python射线法判断一个点在图形区域内外


Posted in Python onJune 28, 2019

用python 实现的代码:判断一个点在图形区域内外,供大家参考,具体内容如下

# -*-encoding:utf-8 -*-
# file:class.py
#
 
"""
信息楼
0 123.425658,41.774177
1 123.425843,41.774166
2 123.425847,41.774119
3 123.42693,41.774062
4 123.426943,41.774099
5 123.427118,41.774089
6 123.427066,41.773548
7 123.426896,41.773544
8 123.426916,41.773920
9 123.425838,41.773965
10 123.425804,41.773585
11 123.425611,41.773595
图书馆
0 123.425649,41.77303
1 123.426656,41.772993
2 123.426611,41.772398
3 123.425605,41.772445
"""
 
 
class Point:
 lat = ''
 lng = ''
 
 def __init__(self,lat,lng):
 self.lat = lat #纬度
 self.lng = lng #经度
 
 def show(self):
 print self.lat," ",self.lng
 
 
#将信息楼的边界点实例化并存储到points1里
point0 = Point(123.425658,41.774177)
point1 = Point(123.425843,41.774166)
point2 = Point(123.425847,41.774119)
point3 = Point(123.42693,41.774062)
point4 = Point(123.426943,41.774099)
point5 = Point(123.427118,41.774089)
point6 = Point(123.427066,41.773548)
point7 = Point(123.426896,41.773544)
point8 = Point(123.426916,41.773920)
point9 = Point(123.425838,41.773961)
point10 = Point(123.425804,41.773585)
point11 = Point(123.425611,41.773595)
 
points1 = [point0,point1,point2,point3,
   point4,point5,point6,point7,
   point8,point9,point10,point11,
  ]
 
 
#将图书馆的边界点实例化并存储到points2里
point0 = Point(123.425649,41.77303)
point1 = Point(123.426656,41.772993)
point2 = Point(123.426611,41.772398)
point3 = Point(123.425605,41.772445)
 
points2 = [point0,point1,point2,point3]
 
 
'''
将points1和points2存储到points里,
points可以作为参数传入
'''
points = [points1,points2]
 
 
'''
输入一个测试点,这个点通过GPS产生
建议输入三个点测试
在信息学馆内的点:123.4263790000,41.7740520000 123.42699,41.773592 
在图书馆内的点: 123.4261550000,41.7726740000 123.42571,41.772499 123.425984,41.772919 
不在二者内的点: 123.4246270000,41.7738130000
在信息学馆外包矩形内,但不在信息学馆中的点:123.4264060000,41.7737860000
'''
#lat = raw_input(please input lat)
#lng = raw_input(please input lng)
lat = 123.42699
lng = 41.773592
point = Point(lat,lng)
 
debug = raw_input("请输入debug")
if debug == '1':
 debug = True
else:
 debug = False
 
#求外包矩形
def getPolygonBounds(points):
 length = len(points)
 #top down left right 都是point类型
 top = down = left = right = points[0]
 for i in range(1,length):
 if points[i].lng > top.lng:
  top = points[i]
 elif points[i].lng < down.lng:
  down = points[i]
 else:
  pass
 if points[i].lat > right.lat:
  right = points[i]
 elif points[i].lat < left.lat:
  left = points[i]
   else:
   pass
 
 point0 = Point(left.lat,top.lng)
 point1 = Point(right.lat,top.lng)
 point2 = Point(right.lat,down.lng)
 point3 = Point(left.lat,down.lng)
 polygonBounds = [point0,point1,point2,point3]
 return polygonBounds
 
#测试求外包矩形的一段函数
if debug:
 poly1 = getPolygonBounds(points[0])
 print "第一个建筑的外包是:"
 for i in range(0,len(poly1)):
 poly1[i].show() 
 poly2 = getPolygonBounds(points[1])
 print "第二个建筑的外包是:"
 for i in range(0,len(poly2)):
 poly2[i].show() 
 
