python均分纸牌_Opencv学习笔记（3）---纸牌数字识别练习实践项目

蔚落 2023-01-01 05:53 173阅读 0赞

import numpy as np

import cv2

import argparse

import pytesseract

import os

from PIL import Image

ap = argparse.ArgumentParser()

ap.add\_argument("-i","--image",required=True,

help="Path to image to be scanned")

ap.add\_argument("-t", "--template", required=True,

help="path to template OCR-A image")

args = vars(ap.parse\_args())

def cv\_show(name,file):

cv2.imshow(name, file)

cv2.waitKey(0)

cv2.destroyAllWindows()

def sort\_contours(cnts, method="left-to-right"):

reverse = False

i = 0

if method == "right-to-left" or method == "bottom-to-top":

reverse = True

if method == "top-to-bottom" or method == "bottom-to-top":

i = 1

boundingBoxes = \[cv2.boundingRect(c) for c in cnts\] \#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==None and height ==None:

return image

if width==None:

r = height / float(h)

dim = (int(w \* r),height)

else:

r = width / float(w)

dim = (w,int(h \* r))

resizes = cv2.resize(image,dim,interpolation=inter)

return resizes

def order\_points(pts):

rect = np.zeros((4,2),dtype="float32")

s = pts.sum(axis=1)

rect\[0\] = pts\[np.argmin(s)\]

rect\[2\] = pts\[np.argmax(s)\]

diff = np.diff(pts,axis=1)

rect\[1\] = pts\[np.argmin(diff)\]

rect\[3\] = pts\[np.argmax(diff)\]

return rect

def four\_point\_transform(image,pts):

rect = order\_points(pts)

(tl,tr,bl,br) = rect

\# 计算输入的w和h的值

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 ))

maxWidth = 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))

maxHeight = max(int(heightA), int(heightB))

\# 变换后对应坐标位置

dst = np.array(\[

\[0,0\],

\[maxWidth - 1,0\],

\[maxWidth - 1,maxHeight - 1\],

\[0,maxHeight - 1\]\],dtype = "float32")

M = cv2.getPerspectiveTransform(rect,dst)

warped = cv2.warpPerspective(image,M,(maxWidth,maxHeight))

return warped

image = cv2.imread(args\["image"\])

\#cv\_show("image",image)

ratio = image.shape\[0\] / 500

orig = image.copy()

image = resize(image.copy(),height=500)

gray = cv2.cvtColor(image,cv2.COLOR\_BGR2GRAY)

edged = cv2.Canny(gray,75,200)

\#cv\_show("edged",edged)

\# 查找轮廓，并且去除扑克牌中间的轮廓，只保留花色和数字

cnts = cv2.findContours(edged.copy(),cv2.RETR\_LIST,cv2.CHAIN\_APPROX\_SIMPLE)\[1\]

cnts = sorted(cnts,key=cv2.contourArea,reverse=True)\[:10\]

draw\_img = image.copy()

for c in cnts:

\# dra = cv2.drawContours(draw\_img,c,-1,(0,0,255),2)

\# cv\_show("res",dra)

peri = cv2.arcLength(c,True)

approx = cv2.approxPolyDP(c,0.02\*peri,True)

if len(approx) ==4:

screenCnt = approx

print(np.array(screenCnt.size))

print(np.array(screenCnt))

break

\# 展示结果

\#cv2.drawContours(image,\[screenCnt\],-1,(0,0,255),2)

\#cv\_show("outline",image)

wraped = four\_point\_transform(orig,screenCnt.reshape(4,2) \* ratio)

\#cv\_show("wraped",wraped)

wraped = resize(wraped,height=500)

wraped = cv2.medianBlur(wraped,3)

cv\_show("gray",gray)

\#cv\_show("img",wraped)

\#gray = cv2.cvtColor(wraped,cv2.COLOR\_BGR2GRAY)

ref = cv2.threshold(gray,100,255,cv2.THRESH\_BINARY)\[1\]

edged = cv2.Canny(ref,75,200)

