from imutils import contours
import numpy as np
import argparse
import cv2
import myutils

# 设置参数
ap = argparse.ArgumentParser()
ap.add_argument("-i", "--image", default='./images/credit_card_01.png', help="path to input image")
ap.add_argument("-t", "--template", default='./ocr_a_reference.png', help="path to template OCR image")
args = vars(ap.parse_args())
# 指定信用卡类型
FIRST_NUMBER = {"3": "American Express", "4": "Visa", "5": "MasterCard", "6": "Discover Card"}

def cv_show(name, img):cv2.imshow(name, img)cv2.waitKey(500)cv2.destroyAllWindows()

# 读取一个模板图像
img = cv2.imread(args["template"])
cv_show('template', img)
template = cv2.imread(args["image"])
cv_show('image', template)
# 灰度图
ref = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
cv_show('template_gray', ref)
# 二值图像
ref = cv2.threshold(ref, 10, 255, cv2.THRESH_BINARY_INV)[1]
cv_show('template_bi', ref)

# 计算轮廓
# cv2.findContours()函数接受的参数为二值图，即黑白的（不是灰度图）,cv2.RETR_EXTERNAL只检测外轮廓，cv2.CHAIN_APPROX_SIMPLE只保留终点坐标
# 返回的list中每个元素都是图像中的一个轮廓
refCnts, hierarchy = cv2.findContours(ref.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)  # cv版本大于3.8的,只有两个返回值cv2.drawContours(img, refCnts, -1, (0, 0, 255), 3)  # 轮廓在二值图上得到, 画要画在原图上
cv_show('template_Contours', img)
print(np.array(refCnts).shape)
refCnts = myutils.sort_contours(refCnts, method="left-to-right")[0]  # 排序，从左到右，从上到下
digits = {}  # 存模板的单个数字
# 遍历每一个轮廓,外接矩形
for (i, c) in enumerate(refCnts):  # c是每个轮廓的终点坐标# 计算外接矩形并且resize成合适大小(x, y, w, h) = cv2.boundingRect(c)# 抠出模板roi = ref[y:y + h, x:x + w]  # 每个roi对应一个数字roi = cv2.resize(roi, (57, 88))  # 太小,调大点#  每一个数字对应每一个模板digits[i] = roi

# myutils.py
import cv2def sort_contours(cnts, method="left-to-right"):reverse = Falsei = 0if method == "right-to-left" or method == "bottom-to-top":reverse = Trueif method == "top-to-bottom" or method == "bottom-to-top":i = 1boundingBoxes = [cv2.boundingRect(c) for c in cnts] #用一个最小的矩形，把找到的形状包起来x,y,h,w(cnts, boundingBoxes) = zip(*sorted(zip(cnts, boundingBoxes),key=lambda b: b[1][i], reverse=reverse))return cnts, boundingBoxes
def resize(image, width=None, height=None, inter=cv2.INTER_AREA):dim = None(h, w) = image.shape[:2]if width is None and height is None:return imageif width is None:r = height / float(h)dim = (int(w * r), height)else:r = width / float(w)dim = (width, int(h * r))resized = cv2.resize(image, dim, interpolation=inter)return resized

#  输入图像处理
#  1.初始化卷积核，根据实际任务指定大小，不一定非要3*3
rectKernel = cv2.getStructuringElement(cv2.MORPH_RECT, (9, 3))
sqKernel = cv2.getStructuringElement(cv2.MORPH_RECT, (3, 3))
#  2.读取输入图像，预处理
image = cv2.imread(args["image"])
cv_show("Input image", image)
image = myutils.resize(image, width=300)
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
cv_show("Input_gray", gray)
#  3.礼帽:原始输入-开运算结果(只剩毛刺)
tophat = cv2.morphologyEx(gray, cv2.MORPH_TOPHAT, rectKernel)
cv_show("Input tophat", tophat)
#  4.x方向上面的Sobel算子，实验表明只加x效果更好一些
gradx = cv2.Sobel(tophat, ddepth=cv2.CV_32F, dx=1, dy=0, ksize=-1)
gradx = np.absolute(gradx)
(minVal, maxVal) = (np.min(gradx), np.max(gradx))
gradx = (255 * ((gradx - minVal) / (maxVal - minVal)))
gradx = gradx.astype("uint8")print(np.array(gradx).shape)
cv_show("Input_Sobel_gradx", gradx)#  5.通过闭操作将数字连在一起，将本是4个数字的4个框膨胀成1个框，就腐蚀不掉了
gradx = cv2.morphologyEx(gradx, cv2.MORPH_CLOSE, rectKernel)
cv_show("Input_close_gradx", gradx)# 6.THRESH_OTSU会自动寻找合适的阈值，适合双峰，需把阈值参数设置为0
thresh = cv2.threshold(gradx, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)[1]
cv_show('Input_thresh', thresh)# 7.再来一个闭操作
thresh = cv2.morphologyEx(thresh, cv2.MORPH_CLOSE, sqKernel)
cv_show('Input_thresh_CLOSE', thresh)# 8.计算轮廓
threshCnts, hierarchy = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)cnts = threshCnts
cur_img = image.copy()
cv2.drawContours(cur_img, cnts, -1, (200, 150, 190), 3)
cv_show('Input_Contours', cur_img)

