用模板匹配進行的銀行卡數字識別

實現功能：在銀行卡上識別出卡號數字部分，並在其上方寫下數字，如下圖所示。

模板圖片如下圖，

原始銀行卡圖片，

完整代碼如下，在Pycharm中編寫。

imutils庫需要提前下載，在cmd中用命令 pip install imutils 下載。

import cv2
import numpy as np
import matplotlib.pyplot as plt
import imutils
from imutils import contours

# 處理模板圖片
img = cv2.imread("C:\\Users\\lenovo\\Pictures\\moban.jpg")
#cv2.imshow("img_moban",img)
img_gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
#cv2.imshow("img_moban_gray ",img_gray)
ret,ref = cv2.threshold(img_gray,20,255,cv2.THRESH_BINARY_INV)
#cv2.imshow("binary", ref)
image,refCnts, hierarchy = cv2.findContours(ref.copy(),cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
contour = cv2.drawContours(img,refCnts,-1,(0,0,255),3)
cv2.namedWindow("img_moban_contours",cv2.WINDOW_NORMAL)
#cv2.imshow("img_moban_contours",contour)

#refCnts = refCnts[0] if imutils.is_cv2() else refCnts[1]
refCnts = contours.sort_contours(refCnts,method = "left-to-right")[0] #from imutils import contours
digits = {}
for (i,c) in enumerate(refCnts):     # enumerate(refCnts)下標從0開始，返回值是0，第一個輪廓。1，第二個輪廓
    (x,y,w,h) = cv2.boundingRect(c)  # x，y是矩陣左上點的坐標，w，h是矩陣的寬和高
    roi = ref[y:y+h,x:x+w]
    roi = cv2.resize(roi,(57,88))
    digits[i] = roi


# 處理待檢測圖片
recKernel = cv2.getStructuringElement(cv2.MORPH_RECT,(10,3))
sqKernel = cv2.getStructuringElement(cv2.MORPH_RECT,(2,2))

image = cv2.imread("C:\\Users\\lenovo\\Pictures\\yinhangka.png")
image = cv2.resize(image,(250,200))

gray = cv2.cvtColor(image,cv2.COLOR_BGR2GRAY)
cv2.imshow("imagesize",gray)
tophat = cv2.morphologyEx(gray,cv2.MORPH_TOPHAT,recKernel) # 頂帽操作：突出明亮區域
cv2.imshow("imagetophat",tophat)
gradx = cv2.Sobel(tophat,ddepth=cv2.CV_32F,dx=1,dy=0,ksize=3)
gradx= np.absolute(gradx)
(minVal,maxVal) = (np.min(gradx),np.max(gradx))
gradx = (255*((gradx-minVal)/(maxVal-minVal)))
gradx = gradx.astype("uint8")

gradx = cv2.morphologyEx(gradx,cv2.MORPH_CLOSE,recKernel)
cv2.imshow("imagegradx",gradx)
ret,thresh = cv2.threshold(gradx,0,255,cv2.THRESH_BINARY|cv2.THRESH_OTSU)
thresh = cv2.morphologyEx(thresh,cv2.MORPH_CLOSE,sqKernel)
imageprint,Cnts,hierarchy = cv2.findContours(thresh.copy(),cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
cnts = Cnts # cnts是所有輪廓
curImage = image.copy()
contours = cv2.drawContours(curImage,cnts,-1,(0,0,255),3)
cv2.imshow("contours",contours)
locs = []    # 儲存符合條件的數字模塊
locsnum = []
for (i,c) in enumerate(cnts):
    (x,y,w,h) = cv2.boundingRect(c)
    ar = w/float(h)
    if (ar>2.5) and (ar<4):
        if h>10:   # 這里的條件要多試幾次
            locs.append((x,y,w,h))
locs = sorted(locs,key=lambda x:x[0]) #排序方式按照中括號[]里面的維度進行排序，[0]按照第一維排序，[2]按照第三維排序
print(len(locs))
output=[]
for (i,(gx,gy,gw,gh)) in enumerate(locs):
    groupOutput = []
    group = gray[gy-5:gy+gh+5,gx-5:gx+gw+5] # 在灰度圖中選取roi
    group = cv2.threshold(group,0,255,cv2.THRESH_BINARY|cv2.THRESH_OTSU)[1]
   # cv2.imshow("group", group)
    imageing,digitCnts,hierarchy = cv2.findContours(group.copy(),cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_SIMPLE)
    # digitCnts = contours.sort_contours(digitCnts,method = "left-to-right")[0]
    for c in digitCnts:
        (x,y,w,h) = cv2.boundingRect(c)
        roi = group[y:y+h,x:x+w]
        roi = cv2.resize(roi,(57,88))
        scores=[]
        for (digit, digitROI) in digits.items():  # 在模板預處理中建立了數值的字典類型,一個為索引、一個為值
            result = cv2.matchTemplate(roi, digitROI, cv2.TM_CCOEFF)  # 匹配，返回與之匹配度最高的數值
            (min_val, score, min_index, max_index) = cv2.minMaxLoc(result)  # 做10次匹配，取最大值（注意：取最大值還是最小值跟選取的模板匹配方法有關）
            scores.append(score)
        groupOutput.append(str(np.argmax(scores))) #argmax返回的是最大數的索引
    cv2.rectangle(image, (gx - 5, gy - 5), (gx + gw + 5, gy + gh + 5), (0, 0, 255), 1) # 第一組的矩形框
    groupOutput.reverse()
    image = cv2.putText(image, str("".join(groupOutput)), (gx, gy - 15), cv2.FONT_HERSHEY_SIMPLEX, 0.65, (0, 0, 255), 2)
    output.extend(groupOutput)
# print("Credit Card Type: {}".format(FIRST_NUMBER[output[0]]))
print("Credit Card #: {}".format("".join(output))) # 將output中的字符用雙引號中間的符號連接起來
cv2.imshow("Image", image)
cv2.waitKey(0)
cv2.destroyAllWindows()
# . join()：    連接字符串數組。將字符串、元組、列表中的元素以指定的字符(分隔符)連接生成一個新的字符串