随着现代图像处理和人工智能技术的快速发展，不少学者尝试讲CV应用到教学领域，能够代替老师去阅卷，将老师从繁杂劳累的阅卷中解放出来，从而进一步有效的推动教学质量上一个台阶。
传统的人工阅卷，工作繁琐，效率低下，进度难以控制且容易出现试卷遗漏未改、登分失误等现象。
现代的“机器阅卷”，工作便捷、效率高、易操作，只需要一个相机(手机)，拍照即可获取成绩，可以导入Excel表格便于存档管理。
下面我们从代码实现的角度来解释一下我们这个简易答题卡识别系统的工作原理。第一步，导入工具包及一系列的预处理
import numpy as npimport argparseimport imutilsimport cv2# 设置参数ap = argparse.ArgumentParser()ap.add_argument('-i', '--image', default='test_01.png')args = vars(ap.parse_args())# 正确答案ANSWER_KEY = {0: 1, 1: 4, 2: 0, 3: 3, 4: 1} #def order_points(pts): # 一共4个坐标点 rect = np.zeros((4, 2), dtype = 'float32') # 按顺序找到对应坐标0,1,2,3分别是 左上，右上，右下，左下 # 计算左上，右下 s = pts.sum(axis = 1) rect[0] = pts[np.argmin(s)] rect[2] = pts[np.argmax(s)] # 计算右上和左下 diff = np.diff(pts, axis = 1) rect[1] = pts[np.argmin(diff)] rect[3] = pts[np.argmax(diff)] return rectdef four_point_transform(image, pts): # 获取输入坐标点 rect = order_points(pts) (tl, tr, br, bl) = rect # 计算输入的w和h值 widthA = np.sqrt(((br[0]-bl[0])** 2) + ((br[1]-bl[1])**2)) widthB = np.sqrt(((tr[0] -tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2)) maxWidth = max(int(widthA), int(widthB)) heightA = np.sqrt(((tr[0]-br[0])**2)+((tr[1]-br[1])**2)) heightB = np.sqrt(((tl[0]-bl[0])**2)+((tl[1]-bl[1])**2)) maxHeight = max(int(heightA), int(heightB)) # 变换后对应坐标位置 dst = np.array([ [0, 0], [maxWidth - 1, 0], [maxWidth - 1, maxHeight - 1], [0, maxHeight - 1]], dtype = 'float32') # 计算变换矩阵 M = cv2.getPerspectiveTransform(rect, dst) warped = cv2.warpPerspective(image, M, (maxWidth, maxHeight)) return warped # 返回变换后结果def sort_contours(cnts, method='left-to-right'): reverse = False i = 0 if method == 'right-to-left' or method == 'bottom-to-top': reverse = True if method == 'top-to-bottom' or method == 'bottom-to-top': i = 1 boundingBoxes = [cv2.boundingRect(c) for c in cnts] (cnts, boundingBoxes) = zip(*sorted(zip(cnts, boundingBoxes), key=lambda b: b[1][i], reverse=reverse)) return cnts, boundingBoxesdef cv_show(name,img): cv2.imshow(name, img) cv2.waitKey(0) cv2.destroyAllWindows() image = cv2.imread(args['image'])contours_img = image.copy()gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)blurred = cv2.GaussianBlur(gray, (5, 5), 0)edged = cv2.Canny(blurred, 75, 200)# 轮廓检测cnts = cv2.findContours(edged.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[1]cv2.drawContours(contours_img,cnts,-1,(0,0,255),3) docCnt = None# 确保检测到了if len(cnts) > 0: # 根据轮廓大小进行排序 cnts = sorted(cnts, key=cv2.contourArea, reverse=True) for c in cnts: # 遍历每一个轮廓 # 近似 peri = cv2.arcLength(c, True) approx = cv2.approxPolyDP(c, 0.02 * peri, True) # 准备做透视变换 if len(approx) == 4: docCnt = approx break# 执行透视变换warped = four_point_transform(gray, docCnt.reshape(4, 2))thresh = cv2.threshold(warped, 0, 255, cv2.THRESH_BINARY_INV | cv2.THRESH_OTSU)[1] thresh_Contours = thresh.copy()# 找到每一个圆圈轮廓cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[1]cv2.drawContours(thresh_Contours,cnts,-1,(0,0,255),3) questionCnts = []for c in cnts:# 遍历 # 计算比例和大小 (x, y, w, h) = cv2.boundingRect(c) ar = w / float(h) # 根据实际情况指定标准 if w >= 20 and h >= 20 and ar >= 0.9 and ar <= 1.1: questionCnts.append(c)# 按照从上到下进行排序questionCnts = sort_contours(questionCnts, method='top-to-bottom')[0]correct = 0# 每排有5个选项for (q, i) in enumerate(np.arange(0, len(questionCnts), 5)): cnts = sort_contours(questionCnts[i:i + 5])[0] bubbled = None for (j, c) in enumerate(cnts): # 遍历每一个结果 # 使用mask来判断结果 mask = np.zeros(thresh.shape, dtype='uint8') cv2.drawContours(mask, [c], -1, 255, -1) #-1表示填充 # 通过计算非零点数量来算是否选择这个答案 mask = cv2.bitwise_and(thresh, thresh, mask=mask) total = cv2.countNonZero(mask) # 通过阈值判断 if bubbled is None or total > bubbled[0]: bubbled = (total, j) # 第二步，与正确答案进行对比 color = (0, 0, 255) k = ANSWER_KEY[q] # 判断正确 if k == bubbled[1]: color = (0, 255, 0) correct += 1 cv2.drawContours(warped, [cnts[k]], -1, color, 3) #绘图 #正确率的文本显示score = (correct / 5.0) * 100print('[INFO] score: {:.2f}%'.format(score))cv2.putText(warped, '{:.2f}%'.format(score), (10, 30), cv2.FONT_HERSHEY_SIMPLEX, 0.9, (0, 0, 255), 2)cv2.imshow('Input', image)cv2.imshow('Output', warped)cv2.waitKey(0)
最终实现的效果如下：
