计算两个旋转矩形的交集——Python

知识准备

cv2.rotatedRectangleIntersection(rect1, rect2)

计算两个旋转矩形的交集，返回值 0,1,2分别表示没有，有，包含；以及交点的坐标的一个array。很遗憾的是这个坐标是不是逆时针也不是顺时针。而cv2.contourArea()需要点是顺时针或者逆时针。

例子程序

# 中心点 矩形的w h, 旋转的theta（角度，不是弧度）
rect1 = ((0,0),(1,1),45)
rect2 = ((1.5,0),(4,3),0)
r1 = cv2.rotatedRectangleIntersection(rect1, rect2)
print r1

多边形的点逆时针排序

Python 的自定义的排序程序只能返回0,1，-1。没有true，false。所以没

return a > b

种写法

主要是根据所求出的重心，按照向量的叉积，然后排序

排序函数如下：

def cmp(a, b):
if a.x >= 0 and b.x < 0:
return -1
if a.x == 0 and b.x == 0:
# return a.y > b.y
if a.y > b.y:
return -1
elif a.y < b.y:
return 1
return 0
det = (a.x - c.x) * (b.y - c.y) - (b.x - c.x) * (a.y - c.y)
if det < 0:
return 1
if det > 0:
return -1
d1 = (a.x - c.x) * (a.x - c.x) + (a.y - c.y) * (a.y - c.y)
d2 = (b.x - c.x) * (b.x - c.x) + (b.y - c.y) * (b.y - c.y)
# return d1 > d2
if d1 > d2:
return -1
elif d1 < d2:
return 1
return 0

cv2.contourArea()

计算点所包围的面积

坑点 dtype=’float32’

# 可行
r = ((2,-2),(-2,-2),(-2,2),(2,2))
r2 = cv2.contourArea(np.array(r))
# 可行
r = [[1,-1],[-1,-1],[-1,1],[1,1]]
r2 = cv2.contourArea(np.array(r))
# 不可行
r = np.full((len_p, 2), 0.0)
for i in range(len(pp)):
print pp[i].x, pp[i].y
r[i][0] = pp[i].x
r[i][1] = pp[i].y
r2 = cv2.contourArea(r)
# 可行
r = np.full((len_p, 2), 0.0, dtype='float32')
for i in range(len(pp)):
print pp[i].x, pp[i].y
r[i][0] = pp[i].x
r[i][1] = pp[i].y
r2 = cv2.contourArea(r)

代码

import cv2
import  numpy as np

class Point(object):
def __init__(self, x, y):
self.x = x
self.y = y

def cmp(a, b, c):
if a.x >= 0 and b.x < 0:
return -1
if a.x == 0 and b.x == 0:
# return a.y > b.y
if a.y > b.y:
return -1
elif a.y < b.y:
return 1
return 0
det = (a.x - c.x) * (b.y - c.y) - (b.x - c.x) * (a.y - c.y)
if det < 0:
return 1
if det > 0:
return -1
d1 = (a.x - c.x) * (a.x - c.x) + (a.y - c.y) * (a.y - c.y)
d2 = (b.x - c.x) * (b.x - c.x) + (b.y - c.y) * (b.y - c.y)
# return d1 > d2
if d1 > d2:
return -1
elif d1 < d2:
return 1
return 0
## centerx, centery w, h, thetta
rect1 = ((0,0),(1,1),45)
rect2 = ((1.5,0),(4,3),0)

r1 = cv2.rotatedRectangleIntersection(rect1, rect2)

x = 0
y = 0
p = []
len_p = r1[1].shape[0]
for i in range(len_p):
p.append(Point(r1[1][i][0][0], r1[1][i][0][1]))
x += r1[1][i][0][0]
y += r1[1][i][0][1]

c = Point(x / len_p, y/len_p)

pp = sorted(p, lambda x,y: cmp(x, y, c))
r = np.full((len_p, 2), 0.0, dtype='float32')
for i in range(len(pp)):
print pp[i].x, pp[i].y
r[i][0] = pp[i].x
r[i][1] = pp[i].y
r2 = cv2.contourArea(r)
print r2