#  遍历轮廓
# 1.遍历轮廓
locs = []  # 存符合条件的轮廓
for i, c in enumerate(threshCnts):# 计算矩形x, y, w, h = cv2.boundingRect(c)ar = w / float(h)# 选择合适的区域，根据实际任务来，这里的基本都是四个数字一组if 2.5 < ar < 4.0:if (40 < w < 55) and (10 < h < 20):# 符合的留下来locs.append((x, y, w, h))
#   append函数，给列表里面添加元素
# 将符合的轮廓从左到右排序
locs = sorted(locs, key=lambda x: x[0])# 2.遍历每一个轮廓中的数字
output = []  # 存正确的数字
for (i, (gx, gy, gw, gh)) in enumerate(locs):  # 遍历每一组大轮廓(包含4个数字)# initialize the list of group digitsgroupOutput = []# 根据坐标提取每一个组(4个值)group = gray[gy - 5:gy + gh + 5, gx - 5:gx + gw + 5]  # 往外扩一点cv_show('group_' + str(i), group)# 2.1 预处理group = cv2.threshold(group, 0, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU)[1]  # 二值化的group# cv_show('group_'+str(i),group)# 计算每一组的轮廓 这样就分成4个小轮廓了digitCnts = cv2.findContours(group.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)[0]# 排序digitCnts = myutils.sort_contours(digitCnts, method="left-to-right")[0]# 2.2 计算并匹配每一组中的每一个数值for c in digitCnts:  # c表示每个小轮廓的终点坐标z = 0# 找到当前数值的轮廓,resize成合适的的大小(x, y, w, h) = cv2.boundingRect(c)  # 外接矩形roi = group[y:y + h, x:x + w]  # 在原图中取出小轮廓覆盖区域,即数字roi = cv2.resize(roi, (57, 88))# cv_show("roi_"+str(z),roi)# 计算匹配得分: 0得分多少,1得分多少...scores = []  # 单次循环中,scores存的是一个数值 匹配 10个模板数值的最大得分# 在模板中计算每一个得分# digits的digit正好是数值0,1,...,9;digitROI是每个数值的特征表示for (digit, digitROI) in digits.items():# 进行模板匹配, res是结果矩阵res = cv2.matchTemplate(roi, digitROI, cv2.TM_CCOEFF)  # 此时roi是X digitROI是0 依次是1,2.. 匹配10次,看模板最高得分多少Max_score = cv2.minMaxLoc(res)[1]  # 返回4个,取第二个最大值Maxscorescores.append(Max_score)  # 10个最大值print("scores：", scores)# 得到最合适的数字groupOutput.append(str(np.argmax(scores)))  # 返回的是输入列表中最大值的位置z = z + 1# 2.3 画出来cv2.rectangle(image, (gx - 5, gy - 5), (gx + gw + 5, gy + gh + 5), (0, 0, 255), 1)  # 左上角,右下角# 2.4 putText参数：图片,添加的文字,左上角坐标,字体,字体大小,颜色,字体粗细cv2.putText(image, "".join(groupOutput), (gx, gy - 15),cv2.FONT_HERSHEY_SIMPLEX, 0.65, (0, 0, 255), 2)# 2.5 得到结果output.extend(groupOutput)print("groupOutput:", groupOutput)

# 打印结果
print("Credit Card Type: {}".format(FIRST_NUMBER[output[0]]))
print("Credit Card #: {}".format("".join(output)))
cv2.imshow("Output_image", image)
cv2.waitKey(0)