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传统的手动评分方法耗时且容易出错,自动化评分可以可以显著提高评分过程的速度和准确性、减少人工成本。
答题卡图片处理效果如下:
# 读图片
img = cv2.imread('./images/test_01.png')
cv_show('img', img)
# 变成黑白图片
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
# 去掉一些噪点
blurred = cv2.GaussianBlur(gray, (5, 5), 0)
cv_show('blurred', blurred)
# 边缘检测
edged = cv2.Canny(blurred, 75, 200)
cv_show('edged', edged)
# 检测轮廓 cnts = cv2.findContours(edged.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[1] # 画轮廓会修改被画轮廓的图. contours_img = img.copy() cv2.drawContours(contours_img, cnts, -1, (0, 0, 255), 3) cv_show('contous_img', contours_img) # 确保我们拿到的轮廓是答题卡的轮廓. if len(cnts) > 0: # 根据轮廓面积对轮廓进行排序. cnts = sorted(cnts, key=cv2.contourArea, reverse=True) # 遍历每一个轮廓 for c in cnts: # 计算周长 perimeter = cv2.arcLength(c, True) # 得到近似的轮廓 approx = cv2.approxPolyDP(c, 0.02 * perimeter, True) # 近似完了之后, 应该只剩下4个角的坐标. # print(c) # print(approx) if len(approx) == 4: # 保存approx docCnt = approx # 找到答题卡近似轮廓, 直接推荐. break
# 把透视变换功能封装成一个函数 def four_point_transform(image, pts): # 对输入的4个坐标排序 rect = order_points(pts) (tl, tr, br, bl) = rect # 空间中两点的距离 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) max_width = 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) max_height = max(int(heightA), int(heightB)) # 构造变换之后的对应坐标位置. dst = np.array([ [0, 0], [max_width - 1, 0], [max_width - 1, max_height - 1], [0, max_height - 1]], dtype='float32') # 计算变换矩阵 M = cv2.getPerspectiveTransform(rect, dst) # 透视变换 warped = cv2.warpPerspective(image, M, (max_width, max_height)) return warped
透视变化后二值化:
thresh = cv2.threshold(warped, 0, 255, cv2.THRESH_BINARY_INV | cv2.THRESH_OTSU)[1]
cv_show('thresh', thresh)
此时图片预处理好后如果需要 OCR 文本识别可借助 tesseract 工具识别文字
import pytesseract
from PIL import Image
# pytesseract要求的image不是opencv读进来的image, 而是pillow这个包, 即PIL,按照 pip install pillow
text = pytesseract.image_to_string(Image.open('./scan.jpg'))
print(text)
# 找到每一个圆圈的轮廓 cnts = cv2.findContours(thresh, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[1] thresh_contours = thresh.copy() cv2.drawContours(thresh_contours, cnts, -1, 255, 3) cv_show('thresh_contours', thresh_contours) plt.imshow(thresh_contours, cmap='gray') # 遍历所有的轮廓, 找到特定宽高和特定比例的轮廓, 即圆圈的轮廓. question_cnts = [] for c in cnts: # 找到轮廓的外接矩形 (x, y, w, h) = cv2.boundingRect(c) # 计算宽高比 ar = w / float(h) # 根据实际情况制定标准. if w >= 20 and h >= 20 and 0.9 <= ar <= 1.1: question_cnts.append(c)
对轮廓按照y轴排序
# 轮廓排序功能封装成函数
def sort_contours(cnts, method='left-to-right'):
reverse = False
# 排序的时候, 取x轴数据, i=0, 取y轴数据i =1
i = 0
if method == 'right-to-left' or method == 'bottom-to-top':
reverse = True
# 按y轴坐标排序
if method == 'top-to-bottom' or method == 'bottom-to-top':
i = 1
# 计算每个轮廓的外接矩形
bounding_boxes = [cv2.boundingRect(c) for c in cnts]
(cnts, bounding_boxes) = zip(*sorted(zip(cnts, bounding_boxes), key=lambda b: b[1][i], reverse=reverse))
return cnts, bounding_boxes
# 按照从上到下的顺序对question_cnts排序.
question_cnts = sort_contours(question_cnts, method='top-to-bottom')[0]
进一步按照x轴排序,构造圆圈轮廓的掩膜得出答案,最后和正确答案比较评判答题卡分数。
# 正确答案 ANSWER_KEY = {0:1, 1:4, 2:0, 3:3, 4:1} correct = 0 for (q, i) in enumerate(np.arange(0, 25, 5)): # print(q, i) # 每次取出5个轮廓, 再按照x轴坐标从小到大排序 cnts = sort_contours(question_cnts[i: i + 5])[0] bubbled = None # 遍历每一个结果 for (j, c) in enumerate(cnts): # 使用掩膜, 即mask mask = np.zeros(thresh.shape, dtype='uint8') cv2.drawContours(mask, [c], -1, 255, -1) # cv_show('mask', mask) # 先做与运算 mask = cv2.bitwise_and(thresh, thresh, mask=mask) # cv_show('mask', mask) # 计算非零个数, 选择的选项, 非零个数比较多, 没选中的选项非零个数少一些 total = cv2.countNonZero(mask) if bubbled is None or total > bubbled[0]: bubbled = (total, j) color = (0, 0, 255) k = ANSWER_KEY[q] # 判断是否做题正确 if k == bubbled[1]: correct += 1 color = (0, 255, 0) # 绘图 cv2.drawContours(warped, [cnts[k]], -1, color, 3) # 计算分数 print(correct) score = (correct / 5.0) * 100 print(f'score: {score:.2f}%') cv2.putText(warped, str(score) + '%', (10, 30), cv2.FONT_HERSHEY_SIMPLEX, 0.9, (0, 0, 255), 2) cv_show('result', warped)
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