2013-08-20 38 views
2

我正在嘗試找到橢圓和線之間的交點,但似乎無法這樣做。我嘗試了兩種方法,一種是通過嘗試找到LineString和LinearRing之間的交集,如下面的代碼所示,但沒有從中得到任何可用的值。其中的一個問題,是elipse總是會偏離中心,並在一個小的或大角度查找橢圓和線之間的交點

# -*- 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() 

我使用fsolve如實施例波紋管也嘗試過,但它找到了錯誤的交叉點(或實際上一個錯點。

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) 

任何幫助將appriceated?

+0

似乎是完全適合http://math.stackexchange.com/ –

+0

不是真的我想到一個問題。假設這不能直接解析,所以你需要一個數值方法,在這種情況下,這是發佈的正確地方。否則math.stackexchange.com是要走的路。 – usethedeathstar

+0

@usethedeathstar ...好點!我沒有這樣想過...... –

回答

0

隨着勻稱的一部分,def intersections(a, line)可以是固定的。首先,有一個錯誤,它引用全局b,而不是使用line參數,這被忽略。因此,比較是在兩個橢圓ab之間,而不是在每個橢圓到line之間,正如我認爲的那樣。另外,兩條線之間的交點可以是幾個結果之一:空集,點(1交點),多點(多於1個交點),線串(如果線串的任何部分重疊)並且彼此平行)或點和線的集合。假設只有前三個結果:

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,現在這些看起來是正確的:

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