Python

Question

我有一个生成 HSV 掩码的脚本，并根据面积检测一些矩形：

我已经包括了一些区域限制，我的目标是只检测最低的底部框（上图中红色的那个）。

这是我的检测代码部分，我想在其中添加只能保留 1 个盒子的新功能：

cnts, hierarchy = cv2.findContours(mask_merged.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
for c in cnts:
        p_box = round((cv2.contourArea(c)/image_area)*100,1)
        print('p_box'+str(p_box))
        if (p_box <= 2.2) and (p_box >= 0.7):
            (x, y, w, h) = cv2.boundingRect(c)
            print(x,y,w,h)
            print(cv2.contourArea(c))
            cv2.imshow('frame_ocr_zone',image_src[y:y+h, x:x+w])
            cv2.rectangle(image_src, (x,y), (x+w,y+h), (0, 255, 0), 2)
            ## BEGIN - draw rotated rectangle
            rect = cv2.minAreaRect(c)
            box = cv2.boxPoints(rect)
            box = np.int0(box)
            cv2.drawContours(image_src,[box],0,(123, 245, 66),5)

            font = cv2.FONT_HERSHEY_SIMPLEX 
            color = (157,252,3) 
            thickness = 2
            fontScale = 1
            org = (x-50,y-20)

我正在寻找此处报告的代码： https://www.pyimagesearch.com/2015/04/20/sorting-contours-using-python-and-opencv/#pyi-pyimagesearch-plus-pricing-modal

我想修改这里的功能（排序）：

def sort_contours(cnts, method="left-to-right"):
    # initialize the reverse flag and sort index
    reverse = False
    i = 0
    # handle if we need to sort in reverse
    if method == "right-to-left" or method == "bottom-to-top":
        reverse = True
    # handle if we are sorting against the y-coordinate rather than
    # the x-coordinate of the bounding box
    if method == "top-to-bottom" or method == "bottom-to-top":
        i = 1
    # construct the list of bounding boxes and sort them from top to
    # bottom
    boundingBoxes = [cv2.boundingRect(c) for c in cnts]
    (cnts, boundingBoxes) = zip(*sorted(zip(cnts, boundingBoxes),
        key=lambda b:b[1][i], reverse=reverse))
    # return the list of sorted contours and bounding boxes
    return (cnts, boundingBoxes)

但我想不通只取回 1 个盒子（从上到下排序，并保持最大 c？）

提前感谢您的任何建议。

###更新我确实在这里找到了 post： Python opencv sorting contours

这段代码实际上能够做到这一点：

def get_contour_precedence(contour, cols):
    tolerance_factor = 10
    origin = cv2.boundingRect(contour)
    return ((origin[1] // tolerance_factor) * tolerance_factor) * cols + origin[0]

im, contours, h = cv2.findContours(img.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)

contours.sort(key=lambda x:get_contour_precedence(x, img.shape[1]))

Answer 1

首先我们将初始 X 和初始 Y 定义为 0，而且我们没有轮廓，因此 none 然后我们遍历轮廓检查 x 坐标和 y 坐标，因为底角有x 和 y 比其他的大

intial_x,intial_y = 0,0
my_cnt =None
for c in cnts:
(x, y, w, h) = cv2.boundingRect(c)
 if x>intial_x and y>intial_y:
   my_cnt = c
   intial_x = x+w
   intial_y =y+h
 else:
  continue

Answer 2

def get_contour_precedence(contour, cols):
    tolerance_factor = 10
    origin = cv2.boundingRect(contour)
    return ((origin[1] // tolerance_factor) * tolerance_factor) * cols + origin[0]

im, contours, h = cv2.findContours(img.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)

contours.sort(key=lambda x:get_contour_precedence(x, img.shape[1]))

Python - OpenCV - return 最低底部矩形

Python - OpenCV - return lowest bottom rectangle

opencv

opencv-python