48

是否有任何辅助方法可以绘制由cv2.minAreaRect()返回的旋转矩形,大概为((x1,y1),(x2,y2),angle)cv2.rectangle()不支持角度。而且由于返回的元组不是“RotatedRect”类(因为它似乎没有在 Python 绑定中实现),所以没有points()方法,如 C++ 教程“为轮廓创建边界旋转框和椭圆¶”中所示。

如何从线中绘制旋转的矩形 - 围绕中心点或给定的第一个点旋转?

4

5 回答 5

77
rect = cv2.minAreaRect(cnt)
box = cv2.cv.BoxPoints(rect) # cv2.boxPoints(rect) for OpenCV 3.x
box = np.int0(box)
cv2.drawContours(im,[box],0,(0,0,255),2)

应该做的伎俩。

来源:

1) http://opencvpython.blogspot.in/2012/06/contours-2-brotherhood.html

2) Python OpenCV Box2D

于 2013-08-13T11:35:25.443 回答
6

我知道这是很久以前问过的,但我想分享一种与接受的答案提出的不同的方法,也许这对其他人有帮助(实际上这已经在 C++ 中完成了,但似乎 python 仍然缺乏RotatedRect)。

这个想法是从一个角度、一个大小(W 和 H)和一个初始点定义一个旋转的矩形。这个初始点是相对左上角(相同大小的矩形的左上角,没有旋转角度)。从这里可以得到四个顶点,这样我们就可以用四条线绘制旋转的矩形。

class RRect:
  def __init__(self, p0, s, ang):
    self.p0 = (int(p0[0]),int(p0[1]))
    (self.W, self.H) = s
    self.ang = ang
    self.p1,self.p2,self.p3 = self.get_verts(p0,s[0],s[1],ang)
    self.verts = [self.p0,self.p1,self.p2,self.p3]

  def get_verts(self, p0, W, H, ang):
    sin = numpy.sin(ang/180*3.14159)
    cos = numpy.cos(ang/180*3.14159)
    P1 = (int(self.H*sin)+p0[0],int(self.H*cos)+p0[1])
    P2 = (int(self.W*cos)+P1[0],int(-self.W*sin)+P1[1])
    P3 = (int(self.W*cos)+p0[0],int(-self.W*sin)+p0[1])
    return [P1,P2,P3]

  def draw(self, image):
    print(self.verts)
    for i in range(len(self.verts)-1):
      cv2.line(image, (self.verts[i][0], self.verts[i][1]), (self.verts[i+1][0],self.verts[i+1][1]), (0,255,0), 2)
    cv2.line(image, (self.verts[3][0], self.verts[3][1]), (self.verts[0][0], self.verts[0][1]), (0,255,0), 2)

(W, H) = (30,60)
ang = 35 #degrees
P0 = (50,50)
rr = RRect(P0,(W,H),ang)
rr.draw(image)
cv2.imshow("Text Detection", image)
cv2.waitKey(200)

我想,可以使用类似的方法来定义旋转矩形的中心,而不是其相对的左上角初始点,但我还没有尝试过。

于 2019-10-23T06:12:04.020 回答
6

这是一个绘制旋转矩形的具体示例。这个想法是获得具有Otsu 阈值的二值图像,然后使用cv2.findContours(). 我们可以使用 获得旋转的矩形,使用 获得cv2.minAreaRect()四个角顶点cv2.boxPoints()。要绘制矩形,我们可以使用cv2.drawContours()cv2.polylines()


输入->输出

代码

import cv2
import numpy as np

# Load image, convert to grayscale, Otsu's threshold for binary image
image = cv2.imread('1.jpg')
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
thresh = cv2.threshold(gray, 0, 255, cv2.THRESH_BINARY_INV + cv2.THRESH_OTSU)[1]

# Find contours, find rotated rectangle, obtain four verticies, and draw 
cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = cnts[0] if len(cnts) == 2 else cnts[1]
rect = cv2.minAreaRect(cnts[0])
box = np.int0(cv2.boxPoints(rect))
cv2.drawContours(image, [box], 0, (36,255,12), 3) # OR
# cv2.polylines(image, [box], True, (36,255,12), 3)

cv2.imshow('image', image)
cv2.waitKey()
于 2020-01-29T03:19:41.920 回答
3

延伸 Tobias Hermann 的回答:如果您没有轮廓,而是由其中心点、尺寸和角度定义的旋转矩形:

import cv2
import numpy as np

# given your rotated rectangle is defined by variables used below

rect = ((center_x, center_y), (dim_x, dim_y), angle)
box = cv2.cv.BoxPoints(rect) # cv2.boxPoints(rect) for OpenCV 3.x
box = np.int0(box)
cv2.drawContours(im,[box],0,(0,0,255),2)
于 2021-05-21T14:18:48.720 回答
0

根据@ smajtks的回答,我根据其中心而不是其相对的左上角初始点来定义旋转的矩形。这是代码:

class RRect_center:
  def __init__(self, p0, s, ang):
    (self.W, self.H) = s # rectangle width and height
    self.d = math.sqrt(self.W**2 + self.H**2)/2.0 # distance from center to vertices    
    self.c = (int(p0[0]+self.W/2.0),int(p0[1]+self.H/2.0)) # center point coordinates
    self.ang = ang # rotation angle
    self.alpha = math.radians(self.ang) # rotation angle in radians
    self.beta = math.atan2(self.H, self.W) # angle between d and horizontal axis
    # Center Rotated vertices in image frame
    self.P0 = (int(self.c[0] - self.d * math.cos(self.beta - self.alpha)), int(self.c[1] - self.d * math.sin(self.beta-self.alpha))) 
    self.P1 = (int(self.c[0] - self.d * math.cos(self.beta + self.alpha)), int(self.c[1] + self.d * math.sin(self.beta+self.alpha))) 
    self.P2 = (int(self.c[0] + self.d * math.cos(self.beta - self.alpha)), int(self.c[1] + self.d * math.sin(self.beta-self.alpha))) 
    self.P3 = (int(self.c[0] + self.d * math.cos(self.beta + self.alpha)), int(self.c[1] - self.d * math.sin(self.beta+self.alpha))) 

    self.verts = [self.P0,self.P1,self.P2,self.P3]

  def draw(self, image):
    # print(self.verts)
    for i in range(len(self.verts)-1):
      cv2.line(image, (self.verts[i][0], self.verts[i][1]), (self.verts[i+1][0],self.verts[i+1][1]), (0,255,0), 2)
    cv2.line(image, (self.verts[3][0], self.verts[3][1]), (self.verts[0][0], self.verts[0][1]), (0,255,0), 2)

(W, H) = (30,60)
ang = 35 #degrees
P0 = (50,50)
rr = RRect_center(P0,(W,H),ang)
rr.draw(image)
cv2.imshow("Text Detection", image)
cv2.waitKey(200

这里,矩形围绕其中心旋转,而不是从初始点 P0 开始旋转。

于 2020-07-16T13:27:15.630 回答