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
from typing import Iterable
def pnpoly(verts, x, y):
#check if x and/or y is iterable
xit, yit = isinstance(x, Iterable), isinstance(y, Iterable)
#if not iterable, make an iterable of length 1
X = x if xit else (x, )
Y = y if yit else (y, )
#store verts length as a range to juggle j
r = range(len(verts))
#final results if x or y is iterable
results = []
#traverse x and y coordinates
for xp in X:
for yp in Y:
c = 0 #reset c at every new position
for i in r:
j = r[i-1] #set j to position before i
#store a few arguments to shorten the if statement
yneq = (verts[i][1] > yp) != (verts[j][1] > yp)
xofs, yofs = (verts[j][0] - verts[i][0]), (verts[j][1] - verts[i][1])
#if we have crossed a line, increment c
if (yneq and (xp < xofs * (yp - verts[i][1]) / yofs + verts[i][0])):
c += 1
#if c is odd store the coordinates
if c%2:
results.append((xp, yp))
#return either coordinates or a bool, depending if x or y was an iterable
return results if (xit or yit) else bool(c%2)
这个python版本是通用的。您可以为True/False结果输入单个x和单个y值，也可以使用x和y的范围来遍历整个点网格。如果使用范围，则返回所有True点的x/y对列表。vertices参数需要一个由x/y对组成的二维Iterable，例如:[(x1,y1)， (x2,y2)，…]
使用示例:
vertices = [(25,25), (75,25), (75,75), (25,75)]
pnpoly(vertices, 50, 50) #True
pnpoly(vertices, range(100), range(100)) #[(25,25), (25,26), (25,27), ...]
实际上，这些都可以。
pnpoly(vertices, 50, range(100)) #check 0 to 99 y at x of 50
pnpoly(vertices, range(100), 50) #check 0 to 99 x at y of 50
其他回答
这似乎在R中工作(为丑陋道歉，希望看到更好的版本!)。
pnpoly <- function(nvert,vertx,verty,testx,testy){
c <- FALSE
j <- nvert
for (i in 1:nvert){
if( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]-vertx[i])*(testy-verty[i])/(verty[j]-verty[i])+vertx[i]))
{c <- !c}
j <- i}
return(c)}
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
from typing import Iterable
def pnpoly(verts, x, y):
#check if x and/or y is iterable
xit, yit = isinstance(x, Iterable), isinstance(y, Iterable)
#if not iterable, make an iterable of length 1
X = x if xit else (x, )
Y = y if yit else (y, )
#store verts length as a range to juggle j
r = range(len(verts))
#final results if x or y is iterable
results = []
#traverse x and y coordinates
for xp in X:
for yp in Y:
c = 0 #reset c at every new position
for i in r:
j = r[i-1] #set j to position before i
#store a few arguments to shorten the if statement
yneq = (verts[i][1] > yp) != (verts[j][1] > yp)
xofs, yofs = (verts[j][0] - verts[i][0]), (verts[j][1] - verts[i][1])
#if we have crossed a line, increment c
if (yneq and (xp < xofs * (yp - verts[i][1]) / yofs + verts[i][0])):
c += 1
#if c is odd store the coordinates
if c%2:
results.append((xp, yp))
#return either coordinates or a bool, depending if x or y was an iterable
return results if (xit or yit) else bool(c%2)
这个python版本是通用的。您可以为True/False结果输入单个x和单个y值，也可以使用x和y的范围来遍历整个点网格。如果使用范围，则返回所有True点的x/y对列表。vertices参数需要一个由x/y对组成的二维Iterable，例如:[(x1,y1)， (x2,y2)，…]
使用示例:
vertices = [(25,25), (75,25), (75,75), (25,75)]
pnpoly(vertices, 50, 50) #True
pnpoly(vertices, range(100), range(100)) #[(25,25), (25,26), (25,27), ...]
实际上，这些都可以。
pnpoly(vertices, 50, range(100)) #check 0 to 99 y at x of 50
pnpoly(vertices, range(100), 50) #check 0 to 99 x at y of 50
简单的解决方案是将多边形划分为三角形，并按这里解释的那样对三角形进行测试
如果你的多边形是凸多边形，可能有更好的方法。把这个多边形看作是无限条线的集合。每一行将空间一分为二。对于每一个点，很容易判断它是在直线的一边还是另一边。如果一个点在所有直线的同一侧，那么它在多边形内。
下面是nirg给出的答案的c#版本，它来自RPI教授。请注意，使用来自RPI源代码的代码需要归属。
在顶部添加了一个边界框复选。然而，正如James Brown所指出的，主代码几乎和边界框检查本身一样快，所以边界框检查实际上会减慢整体操作，因为您正在检查的大多数点都在边界框内。所以你可以让边界框签出，或者另一种选择是预先计算多边形的边界框，如果它们不经常改变形状的话。
public bool IsPointInPolygon( Point p, Point[] polygon )
{
double minX = polygon[ 0 ].X;
double maxX = polygon[ 0 ].X;
double minY = polygon[ 0 ].Y;
double maxY = polygon[ 0 ].Y;
for ( int i = 1 ; i < polygon.Length ; i++ )
{
Point q = polygon[ i ];
minX = Math.Min( q.X, minX );
maxX = Math.Max( q.X, maxX );
minY = Math.Min( q.Y, minY );
maxY = Math.Max( q.Y, maxY );
}
if ( p.X < minX || p.X > maxX || p.Y < minY || p.Y > maxY )
{
return false;
}
// https://wrf.ecse.rpi.edu/Research/Short_Notes/pnpoly.html
bool inside = false;
for ( int i = 0, j = polygon.Length - 1 ; i < polygon.Length ; j = i++ )
{
if ( ( polygon[ i ].Y > p.Y ) != ( polygon[ j ].Y > p.Y ) &&
p.X < ( polygon[ j ].X - polygon[ i ].X ) * ( p.Y - polygon[ i ].Y ) / ( polygon[ j ].Y - polygon[ i ].Y ) + polygon[ i ].X )
{
inside = !inside;
}
}
return inside;
}
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