 
#判断点是否在外包矩形外
def isPointInRect(point,polygonBounds):
 if point.lng >= polygonBounds[3].lng and \
  point.lng <= polygonBounds[0].lng and \
  point.lat >= polygonBounds[3].lat and \
  point.lat <= polygonBounds[2].lat:\
  return True
 else:
 return False
 
#测试是否在外包矩形外的代码
if debug:
 if(isPointInRect(point,poly1)):
 print "在信息外包矩形内"
 else:
 print "在信息外包矩形外"
 
 if(isPointInRect(point,poly2)):
 print "在图书馆外包矩形内"
 else:
 print "在图书馆外包矩形外"
 
 
 
#采用射线法,计算测试点是否任意一个建筑内
def isPointInPolygon(point,points):
 #定义在边界上或者在顶点都建筑内
 Bound = Vertex = True
 count = 0
 precision = 2e-10
 
 #首先求外包矩形
 polygonBounds = getPolygonBounds(points)
 
 #然后判断是否在外包矩形内,如果不在,直接返回false
 if not isPointInRect(point, polygonBounds):
 if debug:
  print "在外包矩形外"
 return False
 else:
 if debug:
  print "在外包矩形内"
 
 length = len(points)
 p = point
 p1 = points[0]
 for i in range(1,length):
 if p.lng == p1.lng and p.lat == p1.lat:
  if debug:
  print "Vertex1"
  return Vertex
 
 p2 = points[i % length]
 if p.lng == p2.lng and p.lat == p2.lat:
  if dubug:
  print "Vertex2"
  return Vertex
 
 if debug: 
  print i-1,i
  print "p:"
  p.show()
  print "p1:"
  p1.show()
  print "p2:"
  p2.show()
 
 if p.lng < min(p1.lng,p2.lng) or \
  p.lng > max(p1.lng,p2.lng) or \
  p.lat > max(p1.lat,p2.lat): 
  p1 = p2
  if debug:
  print "Outside"
  continue
 
 elif p.lng > min(p1.lng,p2.lng) and \
  p.lng < max(p1.lng,p2.lng):
  if p1.lat == p2.lat:
  if p.lat == p1.lat and \
   p.lng > min(p1.lng,p2.lng) and \
   p.lng < max(p1.lng,p2.lng):
   return Bound
  else:
   count = count + 1
   if debug:
   print "count1:",count
   continue
  if debug:
  print "into left or right"   
 
  a = p2.lng - p1.lng
  b = p1.lat - p2.lat
  c = p2.lat * p1.lng - p1.lat * p2.lng
  d = a * p.lat + b * p.lng + c
  if p1.lng < p2.lng and p1.lat > p2.lat or \
   p1.lng < p2.lng and p1.lat < p2.lat: 
  if d < 0:
   count = count + 1
   if debug:
   print "count2:",count
  elif d > 0:
   p1 = p2
   continue
  elif abs(p.lng-d) < precision :
   return Bound
  else :    
  if d < 0:
   p1 = p2
   continue
  elif d > 0:
   count = count + 1
   if debug:
   print "count3:",count
  elif abs(p.lng-d) < precision :
   return Bound
 else:
  if p1.lng == p2.lng:
  if p.lng == p1.lng and \
   p.lat > min(p1.lat,p2.lat) and \
   p.lat < max(p1.lat,p2.lat):
    return Bound
  else:
  p3 = points[(i+1) % length]
  if p.lng < min(p1.lng,p3.lng) or \
   p.lng > max(p1.lng,p3.lng):
   count = count + 2
   if debug:
   print "count4:",count
  else:
   count = count + 1
   if debug:
   print "count5:",count 
 p1 = p2
 if count % 2 == 0 :
 return False
 else :
 return True
 
 
 
length = len(points)
flag = 0
for i in range(length):
 if isPointInPolygon(point,points[i]):
 print "你刚才输入的点在第 %d 个建筑里" % (i+1)
 print "然后根据i值,可以读出建筑名,或者修改传入的points参数"
 break
 else:
 flag = flag + 1
 
if flag == length:
 print "在头 %d 建筑外" % (i+1)

