我试图创建一个快速的2D点内多边形算法,用于命中测试(例如多边形.contains(p:点))。对有效技术的建议将不胜感激。
当前回答
nirg回答的Swift版本:
extension CGPoint {
func isInsidePolygon(vertices: [CGPoint]) -> Bool {
guard !vertices.isEmpty else { return false }
var j = vertices.last!, c = false
for i in vertices {
let a = (i.y > y) != (j.y > y)
let b = (x < (j.x - i.x) * (y - i.y) / (j.y - i.y) + i.x)
if a && b { c = !c }
j = i
}
return c
}
}
其他回答
令人惊讶的是之前没有人提出这个问题,但是对于需要数据库的实用主义者来说:MongoDB对Geo查询提供了出色的支持,包括这个查询。
你需要的是:
db.neighborhoods。findOne({geometry: {$geoIntersects: {$geometry: { type: "Point",坐标:["经度","纬度"]}}} })
communities是存储一个或多个标准GeoJson格式多边形的集合。如果查询返回null,则表示不相交,否则为。
这里有详细的记录: https://docs.mongodb.com/manual/tutorial/geospatial-tutorial/
在330个不规则多边形网格中,超过6000个点分类的性能不到一分钟,没有任何优化,包括用各自的多边形更新文档的时间。
这个问题很有趣。我有另一个可行的想法,不同于这篇文章的其他答案。其原理是利用角度之和来判断目标是在内部还是外部。也就是圈数。
设x为目标点。让数组[0,1,....N]是该区域的所有点。用一条线将目标点与每一个边界点连接起来。如果目标点在这个区域内。所有角的和是360度。如果不是,角度将小于360度。
参考这张图来对这个概念有一个基本的了解:
我的算法假设顺时针是正方向。这是一个潜在的输入:
[[-122.402015, 48.225216], [-117.032049, 48.999931], [-116.919132, 45.995175], [-124.079107, 46.267259], [-124.717175, 48.377557], [-122.92315, 47.047963], [-122.402015, 48.225216]]
下面是实现这个想法的python代码:
def isInside(self, border, target):
degree = 0
for i in range(len(border) - 1):
a = border[i]
b = border[i + 1]
# calculate distance of vector
A = getDistance(a[0], a[1], b[0], b[1]);
B = getDistance(target[0], target[1], a[0], a[1])
C = getDistance(target[0], target[1], b[0], b[1])
# calculate direction of vector
ta_x = a[0] - target[0]
ta_y = a[1] - target[1]
tb_x = b[0] - target[0]
tb_y = b[1] - target[1]
cross = tb_y * ta_x - tb_x * ta_y
clockwise = cross < 0
# calculate sum of angles
if(clockwise):
degree = degree + math.degrees(math.acos((B * B + C * C - A * A) / (2.0 * B * C)))
else:
degree = degree - math.degrees(math.acos((B * B + C * C - A * A) / (2.0 * B * C)))
if(abs(round(degree) - 360) <= 3):
return True
return False
VBA版本:
注意:请记住,如果你的多边形是地图中的一个区域,纬度/经度是Y/X值,而不是X/Y(纬度= Y,经度= X),因为从我的理解来看,这是历史含义,因为经度不是一个测量值。
类模块:CPoint
Private pXValue As Double
Private pYValue As Double
'''''X Value Property'''''
Public Property Get X() As Double
X = pXValue
End Property
Public Property Let X(Value As Double)
pXValue = Value
End Property
'''''Y Value Property'''''
Public Property Get Y() As Double
Y = pYValue
End Property
Public Property Let Y(Value As Double)
pYValue = Value
End Property
模块:
Public Function isPointInPolygon(p As CPoint, polygon() As CPoint) As Boolean
Dim i As Integer
Dim j As Integer
Dim q As Object
Dim minX As Double
Dim maxX As Double
Dim minY As Double
Dim maxY As Double
minX = polygon(0).X
maxX = polygon(0).X
minY = polygon(0).Y
maxY = polygon(0).Y
For i = 1 To UBound(polygon)
Set q = polygon(i)
minX = vbMin(q.X, minX)
maxX = vbMax(q.X, maxX)
minY = vbMin(q.Y, minY)
maxY = vbMax(q.Y, maxY)
Next i
If p.X < minX Or p.X > maxX Or p.Y < minY Or p.Y > maxY Then
isPointInPolygon = False
Exit Function
End If
' SOURCE: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html
isPointInPolygon = False
i = 0
j = UBound(polygon)
Do While i < UBound(polygon) + 1
If (polygon(i).Y > p.Y) Then
If (polygon(j).Y < p.Y) Then
If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
isPointInPolygon = True
Exit Function
End If
End If
ElseIf (polygon(i).Y < p.Y) Then
If (polygon(j).Y > p.Y) Then
If p.X < (polygon(j).X - polygon(i).X) * (p.Y - polygon(i).Y) / (polygon(j).Y - polygon(i).Y) + polygon(i).X Then
