如何在轮廓区域内找到并绘制最大的矩形？

Question

我想使用

比较具有特定像素的图像

(score, diff) = compare_ssim(grayA[y:y+h, x:x+w], grayB[y:y+h, x:x+w], full=True)

但是该函数只支持矩形ROI。我的投资回报率是一个轮廓。为了比较，我需要轮廓内最大的矩形。如何找到轮廓区域内最大的矩形？

示例图片

Answer 1

根据您的OP，我建议使用warpAffine 将ROI 旋转为矩形，因为ROI 已经是矩形但旋转了。这是一个简单的示例：

import cv2
import numpy as np

img = cv2.imread("1.png")
(H,W,c) = img.shape
print("shape = {},{}".format(H,W))
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_,thresh = cv2.threshold(gray,128,255,cv2.THRESH_BINARY_INV)

_,contours,_ = cv2.findContours(thresh,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)

res = np.zeros_like(img)
c = np.squeeze(contours[0])

# find rectangle's conner points
x = sorted(c, key=lambda a:a[0])
left = x[0]
right = x[-1]
y= sorted(c, key=lambda a:a[1])
top = y[0]
bottom = y[-1]

cv2.circle(img, (left[0],left[1]), 4, (0, 0, 255), -1)
cv2.circle(img, (right[0],right[1]), 4, (0, 0, 255), -1)
cv2.circle(img, (top[0],top[1]), 4, (0, 0, 255), -1)
cv2.circle(img, (bottom[0],bottom[1]), 4, (0, 0, 255), -1)

#calculate rectangle's shape
roi_w = int(np.sqrt((top[0]-right[0])*(top[0]-right[0])+(top[1]-right[1])*(top[1]-right[1])))
roi_h = int(np.sqrt((top[0]-left[0])*(top[0]-left[0])+(top[1]-left[1])*(top[1]-left[1])))

pts1 = np.float32([top,right,left])

# keep the top coords and calculate new coords for left and right
new_top = top
new_right = [top[0] + roi_w, top[1]]
new_left = [top[0], top[1] + roi_h]
pts2 = np.float32([new_top,new_right,new_left])

#rotate 
matrix = cv2.getAffineTransform(pts1, pts2)
result = cv2.warpAffine(img, matrix, (W,H))
cv2.drawContours(res, [contours[0]], 0, (0,255,0), 3)

# extract roi
roi = result[new_top[1]:new_left[1],new_top[0]:new_right[0]]

cv2.imshow("img",img)
cv2.imshow("result",result)
cv2.waitKey(0)

如何在轮廓区域内找到并绘制最大的矩形？

How to find and draw largest rectangle inside contour area?

python

opencv

boundary

drawrectangle

opencv-contour