圖像輪廓最大內接矩形的求法

平時工作中圖像處理經常會用到圖像最大輪廓及最小外接矩形的獲取：

計算過程如下：

 

img = cv2.imread(path)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
blurred = cv2.blur(gray, (9, 9))
_, thresh = cv2.threshold(blurred, 155, 255, cv2.THRESH_BINARY)
_, cnts, _ = cv2.findContours( thresh.copy(), cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
c = sorted(cnts, key=cv2.contourArea, reverse=True)[0]
rect = cv2.minAreaRect(c)
#box即為最小外接矩形坐標
box = np.int0(cv2.boxPoints(rect))
cv2.drawContours(img, [box], -1, (0, 255, 0), 3)
cv2.imshow("Image", img)
cv2.imwrite("pic.jpg", img)
cv2.waitKey(0)

 

然而有時候我們需要的是最大內接矩形：

從輪廓中所有坐標中獲取其中4個坐標即可：

獲取過程如下：

 

def order_points(pts):
    # pts為輪廓坐標
    # 列表中存儲元素分別為左上角，右上角，右下角和左下角
    rect = np.zeros((4, 2), dtype = "float32")
    # 左上角的點具有最小的和，而右下角的點具有最大的和
    s = pts.sum(axis = 1)
    rect[0] = pts[np.argmin(s)]
    rect[2] = pts[np.argmax(s)]
    # 計算點之間的差值
    # 右上角的點具有最小的差值,
    # 左下角的點具有最大的差值
    diff = np.diff(pts, axis = 1)
    rect[1] = pts[np.argmin(diff)]
    rect[3] = pts[np.argmax(diff)]
    # 返回排序坐標(依次為左上右上右下左下)
    return rect

 

　

img = cv2.imread(path)
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
blurred = cv2.blur(gray, (9, 9))
_, thresh = cv2.threshold(blurred, 155, 255, cv2.THRESH_BINARY)
_, cnts, _ = cv2.findContours( thresh.copy(), cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
c = sorted(cnts, key=cv2.contourArea, reverse=True)[0]
先找出輪廓點

rect = order_points(c.reshape(c.shape[0], 2))
print(rect)
xs = [i[0] for i in rect]
ys = [i[1] for i in rect]
xs.sort()
ys.sort()
#內接矩形的坐標為
print(xs[1],xs[2],ys[1],ys[2])