isPointInPolygon = True
Exit Function
End If
End If
End If
j = i
i = i + 1
Loop
End Function
Function vbMax(n1, n2) As Double
vbMax = IIf(n1 > n2, n1, n2)
End Function
Function vbMin(n1, n2) As Double
vbMin = IIf(n1 > n2, n2, n1)
End Function
Sub TestPointInPolygon()
Dim i As Integer
Dim InPolygon As Boolean
' MARKER Object
Dim p As CPoint
Set p = New CPoint
p.X = <ENTER X VALUE HERE>
p.Y = <ENTER Y VALUE HERE>
' POLYGON OBJECT
Dim polygon() As CPoint
ReDim polygon(<ENTER VALUE HERE>) 'Amount of vertices in polygon - 1
For i = 0 To <ENTER VALUE HERE> 'Same value as above
Set polygon(i) = New CPoint
polygon(i).X = <ASSIGN X VALUE HERE> 'Source a list of values that can be looped through
polgyon(i).Y = <ASSIGN Y VALUE HERE> 'Source a list of values that can be looped through
Next i
InPolygon = isPointInPolygon(p, polygon)
MsgBox InPolygon
End Sub
Obj-C版本nirg的答案与样本方法测试点。Nirg的回答对我很有效。
- (BOOL)isPointInPolygon:(NSArray *)vertices point:(CGPoint)test {
NSUInteger nvert = [vertices count];
NSInteger i, j, c = 0;
CGPoint verti, vertj;
for (i = 0, j = nvert-1; i < nvert; j = i++) {
verti = [(NSValue *)[vertices objectAtIndex:i] CGPointValue];
vertj = [(NSValue *)[vertices objectAtIndex:j] CGPointValue];
if (( (verti.y > test.y) != (vertj.y > test.y) ) &&
( test.x < ( vertj.x - verti.x ) * ( test.y - verti.y ) / ( vertj.y - verti.y ) + verti.x) )
c = !c;
}
return (c ? YES : NO);
}
- (void)testPoint {
NSArray *polygonVertices = [NSArray arrayWithObjects:
[NSValue valueWithCGPoint:CGPointMake(13.5, 41.5)],
[NSValue valueWithCGPoint:CGPointMake(42.5, 56.5)],
[NSValue valueWithCGPoint:CGPointMake(39.5, 69.5)],
[NSValue valueWithCGPoint:CGPointMake(42.5, 84.5)],
[NSValue valueWithCGPoint:CGPointMake(13.5, 100.0)],
[NSValue valueWithCGPoint:CGPointMake(6.0, 70.5)],
nil
];
CGPoint tappedPoint = CGPointMake(23.0, 70.0);
if ([self isPointInPolygon:polygonVertices point:tappedPoint]) {
NSLog(@"YES");
} else {
NSLog(@"NO");
}
}
当我还是Michael Stonebraker手下的一名研究员时,我做了一些关于这方面的工作——你知道,就是那位提出了Ingres、PostgreSQL等的教授。
我们意识到最快的方法是首先做一个边界框,因为它非常快。如果它在边界框之外,它就在外面。否则,你就得做更辛苦的工作……
如果你想要一个伟大的算法,看看开源项目PostgreSQL的源代码的地理工作…
我想指出的是,我们从来没有深入了解过左撇子和右撇子(也可以表达为“内”和“外”的问题……
更新
BKB's link provided a good number of reasonable algorithms. I was working on Earth Science problems and therefore needed a solution that works in latitude/longitude, and it has the peculiar problem of handedness - is the area inside the smaller area or the bigger area? The answer is that the "direction" of the verticies matters - it's either left-handed or right handed and in this way you can indicate either area as "inside" any given polygon. As such, my work used solution three enumerated on that page.
此外,我的工作使用单独的函数进行“在线”测试。
...因为有人问:我们发现当垂直的数量超过某个数字时,边界盒测试是最好的——如果有必要,在做更长的测试之前做一个非常快速的测试……边界框是通过简单地将最大的x,最小的x,最大的y和最小的y放在一起,组成一个框的四个点来创建的……
另一个提示是:我们在网格空间中进行了所有更复杂的“调光”计算,都是在平面上的正点上进行的,然后重新投影到“真实”的经度/纬度上,从而避免了在经度180线交叉时和处理极地时可能出现的环绕错误。工作好了!
推荐文章
- 确定记录是否存在的最快方法
- 阅读GHC核心
- Python: List vs Dict用于查找表
- 为什么MATLAB的矩阵乘法运算这么快?
- for循环和for-each循环在性能上有区别吗?
- 就性能而言,使用std::memcpy()还是std::copy()更好?
- 什么时候我应该(不)想要在我的代码中使用熊猫apply() ?
- 如何加速gwt编译器?
- MySQL OR与IN性能
- 应该……接住环内还是环外?
- 哪个更快/最好?SELECT *或SELECT columnn1, colum2, column3等
- 加快R中的循环操作
- INT和VARCHAR主键之间有真正的性能差异吗?
- c++标准是否要求iostreams的性能很差,或者我只是在处理一个糟糕的实现?
- 大概的成本访问各种缓存和主存储器?