计算椭圆内的点

Question

我试图计算每个椭圆环内的给定数据点:

问题是我有一个函数来检查:所以对于每个椭圆,要确定一个点是否在其中,必须计算三个输入:

def get_focal_point(r1,r2,center_x):
    # f = square root of r1-squared - r2-squared
    focal_dist = sqrt((r1**2) - (r2**2))
    f1_x = center_x - focal_dist
    f2_x = center_x + focal_dist
    return f1_x, f2_x

def get_distance(f1,f2,center_y,t_x,t_y):
    d1 = sqrt(((f1-t_x)**2) + ((center_y - t_y)**2)) 
    d2 = sqrt(((f2-t_x)**2) + ((center_y - t_y)**2))
    return d1,d2

def in_ellipse(major_ax,d1,d2):
    if (d1+d2) <= 2*major_ax:
        return True
    else:
        return False

现在我正在通过以下方式检查它是否在椭圆形中:

for i in range(len(data.latitude)):
    t_x = data.latitude[i] 
    t_y = data.longitude[i] 
    d1,d2 = get_distance(f1,f2,center_y,t_x,t_y)
    d1_array.append(d1)
    d2_array.append(d2)
    if in_ellipse(major_ax,d1,d2) == True:
        core_count += 1
        # if the point is not in core ellipse 
        # check the next ring up
    else:
        for i in range(loop):
            .....

但是我必须计算外部循环的每对焦点.是否有更有效和/或更聪明的方法来做到这一点？

Answer 1

这可能与您正在做的类似.我只是想看看f(x,y)= x ^ 2/r1 ^ 2 + y ^ 2/r2 ^ 2 = 1.

当f(x,y)大于1时,点x,y在椭圆之外.当它较小时,它在椭圆内.当f(x,y)小于1时,我遍历每个椭圆以找到一个椭圆.

该代码也没有考虑以原点为中心的椭圆.包含此功能只是一个小小的改动.

import matplotlib.pyplot as plt
import matplotlib.patches as patches
import numpy as np

def inWhichEllipse(x,y,rads):
    '''
    With a list of (r1,r2) pairs, rads, return the index of the pair in which
    the point x,y resides. Return None as the index if it is outside all 
    Ellipses.
    '''
    xx = x*x
    yy = y*y

    count = 0
    ithEllipse =0
    while True:
        rx,ry = rads[count]
        ellips = xx/(rx*rx)+yy/(ry*ry)
        if ellips < 1:
            ithEllipse = count
            break
        count+=1
        if count >= len(rads):
            ithEllipse = None
            break

    return ithEllipse

rads = zip(np.arange(.5,10,.5),np.arange(.125,2.5,.25))

fig = plt.figure()
ax = fig.add_subplot(111)
ax.set_xlim(-15,15)
ax.set_ylim(-15,15)

# plot Ellipses
for rx,ry in rads:
    ellipse = patches.Ellipse((0,0),rx*2,ry*2,fc='none',ec='red')    
    ax.add_patch(ellipse)

x=3.0
y=1.0
idx = inWhichEllipse(x,y,rads)
rx,ry = rads[idx]
ellipse = patches.Ellipse((0,0),rx*2,ry*2,fc='none',ec='blue')    
ax.add_patch(ellipse)

if idx != None:
    circle = patches.Circle((x,y),.1)
    ax.add_patch(circle)

plt.show()

此代码生成下图:

请记住,这只是一个起点.首先,您可以更改inWhichEllipse为接受r1和r2的平方列表,即(r1*r1,r2*r2)对,这将进一步减少计算量.