blocks|key|259877|text|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)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|259878|应该能行得通。|unstyled|259879|资料来源：|259880|1)+http://opencvpython.blogspot.in/2012/06/contours-2-brotherhood.html|offset|length|259881|2)+Python+OpenCV+Box2D|259882|entityMap|0|LINK|mutability|MUTABLE|url|http://opencvpython.blogspot.in/2012/06/contours-2-brotherhood.html|1|https://stackoverflow.com/questions/11779100/python-opencv-box2d^0|0|0|0|3|1V|0|0|3|J|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|11|8|@]|9|@$K|12|L|13|1|14]]|A|$]]|$1|M|3|N|5|F|7|15|8|@]|9|@$K|16|L|17|1|18]]|A|$]]|$1|O|3|-4|5|F|7|19|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]]]

<pre><code>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)
</code></pre>

should do the trick.

sources:

1) <a href="http://opencvpython.blogspot.in/2012/06/contours-2-brotherhood.html" rel="noreferrer">http://opencvpython.blogspot.in/2012/06/contours-2-brotherhood.html</a>

2) <a href="https://stackoverflow.com/questions/11779100/python-opencv-box2d">Python OpenCV Box2D</a>

blocks|key|43203|text|我知道这在很久以前就被问到了，但是我想分享一种不同于公认答案所提出的方法，也许这可以对其他人有所帮助(实际上在C%2B%2B中已经这样做了，但是似乎python仍然缺乏RotatedRect类)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|43204|这个想法是从一个角度，一个尺寸(W和H)和一个初始点定义一个旋转的矩形。此初始点是相对左上角(没有旋转角度的相同大小矩形的左上角)。从这里，可以获得四个顶点，这允许我们绘制具有四条线的旋转矩形。|43205|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)%2Bp0[0],int(self.H*cos)%2Bp0[1])
++++P2+=+(int(self.W*cos)%2BP1[0],int(-self.W*sin)%2BP1[1])
++++P3+=+(int(self.W*cos)%2Bp0[0],int(-self.W*sin)%2Bp0[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%2B1][0],self.verts[i%2B1][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)|code-block|syntax|javascript|43206|我猜，可以使用类似的方法来定义旋转矩形的中心，而不是其相对左上角的初始点，但我还没有尝试过。|43207|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.opencv.org/4.1.2/db/dd6/classcv_1_1RotatedRect.html^0|28|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

I know this was asked long ago, but I would like to share a different approach as the one proposed by the accepted answer, maybe this could be helpful for someone else (actually this has been done before in C++, but it seems python still lacks of <a href="https://docs.opencv.org/4.1.2/db/dd6/classcv_1_1RotatedRect.html" rel="nofollow noreferrer">RotatedRect</a> class).

The idea is to define a rotated rectangle from an angle, a size (W and H) and an initial point. This initial point is the relative top-left corner (the top-left corner of the same size rectangle with no rotation angle). From here, the four vertices can be obtained, which allows us to draw the rotated rectangle with four lines. 

<pre><code>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)
</code></pre>

I guess, a similar approach can be used to define the rotated rectangle in terms of its center instead of its relative top-left initial point, but I haven't tried it yet.

blocks|key|260035|text|下面是一个绘制旋转矩形的具体示例。其思想是使用Otsu's+threshold获取二进制图像，然后使用cv2.findContours()查找轮廓。我们可以使用cv2.minAreaRect()获得旋转的矩形，使用cv2.boxPoints()获得四个角点。要绘制矩形，可以使用cv2.drawContours()或cv2.polylines()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|260036|260037|输入->输出|260038|📷|atomic|260039|260040|代码|260041|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+%2B+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()|code-block|syntax|javascript|260042|entityMap|0|LINK|mutability|MUTABLE|url|https://opencv24-python-tutorials.readthedocs.io/en/latest/py_tutorials/py_imgproc/py_thresholding/py_thresholding.html#otsus-binarization|1|https://docs.opencv.org/2.4/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html#findcontours|2|https://opencv24-python-tutorials.readthedocs.io/en/latest/py_tutorials/py_imgproc/py_contours/py_contours_begin/py_contours_begin.html|3|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/094901c3b587a3b58109f452f3110b62.jpeg|imageAlt|4|https://ask.qcloudimg.com/http-save/yehe-900000/ea65c68bde6dac9151949899b0cbbd9e.png^0|1F|I|28|H|2Z|F|3V|I|4E|F|N|G|0|1F|I|1|3V|I|2|0|0|2|2|0|0|1|3|0|0|1|4|0|0|0^^$0|@$1|2|3|4|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]]|D|@$9|1O|A|1P|1|1Q]|$9|1R|A|1S|1|1T]|$9|1U|A|1V|1|1W]]|E|$]]|$1|F|3|-4|5|6|7|1X|8|@]|D|@]|E|$]]|$1|G|3|H|5|6|7|1Y|8|@$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|I|3|J|5|K|7|21|8|@]|D|@$9|22|A|23|1|24]]|E|$]]|$1|L|3|J|5|K|7|25|8|@]|D|@$9|26|A|27|1|28]]|E|$]]|$1|M|3|N|5|6|7|29|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|2A|8|@]|D|@]|E|$R|S]]|$1|T|3|-4|5|6|7|2B|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10]]|11|$5|W|X|Y|E|$Z|12]]|13|$5|W|X|Y|E|$Z|14]]|15|$5|16|X|17|E|$18|19|1A|-4]]|1B|$5|16|X|17|E|$18|1C|1A|-4]]]]

