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我有几组线段AB1, AB2, ... ABn。每个都有(Ax, Ay), (Bx, By)坐标。然后，我有一个中心坐标（Cx，Cy）和r（半径）的圆。

问题： 如何检测哪条线段位于圆上（如图）？图中描述.

我试图用 Python 来表达我的想法：

import numpy as np
import pylab as plt

def detect(A,B, C, r):
    '''
    Returns 'True' if line is inside or intersected the circle, otherwise 'False'.
    Inputs:
       - A - segment line coordinate (Ax, Ay)
       - B - segment line coordinate (Bx, By)
       - C - circle coordinate (Cx, Cy)
       - r - circle radius
    ''' 
    # Do process for detection
    return (boolean)

def plot_detected(An, Bn, C, r):
    '''
    Plots newly detected line segments with red color 
    while the rest remains with blue color 
    '''
    plt.figure(1)
    plt.subplot(111)
    for A, B in zip(An, Bn):
        if detect(A, B, C, r):
              line1, = plt.plot([ A[0], B[0] ], [ A[1], B[1] ], 'ro-')
        else:
              line2, = plt.plot([ A[0], B[0] ], [ A[1], B[1] ], 'bo-')
    pl.legend([line1, line2], ('detected','un-detected'))
    plt.show()

def main():
    C = [18.5, 18.5]
    r = 2.4
    Ax = np.array([16.2, 17.2, 22.2, 18.2, 23.8, 18.8])
    Ay = np.array([22.8, 20.6, 23.8, 18.4, 20.8, 22.8])
    Bx = np.array([21.8, 19.8, 18.2, 19.8, 17.2, 22.8])
    By = np.array([17.8, 17.2, 19.2, 19.2, 16.8, 20.8])
    An = np.vstack([Ax, Ay]).T
    Bn = np.vstack([Bx, By]).T

    plot_detected(An, Bn, C, r)

if __name__ == '__main__':
    main()

提前谢谢你的帮助。

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2 回答 2

2

对于每条线，您应该能够计算线上距离圆心最小的点。为此，您将中心的位置向量投影到线的方向向量上。称其为最小距离点 P。如果 P 在圆内（即其坐标的平方和的 sqrt 小于圆半径）并且 P 也在线段的端点之间，则线段与圆圈。

您还必须检查线端点本身是否在圆内。

def detect( A, B, C, r ):

    # First, let's express each vector as a complex number.
    # This simplifies the rest of the code because we can then subtract them
    # from each other in one statement, or find their length with one statement.
    # (Downside: it does not allow us to generalize the code to spheres in 3D.)
    OA = complex( *A )
    OB = complex( *B )
    OC = complex( *C )

    # Now let's translate into a coordinate system where A is the origin
    AB = OB - OA
    AC = OC - OA

    # Before we go further let's cover one special case:  if either A or B is actually in
    # the circle,  then mark it as a detection
    BC = OC - OB
    if abs( BC ) < r or abs( AC ) < r: return True

    # Project C onto the line to find P, the point on the line that is closest to the circle centre
    AB_normalized = AB / abs( AB )
    AP_distance = AC.real * AB_normalized.real  +  AC.imag * AB_normalized.imag    # dot product (scalar result)
    AP = AP_distance * AB_normalized   # actual position of P relative to A (vector result)

    # If AB intersects the circle, and neither A nor B itself is in the circle,
    # then P, the point on the extended line that is closest to the circle centre, must be...

    # (1) ...within the segment AB:
    AP_proportion = AP_distance / abs( AB )   # scalar value: how far along AB is P?
    in_segment =   0 <= AP_proportion <= 1

    # ...and (2) within the circle:
    CP = AP - AC
    in_circle = abs( CP ) < r

    detected = in_circle and in_segment


    #OP = OA + AP
    #plt.plot( [OC.real, OP.real], [OC.imag, OP.imag], {True:'rs--', False:'bs--'}[detected] )

    return detected



def plot_detected(An, Bn, C, r):
    '''
    Plots newly detected line segments with red color 
    while the rest remains with blue color 
    '''
    plt.figure(1)
    plt.clf()
    plt.subplot(111)
    for A, B in zip(An, Bn):
        if detect(A, B, C, r):
              line1, = plt.plot([ A[0], B[0] ], [ A[1], B[1] ], 'ro-')
        else:
              line2, = plt.plot([ A[0], B[0] ], [ A[1], B[1] ], 'bo-')
    plt.legend([line1, line2], ('detected','un-detected'))
    circle = mpatches.Circle( C, r, fc="none", ec='k' )
    plt.gca().add_patch(circle)
    plt.gca().set_aspect('equal')
于 2014-11-04T00:42:35.290 回答
0

第一个：任务定义的消除歧义拓扑

术语定义：

给出了所有“线段”。

每个线段由其[x,y]每个端点的笛卡尔对定义。

关于拓扑条件的初始问题陈述指出：

检测哪条线段 位于圆上（如图）？

作者自己的评论补充说：

找出哪些线段不仅相交而且位于圆内。

问题标题指出：

拾取在圆内或与圆相交的线段

从标题和评论看来，为了满足“线段”到“位于圆上”的拓扑条件，如果它的任何端点在圆内就{ 1 | 2 }足够了。

如果此拓扑假设失败，则 jez-posted 矢量法线距离解是正确的。

解决方案：

基于消除歧义的术语定义和拓扑条件，满足定义约束的“线段”集合通过以下方式获得：

def detect( A, B, C, r ):
    '''
    Returns 'True' if line is inside or intersected the circle, otherwise 'False'.
    Inputs:
       - A - segment line coordinate (Ax, Ay)
       - B - segment line coordinate (Bx, By)
       - C - circle coordinate (Cx, Cy)
       - r - circle radius
    '''
    # Do process for detection

    aBooleanANSWER = ( aCartesianDISTANCE( A, C ) <= r )
                     or
                     ( aCartesianDISTANCE( B, C ) <= r )

    return ( aBooleanANSWER )
于 2014-11-04T16:07:11.957 回答