Question

这个问题有点类似于this.我已经比OP更远了，而且我在Python 2中（不知道他在用什么）。

我有一个Python函数，可以确定从凸多边形内的点到多边形周长的规则定义间隔的距离。问题是它返回了我需要消除的“额外”距离。（请注意 - 我怀疑这对矩形不起作用。我还没完成。）首先，代码：

#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
#  t1.py
# 
#  Copyright 2015 FRED <fred@matthew24-25>
# 
#  THIS IS TESTING CODE ONLY. IT WILL BE MOVED INTO THE CORRECT MODULE
#  UPON COMPLETION.
# 
from __future__ import division
import math
import matplotlib.pyplot as plt
def Dist(center_point, Pairs, deg_Increment):

# I want to set empty lists to store the values of m_lnsgmnt and b_lnsgmnts
# for every iteration of the for loop.
    m_linesegments = []
    b_linesegments = []

# Scream and die if Pairs[0] is the same as the last element of Pairs--i.e.
# it has already been run once.
#if Pairs[0] == Pairs[len(Pairs)-1]:
  ##print "The vertices contain duplicate points!"

## Creates a new list containing the original list plus the first element. I did this because, due
## to the way the for loop is set up, the last iteration of the loop subtracts the value of the
## last value of Pairs from the first value. I therefore duplicated the first value.
#elif:
    new_Pairs = Pairs + [Pairs[0]]

    # This will calculate the slopes and y-intercepts of the linesegments of the polygon.
    for a in range(len(Pairs)):

        # This calculates the slope of each line segment and appends it to m_linesegments.
        m_lnsgmnt = (new_Pairs[a+1][2] - new_Pairs[a][3]) / (new_Pairs[a+1][0] - new_Pairs[a][0])
        m_linesegments.append(m_lnsgmnt)

        # This calculates the y-intercept of each line segment and appends it to b_linesegments.
        b_lnsgmnt = (Pairs[a][4]) - (m_lnsgmnt * Pairs[a][0])
        b_linesegments.append(b_lnsgmnt)

    # These are temporary testing codes.
    print "m_linesegments =", m_linesegments
    print "b_linesegments =", b_linesegments

    # I want to set empty lists to store the value of m_rys and b_rys for every
    # iteration of the for loop.
    m_rays = []
    b_rays = []

    # I need to set a range of degrees the intercepts will be calculated for.
    theta = range(0, 360, deg_Increment)

    # Temporary testing line.
    print "theta =", theta

    # Calculate the slope and y-intercepts of the rays radiating from the center_point.
    for b in range(len(theta)):

        m_rys = math.tan(math.radians(theta[b]))
        m_rays.append(m_rys)

        b_rys = center_point[1] - (m_rys * center_point[0])
        b_rays.append(b_rys)

    # Temporary testing lines.   
    print "m_rays =", m_rays
    print "b_rays =", b_rays


    # Set empty matrix for Intercepts.
    Intercepts = []
    angle = []

    # Calculate the intersections of the rays with the line segments.
    for c in range((360//deg_Increment)):

        for d in range(len(Pairs)):

            # Calculate the x-coordinates and the y-coordinates of each
            # intersection
            x_Int = (b_rays[c] - b_linesegments[d]) / (m_linesegments[d] - m_rays[c])
            y_Int = ((m_linesegments[d] * x_Int) + b_linesegments[d])


            Intercepts.append((x_Int, y_Int))

            # Calculates the angle of the ray. Rounding is necessary to 
            # compensate for binary-decimal errors. 
            a_ngle = round(math.degrees(math.atan2((y_Int - center_point[1]), (x_Int - center_point[0]))))

            # Substitutes positive equivalent for every negative angle, 
            # i.e. -270 degrees equals 90 degrees.
            if a_ngle < 0:
                a_ngle = a_ngle + 360

            # Selects the angles that correspond to theta
            if a_ngle == theta[c]:
                angle.append(a_ngle)


    print "INT1=", Intercepts
    print "angle=", angle

    dist = []

    # Calculates distance.
    for e in range(len(Intercepts) - 1):
        distA = math.sqrt(((Intercepts[e][0] - center_point[0])**2) +  ((Intercepts[e][5]- center_point[1])**2))
        dist.append(distA)

    print "dist=", dist

if __name__ == "__main__":
    main()

现在，关于它是如何工作的：代码需要3个输入：center_point（包含在多边形中的点，以（x，y）坐标给出），Pairs（多边形的顶点，也在（x，y）坐标中给出）和deg_Increment（定义如何）经常计算距离）。我们假设center_point = (4,5)，Pairs = [(1, 4), (3, 8), (7, 2)]和deg_Increment = 20。这意味着创建了一个多边形（有点），其顶点为Pairs，center_point是包含在多边形内的点。现在光线设置为每20度从center_point辐射（deg_Increment）。确定光线与多边形周长的交点，并使用距离公式计算距离。唯一的问题是我的距离太远了。 :(在上面的例子中，正确的距离是 1.00000 0.85638 0.83712 0.92820 1.20455 2.07086 2.67949 2.29898 2.25083 2.50000 3.05227 2.22683 1.93669 1.91811 2.15767 2.85976 2.96279 1.40513

但是我的代码又回来了 dist= [2.5, 1.0, 6.000000000000001, 3.2523178818773006, 0.8563799085248148, 3.0522653889161626, 5.622391569468206, 0.8371216462519347, 2.226834844885431, 37.320508075688686, 0.9282032302755089, 1.9366857335569072, 7.8429970322236064, 1.2045483557883576, 1.9181147622136665, 3.753460385470896, 2.070863609380179, 2.157671808913309, 2.6794919243112276, 12.92820323027545, 2.85976265663383, 2.298981118867903, 2.962792920643178, 5.162096782237789, 2.250827351906659, 1.4051274947736863, 69.47032761621092, 2.4999999999999996, 1.0, 6.000000000000004, 3.2523178818773006, 0.8563799085248148, 3.0522653889161626, 5.622391569468206, 0.8371216462519347, 2.226834844885431, 37.32050807568848, 0.9282032302755087, 1.9366857335569074, 7.842997032223602, 1.2045483557883576, 1.9181147622136665, 3.7534603854708997, 2.0708636093801767, 2.1576718089133085, 2.679491924311227, 12.928203230275532, 2.85976265663383, 2.298981118867903, 2.9627929206431776, 5.162096782237789, 2.250827351906659, 1.4051274947736847]

如果有人能帮助我获得正确的距离，我会非常感激。谢谢！

仅供参考，以下是我的示例仅正确的距离： DISTANCE

Answer 1

你在Intercepts中获得了太多的值，因为它被附加到第二个for循环[for d in range(len(Pairs))]内。

通过外部for循环[for c in range((360//deg_Increment))]，每个步骤只需要一个Intercept中的值，因此Intercept的追加需要在此循环中。

我不确定你在使用内循环做什么，但是你似乎正在为构成多边形边的每条线计算一个单独的截距。但是你只想要那个你朝那个方向前进的“第一”。

你必须添加一些代码来弄清楚你实际上要遇到的多边形的3个边（在这种情况下）中的哪一个。

从列表中选择正确的距离

1 个答案: