dcd>", line 1, in <module>
f0(2)
File "<string>", line 1, in <lambda>
ValueError: math domain error
In [7]:
域错误将需要尝试/执行才能感觉"到有效的x范围或更多数学运算
The domain error would require a try/execpt to "feel out" the valid x range or some more math
像下面的try/except一样:(编辑为关闭"图纸，重新注释)
like the try/except below: (Edited to "close" drawing re comment )
def feeloutXrange(f, midx, endx):
fxs = []
x = midx
while True:
try: f(x)
except:
break
fxs.append(x)
x += (endx - midx)/100
return fxs
midx = (min(xs) + max(xs))/2
xpos = feeloutXrange(f0, midx, max(xs))
xnegs = feeloutXrange(f0, midx, min(xs))
xs_ellipse = xnegs[::-1] + xpos[1:]
y0s = [f0(x) for x in xs_ellipse]
y1s = [f1(x) for x in xs_ellipse]
ys_ellipse = y0s + y1s[::-1] + [y0s[0]] # add y start point to end to close drawing
xs_ellipse = xs_ellipse + xs_ellipse[::-1] + [xs_ellipse[0]] # added x start point
plt.scatter(xs, ys)
plt.scatter(xs, y0, s=100, color = 'red', marker='+')
plt.scatter(xs, y1, s=100, color = 'green', marker='+')
plt.plot(xs_ellipse, ys_ellipse)
plt.show()
在椭圆点列表的末尾添加重复的起点以关闭绘图
added duplicate start point to the end of ellipse point lists to close the plot figure
ys_ellipse = y0s + y1s[::-1] + [y0s[0]] # add y start point to end to close drawing
xs_ellipse = xs_ellipse + xs_ellipse[::-1] + [xs_ellipse[0]] # added x start point
这篇关于在Python中绘制椭圆形轨道(使用numpy，matplotlib)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！