找到椭圆和直线之间的交点

时间:2013-08-20 04:18:32

标签: python numpy shapely

我试图找到一个elipse和一条线之间的交点,但似乎无法这样做。我尝试了两种方法,一种是通过尝试找到LineString和LinearRing之间的交集,如下面的代码中那样,但是没有得到任何可用的值。其中一个问题是,椭圆总是偏离中心并且处于小角度或高角度

  # -*- coding: utf-8 -*-
"""
Created on Mon Aug 19 17:38:55 2013

@author: adudchenko
"""
from pylab import *
import numpy as np
from shapely.geometry.polygon import LinearRing
from shapely.geometry import LineString
def ellipse_polyline(ellipses, n=100):
    t = np.linspace(0, 2*np.pi, n, endpoint=False)
    st = np.sin(t)
    ct = np.cos(t)
    result = []
    for x0, y0, a, b, angle in ellipses:
        angle = np.deg2rad(angle)
        sa = np.sin(angle)
        ca = np.cos(angle)
        p = np.empty((n, 2))
        p[:, 0] = x0 + a * ca * ct - b * sa * st
        p[:, 1] = y0 + a * sa * ct + b * ca * st
        result.append(p)
    return result

def intersections(a, line):
    ea = LinearRing(a)
    eb = LinearRing(b)
    mp = ea.intersection(eb)
    print mp
    x = [p.x for p in mp]
    y = [p.y for p in mp]
    return x, y

ellipses = [(1, 1, 2, 1, 45), (2, 0.5, 5, 1.5, -30)]
a, b = ellipse_polyline(ellipses)
line=LineString([[0,0],[4,4]])
x, y = intersections(a, line)
figure()
plot(x, y, "o")
plot(a[:,0], a[:,1])
plot(b[:,0], b[:,1])
show()

我也尝试过使用fsolve,如下例所示,但它找到了错误的交叉点(实际上是一个错误的点。

from pylab import *
from scipy.optimize import fsolve
import numpy as np

def ellipse_polyline(ellipses, n=100):
    t = np.linspace(0, 2*np.pi, n, endpoint=False)
    st = np.sin(t)
    ct = np.cos(t)
    result = []
    for x0, y0, a, b, angle in ellipses:
        angle = np.deg2rad(angle)
        sa = np.sin(angle)
        ca = np.cos(angle)
        p = np.empty((n, 2))
        p[:, 0] = x0 + a * ca * np.cos(t) - b * sa * np.sin(t)
        p[:, 1] = y0 + a * sa * np.cos(t) + b * ca * np.sin(t)
        result.append(p)
    return result
def ellipse_line(txy):
    t,x,y=txy
    x0, y0, a, b, angle,m,lb=1, 1, 2, 1, 45,0.5,0
    sa = np.sin(angle)
    ca = np.cos(angle)
    return (x0 + a * ca * np.cos(t) - b * sa * np.sin(t)-x,y0 + a * sa * np.cos(t) + b * ca * np.sin(t)-y,m*x+lb-y) 
a,b= ellipse_polyline([(1, 1, 2, 1, 45), (2, 0.5, 5, 1.5, -30)])
t,y,x=fsolve(ellipse_line,(0,0,0))
print t,y,x
#print a[:,0]
m=0.5
bl=0
xl,yl=[],[]
for i in range(10):
    xl.append(i)
    yl.append(m*i+bl)
figure()
plot(x, y, "o")
plot(a[:,0], a[:,1])
plot(xl,yl)

任何帮助都会被评估?

1 个答案:

答案 0 :(得分:0)

使用Shapely部分,可以修复def intersections(a, line)。首先,存在一个错误,它引用全局b而不是使用line参数,该参数将被忽略。因此,比较是在两个省略号ab之间,而不是在每个椭圆到line之间,正如我所想的那样。

此外,两条线之间的交点可以是以下几种结果之一:空集,点(1交点),多点(多于1个交点),线串(如果线串的任何部分重叠,彼此平行),或点和线的集合。假设只有前三个结果:

def intersections(a, line):
    ea = LinearRing(a)
    mp = ea.intersection(line)
    if mp.is_empty:
        print('Geometries do not intersect')
        return [], []
    elif mp.geom_type == 'Point':
        return [mp.x], [mp.y]
    elif mp.geom_type == 'MultiPoint':
        return [p.x for p in mp], [p.y for p in mp]
    else:
        raise ValueError('something unexpected: ' + mp.geom_type)

所以现在这些看起来是正确的:

>>> intersections(a, line)
([2.414213562373095], [2.414213562373095])
>>> intersections(b, line)
([0.0006681263405436677, 2.135895843256409], [0.0006681263405436642, 2.135895843256409])