利用Python开发智能阅卷系统

前言

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作者： 机器学习与统计学

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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

  1 import numpy as np
  2 import argparse
  3 import imutils
  4 import cv2
  5 # 设置参数
  6 ap = argparse.argumentparser()
  7 ap.add_argument("-i", "--image", default="test_01.png")
  8 args = vars(ap.parse_args())
  9 # 正确答案
 10 answer_key = {0: 1, 1: 4, 2: 0, 3: 3, 4: 1} #
 11 def order_points(pts):
 12    # 一共4个坐标点
 13    rect = np.zeros((4, 2), dtype = "float32")
 14 ​
 15    # 按顺序找到对应坐标0,1,2,3分别是 左上，右上，右下，左下
 16    # 计算左上，右下
 17    s = pts.sum(axis = 1)
 18    rect[0] = pts[np.argmin(s)]
 19    rect[2] = pts[np.argmax(s)]
 20    # 计算右上和左下
 21    diff = np.diff(pts, axis = 1)
 22    rect[1] = pts[np.argmin(diff)]
 23    rect[3] = pts[np.argmax(diff)]
 24    return rect
 25 ​
 26 def four_point_transform(image, pts):
 27    # 获取输入坐标点
 28    rect = order_points(pts)
 29    (tl, tr, br, bl) = rect
 30    # 计算输入的w和h值
 31    widtha = np.sqrt(((br[0]-bl[0])** 2) + ((br[1]-bl[1])**2))
 32    widthb = np.sqrt(((tr[0] -tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2))
 33    maxwidth = max(int(widtha), int(widthb))
 34    heighta = np.sqrt(((tr[0]-br[0])**2)+((tr[1]-br[1])**2))
 35    heightb = np.sqrt(((tl[0]-bl[0])**2)+((tl[1]-bl[1])**2))
 36    maxheight = max(int(heighta), int(heightb))
 37    # 变换后对应坐标位置
 38    dst = np.array([
 39       [0, 0],
 40       [maxwidth - 1, 0],
 41       [maxwidth - 1, maxheight - 1],
 42       [0, maxheight - 1]], dtype = "float32")
 43    # 计算变换矩阵
 44    m = cv2.getperspectivetransform(rect, dst)
 45    warped = cv2.warpperspective(image, m, (maxwidth, maxheight))
 46    return warped # 返回变换后结果
 47 ​
 48 def sort_contours(cnts, method="left-to-right"):
 49     reverse = false
 50     i = 0
 51     if method == "right-to-left" or method == "bottom-to-top":
 52         reverse = true
 53     if method == "top-to-bottom" or method == "bottom-to-top":
 54         i = 1
 55     boundingboxes = [cv2.boundingrect(c) for c in cnts]
 56     (cnts, boundingboxes) = zip(*sorted(zip(cnts, boundingboxes),
 57                                         key=lambda b: b[1][i], reverse=reverse))
 58     return cnts, boundingboxes
 59 def cv_show(name,img):
 60         cv2.imshow(name, img)
 61         cv2.waitkey(0)
 62         cv2.destroyallwindows()  
 63 ​
 64 ​
 65 image = cv2.imread(args["image"])
 66 contours_img = image.copy()
 67 gray = cv2.cvtcolor(image, cv2.color_bgr2gray)
 68 blurred = cv2.gaussianblur(gray, (5, 5), 0)
 69 edged = cv2.canny(blurred, 75, 200)
 70 # 轮廓检测
 71 cnts = cv2.findcontours(edged.copy(), cv2.retr_external,
 72    cv2.chain_approx_simple)[1]
 73 cv2.drawcontours(contours_img,cnts,-1,(0,0,255),3) 
 74 doccnt = none
 75 ​
 76 # 确保检测到了
 77 if len(cnts) > 0:
 78    # 根据轮廓大小进行排序
 79    cnts = sorted(cnts, key=cv2.contourarea, reverse=true)
 80    for c in cnts:   # 遍历每一个轮廓
 81       # 近似
 82       peri = cv2.arclength(c, true)
 83       approx = cv2.approxpolydp(c, 0.02 * peri, true)
 84       # 准备做透视变换
 85       if len(approx) == 4:
 86          doccnt = approx
 87          break
 88 # 执行透视变换
 89 warped = four_point_transform(gray, doccnt.reshape(4, 2))
 90 ​
 91 thresh = cv2.threshold(warped, 0, 255,
 92    cv2.thresh_binary_inv | cv2.thresh_otsu)[1] 
 93 thresh_contours = thresh.copy()
 94 # 找到每一个圆圈轮廓
 95 cnts = cv2.findcontours(thresh.copy(), cv2.retr_external,
 96    cv2.chain_approx_simple)[1]
 97 cv2.drawcontours(thresh_contours,cnts,-1,(0,0,255),3) 
 98 questioncnts = []
 99 for c in cnts:# 遍历
100    # 计算比例和大小
101    (x, y, w, h) = cv2.boundingrect(c)
102    ar = w / float(h)
103    # 根据实际情况指定标准
104    if w >= 20 and h >= 20 and ar >= 0.9 and ar <= 1.1:
105       questioncnts.append(c)
106 # 按照从上到下进行排序
107 questioncnts = sort_contours(questioncnts,
108    method="top-to-bottom")[0]
109 correct = 0
110 # 每排有5个选项
111 for (q, i) in enumerate(np.arange(0, len(questioncnts), 5)):
112    cnts = sort_contours(questioncnts[i:i + 5])[0]
113    bubbled = none
114    for (j, c) in enumerate(cnts):  # 遍历每一个结果
115       # 使用mask来判断结果
116       mask = np.zeros(thresh.shape, dtype="uint8")
117       cv2.drawcontours(mask, [c], -1, 255, -1) #-1表示填充
118       # 通过计算非零点数量来算是否选择这个答案
119       mask = cv2.bitwise_and(thresh, thresh, mask=mask)
120       total = cv2.countnonzero(mask)
121       # 通过阈值判断
122       if bubbled is none or total > bubbled[0]:
123          bubbled = (total, j)
124    # 第二步，与正确答案进行对比
125    color = (0, 0, 255)
126    k = answer_key[q]
127    # 判断正确
128    if k == bubbled[1]:
129       color = (0, 255, 0)
130       correct += 1
131    cv2.drawcontours(warped, [cnts[k]], -1, color, 3) #绘图
132 ​
133    #正确率的文本显示
134 score = (correct / 5.0) * 100
135 print("[info] score: {:.2f}%".format(score))
136 cv2.puttext(warped, "{:.2f}%".format(score), (10, 30),
137    cv2.font_hershey_simplex, 0.9, (0, 0, 255), 2)
138 cv2.imshow("input", image)
139 cv2.imshow("output", warped)
140 cv2.waitkey(0)

 

最终实现的效果如下：

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