检测复选框是否被选中的最佳方法 [英] Best way to detect if checkbox is ticked

问题描述

我的工作

1. 扫描纸张


2. 检查水平线和垂直线


3. 检测复选框


4. 如何知道复选框是否已选中

1. Scan the paper
2. Check horizontal and vertical line
3. Detect checkbox
4. How to know checkbox is ticked or not

在这一点上，我认为我可以通过使用层次和轮廓来找到它：以下是我的工作

At this point, I thought I could find it by using Hierarchical and Contours: Below is my work

for i in range (len( contours_region)):     #I already have X,Y,W,H of the checkbox through 
    #print(i)                               #cv2.connectedComponentsWithStats
    x = contours_region[i][0][1]            #when detecting checkbox
    x_1 = contours_region[i][2][1]
    y = contours_region[i][0][0]
    y_1 = contours_region[i][2][0]

    image_copy= image.copy()

    X,Y,W,H = contours_info[i]
    cv2.drawContours(image_copy, [numpy.array([[[X,Y]],[[X+W,Y]],[[X+W,Y+H]],[[X,Y+H]]])], 0, (0,0,255),2)
    gray = cv2.cvtColor(image_copy, cv2.COLOR_BGR2GRAY)
    ret,bw = cv2.threshold(gray,220,255,cv2.THRESH_BINARY_INV)
    
    contours,hierarchy = cv2.findContours(bw[x:x_1, y:y_1], cv2.RETR_CCOMP,1)
    
    print('-----Hierarchy-----')
    print(hierarchy)
    print('-----Number of Contours : '+ str(len(contours)))
    cv2.imshow('a', image_copy)
    cv2.waitKey(0)

我得到了这个结果（一些高轮廓，一些高层次）

I got this result (some high contours, some high hierarchy)

-----Hierarchy-----
[[[-1 -1  1 -1]
  [ 2 -1 -1  0]
  [ 3  1 -1  0]
  [ 4  2 -1  0]
  [ 5  3 -1  0]
  [ 6  4 -1  0]
  [ 7  5 -1  0]
  [-1  6 -1  0]]]
-----Number of Contours : 8

---

另一个结果：

---

Another result:

轮廓低，层次低

-----Hierarchy-----
[[[-1 -1  1 -1]
  [ 2 -1 -1  0]
  [-1  1 -1  0]]]
-----Number of Contours : 3

---

但是，在某些情况下，它没有被打勾，但仍然获得了很高的结果，这并不完美

---

However, it's not perfect some case where it's not ticked but still got a really high result

[[[-1 -1  1 -1]
  [ 2 -1 -1  0]
  [ 3  1 -1  0]
  [ 4  2 -1  0]
  [ 5  3 -1  0]
  [-1  4 -1  0]]]
-----Number of Contours : 6

通常，在检查了整个数据之后，刻度和未刻度之间的差距并不令人信服。大约30％的盒子，给出了错误的结果。因此，我真的希望有一个更好的方法。

In general, After review the whole data, the gap is not convincing between ticked and not ticked. Around 30% of boxes, giving the wrong result. Therefore, really wish to have a better method.

推荐答案

我认为腐蚀函数可以为您提供帮助。使用侵蚀使刻度线更大，然后计算非零像素。
在这里您可以找到基本知识：

I think erode function help you. Use erosion to make the ticks bigger then count the non zero pixels. Here You can find the basics:

import cv2 
import numpy as np
from google.colab.patches import cv2_imshow
img = cv2.imread("image.png");
cv2_imshow(img)
kernel = np.ones((3, 3), np.uint8) 

better_image = cv2.erode(img,kernel)
cv2_imshow(better_image)

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