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实现保存网页到本地示例
Mar 16 Python
简单谈谈Python中的闭包
Nov 30 Python
浅谈Scrapy框架普通反爬虫机制的应对策略
Dec 28 Python
Python字典操作详细介绍及字典内建方法分享
Jan 04 Python
python将回车作为输入内容的实例
Jun 23 Python
使用python实现简单五子棋游戏
Jun 18 Python
OpenCV3.0+Python3.6实现特定颜色的物体追踪
Jul 23 Python
python带参数打包exe及调用方式
Dec 21 Python
如何基于线程池提升request模块效率
Apr 18 Python
pycharm 关掉syntax检查操作
Jun 09 Python
Python如何实现自带HTTP文件传输服务
Jul 08 Python
用Python进行栅格数据的分区统计和批量提取
May 27 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
PHP面向对象分析设计的61条军规小结
2010/07/17 PHP
php输出echo、print、print_r、printf、sprintf、var_dump的区别比较
2013/06/21 PHP
php中mail函数发送邮件失败的解决方法
2014/12/24 PHP
PHP微信开发之查询微信精选文章
2016/06/23 PHP
php插件Xajax使用方法详解
2017/08/31 PHP
php设计模式之备忘模式分析【星际争霸游戏案例】
2020/03/24 PHP
jQuery each()小议
2010/03/18 Javascript
js获取元素到文档区域document的(横向、纵向)坐标的两种方法
2013/05/17 Javascript
javascript中对Attr(dom中属性)的操作示例讲解
2013/12/02 Javascript
jquery解决客户端跨域访问问题
2015/01/06 Javascript
最新最热最实用的15个jQuery插件汇总
2015/07/05 Javascript
JavaScript实现鼠标滚轮控制页面图片切换功能示例
2017/10/14 Javascript
详解vue-cli脚手架build目录中的dev-server.js配置文件
2017/11/24 Javascript
解决Mac下安装nmp的淘宝镜像失败问题
2018/05/16 Javascript
layui select获取自定义属性方法
2018/08/15 Javascript
js实现json数组分组合并操作示例
2019/02/12 Javascript
JavaScript面试技巧之数组的一些不low操作
2019/03/22 Javascript
jQuery实现input输入框获取焦点与失去焦点时提示的消失与显示功能示例
2019/05/27 jQuery
JS通过ajax + 多列布局 + 自动加载实现瀑布流效果
2019/05/30 Javascript
[19:15]DK战队纪录片
2014/09/02 DOTA
Python标准库之Sys模块使用详解
2015/05/23 Python
Python实现线程状态监测简单示例
2018/03/28 Python
python re库的正则表达式入门学习教程
2019/03/08 Python
在python中利用numpy求解多项式以及多项式拟合的方法
2019/07/03 Python
python求最大值,不使用内置函数的实现方法
2019/07/09 Python
Python 元组操作总结
2019/09/18 Python
Python中用pyinstaller打包时的图标问题及解决方法
2020/02/17 Python
欧洲最大的球衣网上商店:Kitbag
2017/11/11 全球购物
C#里面如何倒序排列一个数组的元素?
2013/06/21 面试题
高一自我鉴定
2013/12/17 职场文书
初中数学教学反思
2014/01/16 职场文书
离婚协议书范文2015
2015/01/26 职场文书
2019升学宴主持词范本5篇
2019/10/09 职场文书
教你用python控制安卓手机
2021/05/13 Python
详细介绍MySQL中limit和offset的用法
2022/05/06 MySQL
python热力图实现的完整实例
2022/06/25 Python