python 实现射线法判断一个点在图形区域内外

本文主要是介绍python 实现射线法判断一个点在图形区域内外，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

关于射线法可以参考下面这个blog ，写得非常详细：http://www.cnblogs.com/mazhenyu/p/3800638.html

下面是我用python 实现的代码：

# -*-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:passif points[i].lat > right.lat:right = points[i]elif points[i].lat < left.lat:left = points[i]else:passpoint0 = 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 Trueelse: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 = Truecount = 0precision = 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falseif not isPointInRect(point, polygonBounds):if debug:print "在外包矩形外"return Falseelse:if debug:print "在外包矩形内"length = len(points)p = pointp1 = points[0]for i in range(1,length):if p.lng == p1.lng and p.lat == p1.lat:if debug:print "Vertex1"return Vertexp2 = points[i % length]if p.lng == p2.lng and p.lat == p2.lat:if dubug:print "Vertex2"return Vertexif debug:	print i-1,iprint "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 = p2if debug:print "Outside"continueelif 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 Boundelse:count = count + 1if debug:print "count1:",countcontinueif debug:print "into left or right"	  		a = p2.lng - p1.lngb = p1.lat - p2.latc = p2.lat * p1.lng - p1.lat * p2.lngd = a * p.lat + b * p.lng + cif p1.lng < p2.lng and p1.lat > p2.lat or \p1.lng < p2.lng and p1.lat < p2.lat:	if d < 0:count = count + 1if debug:print "count2:",countelif d > 0:p1 = p2continueelif abs(p.lng-d) < precision :return Boundelse :	     	if d < 0:p1 = p2continueelif d > 0:count = count + 1if debug:print "count3:",countelif abs(p.lng-d) < precision :return Boundelse: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 Boundelse:p3 = points[(i+1) % length]if p.lng < min(p1.lng,p3.lng) or \p.lng > max(p1.lng,p3.lng):count = count + 2if debug:print "count4:",countelse:count = count + 1if debug:print "count5:",count	 p1 = p2if count % 2 == 0 :return Falseelse :return Truelength = 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"breakelse:flag = flag + 1if flag == length:print "在头 %d 建筑外" % (i+1)