 以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持三水点靠木。

Python 相关文章推荐
Python之PyUnit单元测试实例
Oct 11 Python
python模块之time模块(实例讲解)
Sep 13 Python
Python数据结构与算法之字典树实现方法示例
Dec 13 Python
Python爬虫实例_利用百度地图API批量获取城市所有的POI点
Jan 10 Python
Python math库 ln(x)运算的实现及原理
Jul 17 Python
python3 求约数的实例
Dec 05 Python
Python3 元组tuple入门基础
Feb 09 Python
Python Flask上下文管理机制实例解析
Mar 16 Python
Matlab中plot基本用法的具体使用
Jul 17 Python
Python常用库Numpy进行矩阵运算详解
Jul 21 Python
Python爬虫设置Cookie解决网站拦截并爬取蚂蚁短租的问题
Feb 22 Python
Python使用永中文档转换服务
May 06 Python
Python OpenCV之图片缩放的实现(cv2.resize)
Jun 28 #Python
如何使用Python 打印各种三角形
Jun 28 #Python
python射线法判断检测点是否位于区域外接矩形内
Jun 28 #Python
python 列表转为字典的两个小方法(小结)
Jun 28 #Python
numpy和pandas中数组的合并、拉直和重塑实例
Jun 28 #Python
使用Python画股票的K线图的方法步骤
Jun 28 #Python
连接pandas以及数组转pandas的方法
Jun 28 #Python
You might like
一个简单的php实现的MySQL数据浏览器
2007/03/11 PHP
Linux下手动编译安装PHP扩展的例子分享
2014/07/15 PHP
PHP面试常用算法(推荐)
2016/07/22 PHP
js滚动条多种样式,推荐
2007/02/05 Javascript
jquery实现弹出窗口效果的实例代码
2013/11/28 Javascript
解决Jquery向页面append新元素之后事件的绑定问题
2015/03/16 Javascript
JavaScript中输出信息的方法(信息确认框-提示输入框-文档流输出)
2016/06/12 Javascript
Javascript点击其他任意地方隐藏关闭DIV实例
2016/06/21 Javascript
JavaScript中Form表单技术汇总(推荐)
2016/06/26 Javascript
基于Angular.js实现的触摸滑动动画实例代码
2017/02/19 Javascript
js和jquery中获取非行间样式
2017/05/05 jQuery
快速解决Vue项目在IE浏览器中显示空白的问题
2018/09/04 Javascript
详解Vue一个案例引发「内容分发slot」的最全总结
2018/12/02 Javascript
浅谈小程序globalData的那些事儿
2019/11/01 Javascript
JavaScript基于用户照片姓名生成海报
2020/05/29 Javascript
[10:18]2018DOTA2国际邀请赛寻真——Fnatic能否笑到最后?
2018/08/14 DOTA
Python获取DLL和EXE文件版本号的方法
2015/03/10 Python
Python中优化NumPy包使用性能的教程
2015/04/23 Python
python自带的http模块详解
2016/11/06 Python
火车票抢票python代码公开揭秘!
2018/03/08 Python
Python 二叉树的层序建立与三种遍历实现详解
2019/07/29 Python
查看Python依赖包及其版本号信息的方法
2019/08/13 Python
Django如何使用第三方服务发送电子邮件
2019/08/14 Python
关于numpy数组轴的使用详解
2019/12/05 Python
python入门之基础语法学习笔记
2020/02/08 Python
关于Python turtle库使用时坐标的确定方法
2020/03/19 Python
将python字符串转化成长表达式的函数eval实例
2020/05/11 Python
pycharm中使用request和Pytest进行接口测试的方法
2020/07/31 Python
使用CSS3设计地图上的雷达定位提示效果
2016/04/05 HTML / CSS
北美三大旅游网站之一:Travelocity加拿大
2016/08/20 全球购物
采购人员的个人自我评价
2014/01/16 职场文书
市场营销战略计划书
2014/05/06 职场文书
书香家庭事迹材料
2014/05/09 职场文书
教师节简报
2015/07/20 职场文书
python利用pandas分析学生期末成绩实例代码
2021/07/09 Python
使用Django框架创建项目
2022/06/10 Python