python射线法判断一个点在图形区域内外_python判断点是否在区域内,python如何判断一个点在一个区域内 - CSDN文库

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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]

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