python射线法判断一个点在图形区域内外 射线法判断一个点在图形区域内外 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 # -*-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我们。