我有一组从多边形派生的点。我正在测试几种解决方案以获得最小面积或矩形。作为基准，我使用的是 ArcGIS (10.1)。
points = [(560036.4495758876, 6362071.890493258),
(560036.4495758876, 6362070.890493258),
(560036.9495758876, 6362070.890493258),
(560036.9495758876, 6362070.390493258),
(560037.4495758876, 6362070.390493258),
(560037.4495758876, 6362064.890493258),
(560036.4495758876, 6362064.890493258),
(560036.4495758876, 6362063.390493258),
(560035.4495758876, 6362063.390493258),
(560035.4495758876, 6362062.390493258),
(560034.9495758876, 6362062.390493258),
(560034.9495758876, 6362061.390493258),
(560032.9495758876, 6362061.390493258),
(560032.9495758876, 6362061.890493258),
(560030.4495758876, 6362061.890493258),
(560030.4495758876, 6362061.390493258),
(560029.9495758876, 6362061.390493258),
(560029.9495758876, 6362060.390493258),
(560029.4495758876, 6362060.390493258),
(560029.4495758876, 6362059.890493258),
(560028.9495758876, 6362059.890493258),
(560028.9495758876, 6362059.390493258),
(560028.4495758876, 6362059.390493258),
(560028.4495758876, 6362058.890493258),
(560027.4495758876, 6362058.890493258),
(560027.4495758876, 6362058.390493258),
(560026.9495758876, 6362058.390493258),
(560026.9495758876, 6362057.890493258),
(560025.4495758876, 6362057.890493258),
(560025.4495758876, 6362057.390493258),
(560023.4495758876, 6362057.390493258),
(560023.4495758876, 6362060.390493258),
(560023.9495758876, 6362060.390493258),
(560023.9495758876, 6362061.890493258),
(560024.4495758876, 6362061.890493258),
(560024.4495758876, 6362063.390493258),
(560024.9495758876, 6362063.390493258),
(560024.9495758876, 6362064.390493258),
(560025.4495758876, 6362064.390493258),
(560025.4495758876, 6362065.390493258),
(560025.9495758876, 6362065.390493258),
(560025.9495758876, 6362065.890493258),
(560026.4495758876, 6362065.890493258),
(560026.4495758876, 6362066.890493258),
(560026.9495758876, 6362066.890493258),
(560026.9495758876, 6362068.390493258),
(560027.4495758876, 6362068.390493258),
(560027.4495758876, 6362068.890493258),
(560027.9495758876, 6362068.890493258),
(560027.9495758876, 6362069.390493258),
(560028.4495758876, 6362069.390493258),
(560028.4495758876, 6362069.890493258),
(560033.4495758876, 6362069.890493258),
(560033.4495758876, 6362070.390493258),
(560033.9495758876, 6362070.390493258),
(560033.9495758876, 6362070.890493258),
(560034.4495758876, 6362070.890493258),
(560034.4495758876, 6362071.390493258),
(560034.9495758876, 6362071.390493258),
(560034.9495758876, 6362071.890493258),
(560036.4495758876, 6362071.890493258)]
一种解决方案是使用cv.MinAreaRect2()OpenCV。
函数 cv.MinAreaRect2 通过为该集合构建凸包并将旋转卡尺技术应用于该包，为 2D 点集合找到一个最小区域的外接矩形。
import cv
# (x,y) - center point of the box
# (w,h) - width and height of the box
# theta - angle of rotation
((x,y),(w,h),th) = cv.MinAreaRect2(points)
print ((x,y),(w,h),th)
((560029.3125, 6362065.5), (10.28591251373291, 18.335756301879883), -63.43495178222656)
# get vertex
box_vtx = cv.BoxPoints(((x,y),(w,h),th))
print box_vtx
((560035.1875, 6362074.0), (560018.8125, 6362066.0), (560023.4375, 6362057.0), (560039.8125, 6362065.0)
当我转换box_vtxshapefile 以便在 ArcGIS 中查看并与最小边界几何（数据管理）进行比较时，我可以看到这种差异，如下图所示，其中：
- 红色 = 多边形的边界
- 蓝色 = ArGIS (10.1) 中面积最小的矩形
- 黄色和黑色 = OpenCV 中面积最小的矩形
使用 OpenCV 与本文中提出的解决方案进行比较：
import osgeo.gdal, ogr
import cv
poly = "...\\polygon.shp"
shp = osgeo.ogr.Open(poly)
layer = shp.GetLayer()
feature = layer.GetFeature(0)
geometry = feature.GetGeometryRef()
pts = geometry.GetGeometryRef(0)
# get point of the polygon border (the points above)
points = []
for p in xrange(pts.GetPointCount()):
points.append((pts.GetX(p), pts.GetY(p)))
# Convex Hull
CH1 = geometry.ConvexHull
# i didn't find a method to extarct the points
print CH1()
# works with openCV
cvxHull = cv.ConvexHull2(points, cv.CreateMemStorage(), return_points=True)
print cvxHull
<cv2.cv.cvseq object at 0x00000000067CCF90>