关于python:使用-OpenCVPython-识别答题卡判卷

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import cv2 as cv
import numpy as np
def cvshow(name, img):

cv.imshow(name, img)
cv.waitKey(0)
cv.destroyAllWindows()

def four_point_transform(img, four_points):

rect = order_points(four_points)
(tl, tr, br, bl) = rect
# 计算输出的 w 和 h 的值
widthA = np.sqrt((tr[0] - tl[0]) ** 2 + (tr[1] - tl[1]) ** 2)
widthB = np.sqrt((br[0] - bl[0]) ** 2 + (br[1] - bl[1]) ** 2)
maxWidth = max(int(widthA), int(widthB))
heightA = np.sqrt((tl[0] - bl[0]) ** 2 + (tl[1] - bl[1]) ** 2)
heightB = np.sqrt((tr[0] - br[0]) ** 2 + (tr[1] - br[1]) ** 2)
maxHeight = max(int(heightA), int(heightB))
# 变换后对应的坐标地位
dst = np.array([[0, 0],
    [maxWidth - 1, 0],
    [maxWidth - 1, maxHeight - 1],
    [0, maxHeight - 1]], dtype='float32')
# 最次要的函数就是 cv2.getPerspectiveTransform(rect, dst) 和 cv2.warpPerspective(image, M, (maxWidth, maxHeight))
M = cv.getPerspectiveTransform(rect, dst)
warped = cv.warpPerspective(img, M, (maxWidth, maxHeight))
return warped

def order_points(points):

res = np.zeros((4, 2), dtype='float32')
# 依照从前往后 0,1,2,3 别离示意左上、右上、右下、左下的程序将 points 中的数填入 res 中
# 将四个坐标 x 与 y 相加,和最大的那个是右下角的坐标,最小的那个是左上角的坐标
sum_hang = points.sum(axis=1)
res[0] = points[np.argmin(sum_hang)]
res[2] = points[np.argmax(sum_hang)]
# 计算坐标 x 与 y 的离散插值 np.diff()
diff = np.diff(points, axis=1)
res[1] = points[np.argmin(diff)]
res[3] = points[np.argmax(diff)]
# 返回 result
return res

def sort_contours(contours, method=”l2r”):

# 用于给轮廓排序,l2r, r2l, t2b, b2t
reverse = False
i = 0
if method == "r2l" or method == "b2t":
    reverse = True
if method == "t2b" or method == "b2t":
    i = 1
boundingBoxes = [cv.boundingRect(c) for c in contours]
(contours, boundingBoxes) = zip(*sorted(zip(contours, boundingBoxes), key=lambda a: a[1][i], reverse=reverse))
return contours, boundingBoxes

正确答案

right_key = {0: 1, 1: 4, 2: 0, 3: 3, 4: 1}

输出图像

img = cv.imread(‘./images/test_01.png’)
img_copy = img.copy()
img_gray = cv.cvtColor(img, cv.COLOR_BGR2GRAY)
cvshow(‘img-gray’, img_gray)

图像预处理

高斯降噪

img_gaussian = cv.GaussianBlur(img_gray, (5, 5), 1)
cvshow(‘gaussianblur’, img_gaussian)

canny 边缘检测

img_canny = cv.Canny(img_gaussian, 80, 150)
cvshow(‘canny’, img_canny)

轮廓辨认——答题卡边缘辨认

cnts, hierarchy = cv.findContours(img_canny, cv.RETR_EXTERNAL, cv.CHAIN_APPROX_SIMPLE)
cv.drawContours(img_copy, cnts, -1, (0, 0, 255), 3)
cvshow(‘contours-show’, img_copy)
docCnt = None

确保检测到了

if len(cnts) > 0:

# 依据轮廓大小进行排序
cnts = sorted(cnts, key=cv.contourArea, reverse=True)
# 遍历每一个轮廓
for c in cnts:
    # 近似
    peri = cv.arcLength(c, True)  # arclength 计算一段曲线的长度或者闭合曲线的周长;# 第一个参数输出一个二维向量,第二个参数示意计算曲线是否闭合
    approx = cv.approxPolyDP(c, 0.02 * peri, True)
    # 用一条顶点较少的曲线 / 多边形来近似曲线 / 多边形,以使它们之间的间隔 <= 指定的精度;# c 是须要近似的曲线,0.02*peri 是精度的最大值,True 示意曲线是闭合的
    # [外汇跟单](https://www.gendan5.com/) 筹备做透视变换
    if len(approx) == 4:
        docCnt = approx
        break

透视变换——提取答题卡主体

docCnt = docCnt.reshape(4, 2)
warped = four_point_transform(img_gray, docCnt)
cvshow(‘warped’, warped)

轮廓辨认——辨认出选项

thresh = cv.threshold(warped, 0, 255, cv.THRESH_BINARY_INV | cv.THRESH_OTSU)[1]
cvshow(‘thresh’, thresh)
thresh_cnts, _ = cv.findContours(thresh, cv.RETR_EXTERNAL, cv.CHAIN_APPROX_SIMPLE)
w_copy = warped.copy()
cv.drawContours(w_copy, thresh_cnts, -1, (0, 0, 255), 2)
cvshow(‘warped_contours’, w_copy)
questionCnts = []

遍历,挑出选项的 cnts

for c in thresh_cnts:

(x, y, w, h) = cv.boundingRect(c)
ar = w / float(h)
# 依据理论状况指定规范
if w >= 20 and h >= 20 and ar >= 0.9 and ar <= 1.1:
    questionCnts.append(c)

查看是否挑出了选项

w_copy2 = warped.copy()
cv.drawContours(w_copy2, questionCnts, -1, (0, 0, 255), 2)
cvshow(‘questionCnts’, w_copy2)

检测每一行抉择的是哪一项,并将后果贮存在元组 bubble 中,记录正确的个数 correct

依照从上到下 t2b 对轮廓进行排序

questionCnts = sort_contours(questionCnts, method=”t2b”)[0]
correct = 0

每行有 5 个选项

for (i, q) in enumerate(np.arange(0, len(questionCnts), 5)):

# 排序
cnts = sort_contours(questionCnts[q:q+5])[0]
bubble = None
# 失去每一个选项的 mask 并填充,与正确答案进行按位与操作取得重合点数
for (j, c) in enumerate(cnts):
    mask = np.zeros(thresh.shape, dtype='uint8')
    cv.drawContours(mask, , -1, 255, -1)
    cvshow('mask', mask)
    # 通过按位与操作失去 thresh 与 mask 重合局部的像素数量
    bitand = cv.bitwise_and(thresh, thresh, mask=mask)
    totalPixel = cv.countNonZero(bitand)
    if bubble is None or bubble[0] < totalPixel:
        bubble = (totalPixel, j)
k = bubble[1]
color = (0, 0, 255)
if k == right_key[i]:
    correct += 1
    color = (0, 255, 0)
# 绘图
cv.drawContours(warped, [cnts[right_key[i]]], -1, color, 3)
cvshow('final', warped)

计算最终得分并在图中标注

score = (correct / 5.0) * 100
print(f”Score: {score}%”)
cv.putText(warped, f”Score: {score}%”, (10, 30), cv.FONT_HERSHEY_SIMPLEX, 0.9, (0, 0, 255), 2)
cv.imshow(“Original”, img)
cv.imshow(“Exam”, warped)
cv.waitKey(0)

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