Here's a concrete example to draw the rotated rectangle. The idea is to obtain a binary image with <a href="https://opencv24-python-tutorials.readthedocs.io/en/latest/py_tutorials/py_imgproc/py_thresholding/py_thresholding.html#otsus-binarization" rel="nofollow noreferrer">Otsu's threshold</a> then find contours using <a href="https://docs.opencv.org/2.4/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html#findcontours" rel="nofollow noreferrer"><code>cv2.findContours()</code></a>. We can obtain the rotated rectangle using <code>cv2.minAreaRect()</code> and the four corner vertices using <code>cv2.boxPoints()</code>. To draw the rectangle we can use <a href="https://opencv24-python-tutorials.readthedocs.io/en/latest/py_tutorials/py_imgproc/py_contours/py_contours_begin/py_contours_begin.html" rel="nofollow noreferrer"><code>cv2.drawContours()</code></a> or <code>cv2.polylines()</code>.
<hr />
Input <code>-&gt;</code> Output
<img src="https://i.stack.imgur.com/2Q1Sj.jpg" width="325">
<img src="https://i.stack.imgur.com/vxLsl.jpg" width="325">
Code
<pre><code>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()
</code></pre>

blocks|key|43329|text|基于@smajtks的回答，我根据旋转矩形的中心而不是相对左上角的初始点来定义它。代码如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|43330|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+%2B+self.H**2)/2.0+#+distance+from+center+to+vertices++++
++++self.c+=+(int(p0[0]%2Bself.W/2.0),int(p0[1]%2Bself.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+%2B+self.alpha)),+int(self.c[1]+%2B+self.d+*+math.sin(self.beta%2Bself.alpha)))+
++++self.P2+=+(int(self.c[0]+%2B+self.d+*+math.cos(self.beta+-+self.alpha)),+int(self.c[1]+%2B+self.d+*+math.sin(self.beta-self.alpha)))+
++++self.P3+=+(int(self.c[0]+%2B+self.d+*+math.cos(self.beta+%2B+self.alpha)),+int(self.c[1]+-+self.d+*+math.sin(self.beta%2Bself.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%2B1][0],self.verts[i%2B1][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|code-block|syntax|javascript|43331|在这里，矩形是围绕其中心旋转的，而不是从初始点P0旋转。|43332|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/2836854/smajtkst^0|3|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Based on @<a href="https://stackoverflow.com/users/2836854/smajtkst">smajtks</a>'s answer I define the rotated rectangle in terms of its center instead of its relative top-left initial point. Here is the code:
<pre><code>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(&quot;Text Detection&quot;, image)
cv2.waitKey(200
</code></pre>
Here, the rectangle is rotated around its center, not from the initial point P0.

blocks|key|260126|text|作为Tobias+Hermann答案的延伸:如果你没有轮廓，而是一个由圆心、尺寸和角度定义的旋转矩形：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|260127|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)|code-block|syntax|javascript|260128|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In extension to Tobias Hermann's answer: in case you don't have a contour, but a rotated rectangle defined by its center point, dimensions and angle:
<pre class="lang-py prettyprint-override"><code>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)
</code></pre>

Are there any helper methods to draw a rotated rectangle that is returned by <a href="http://docs.opencv.org/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html?highlight=minarearect#cv2.minAreaRect" rel="noreferrer">cv2.minAreaRect()</a> presumably as <code>((x1,y1),(x2,y2),angle)</code>? <a href="http://docs.opencv.org/modules/core/doc/drawing_functions.html#rectangle" rel="noreferrer">cv2.rectangle()</a> does not support an angle.
And since the tuple returned is not of the "RotatedRect" class (because it seems to not be implemented in the Python bindings) there is no <code>points()</code> method, as shown in the C++ tutorial <a href="http://docs.opencv.org/doc/tutorials/imgproc/shapedescriptors/bounding_rotated_ellipses/bounding_rotated_ellipses.html#bounding-rotated-ellipses" rel="noreferrer">"Creating Bounding rotated boxes and ellipses for contours¶"</a>.

How could a rotated rectangle be drawn from lines - rotate about the center point or the first point given?

OpenCV Python: Draw minAreaRect ( RotatedRect not implemented)

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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有没有什么帮助器方法来绘制一个旋转的矩形，这个矩形大概是由返回的，比如((x1,y1),(x2,y2),angle)？不支持角度。由于返回的元组不属于"RotatedRect“类(因为它似乎不是在Python绑定中实现的)，因此没有points()方法，如C++教程中所示。如何从直线绘制旋转的矩形-围绕中心点或给定的第...

问OpenCV Python:绘制minAreaRect (未实现RotatedRect )
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问OpenCV Python:绘制minAreaRect (未实现RotatedRect )EN