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服务器之家 - 脚本之家 - Python - python射线法判断一个点在图形区域内外

python射线法判断一个点在图形区域内外

2021-07-26 00:04两鬓已不能斑白 Python

这篇文章主要为大家详细介绍了python射线法判断一个点在图形区域内外，具有一定的参考价值，感兴趣的小伙伴们可以参考一下

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

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# -*-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服务器之家。

原文链接：https://blog.csdn.net/u010429424/article/details/44830379

  • Python
  • 区域
  • 射线法

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