\#cv\_show("ref",edged)

cnts = cv2.findContours(edged,cv2.RETR\_LIST,cv2.CHAIN\_APPROX\_SIMPLE)\[1\]

cnts = sorted(cnts,key=cv2.contourArea,reverse=True)\[:10\]

draw\_img = wraped.copy()

peri = cv2.arcLength(cnts\[0\],True)

approx = cv2.approxPolyDP(cnts\[0\],0.02 \* peri,True)

\#cv2.drawContours(draw\_img,\[approx\],-1,(0,0,255),2)

print("---------------------")

\#print(\[approx\])

\#cv\_show("Outline",draw\_img)

print(draw\_img.shape\[:2\])

print(approx) \# 根据这个输出的结果确定mask

mask\_img = draw\_img

mask\_img\[96:412,64:218\] = 255 \# 这个范围是根据approx的结果得出来的大概值，把卡牌中间花的部位变为白色

mask\_img = mask\_img\[:int(mask\_img.shape\[0\]/2),:\]

cv\_show("1",mask\_img)

mask\_img = cv2.cvtColor(mask\_img,cv2.COLOR\_BGR2GRAY)

clahe = cv2.createCLAHE(clipLimit=2.0, tileGridSize=(8,8))

mask\_img = clahe.apply(mask\_img)

ref = cv2.threshold(mask\_img,70,255,cv2.THRESH\_BINARY)\[1\] \# 大多数牌为75的时候就行，但是有些需要阈值设为100，或者125等。并且有的时候75 与 70得到的结果非常不相同

ref = cv2.medianBlur(ref,3)

cv\_show("ref",ref)

\#ref = cv2.Canny(ref,55,200)

kernel = np.ones((3,3),np.uint8)

ero = cv2.erode(ref,kernel,iterations = 4)

cv\_show("ref",ero)

cnts = cv2.findContours(ero,cv2.RETR\_LIST,cv2.CHAIN\_APPROX\_SIMPLE)\[1\]

cnts = sorted(cnts,key=cv2.contourArea,reverse=True)\[:5\]

\#cv2.drawContours(draw\_img,cnts,-1,(0,0,255),2)

\#cv\_show("the page",draw\_img)

draw\_img = ref.copy()

save\_img1 = \[\]

save\_img2 = \[\]

for c in cnts:

x,y,w,h = cv2.boundingRect(c)

area = w \* h \# 通过面积来寻找要求的轮廓 5917左右为数字 3366左右为花色信息

\# if ar > 0.56 and ar < 0.63:

draw\_img = cv2.rectangle(draw\_img, (x, y), (x + w, y + h), (0, 255, 0), 3)

cv\_show("draw\_img", draw\_img)

if area > 5600 and area < 7000:

draw\_img = cv2.rectangle(draw\_img,(x,y),(x+w,y+h),(0,255,0),2)

cv\_show("draw\_img",draw\_img)

\#print("draw\_img",draw\_img\[y:y+h,x:x+w\])

save\_img1 =draw\_img\[y:y+h,x:x+w\]

\#cv2.imwrite("2.png",save\_img1)

if area > 2700 and area < 4600:

draw\_img = cv2.rectangle(draw\_img, (x, y), (x + w, y + h), (0, 255, 0), 1)

cv\_show("draw\_img", draw\_img)

save\_img2 =draw\_img\[y:y+h,x:x+w\]

\#cv2.imwrite("2.png",save\_img2)

'''

下面的代码是开始进行模板匹配了，上面的代码如果用来生成模板的话，取消cv2.imwrite的注释即可，然后注释掉下面的代码

模板匹配时，现在通过上面的代码已经得到了当前要检测的特征存放在save\_img1中数字特征，save\_img2中花色特征

'''

\#cv\_show("data",save\_img1)

template = cv2.imread(args\["template"\])

ref = cv2.cvtColor(template,cv2.COLOR\_BGR2GRAY)

ref = cv2.threshold(ref,10,255,cv2.THRESH\_BINARY)\[1\]

cv\_show("ref",ref)

cnts = cv2.findContours(ref.copy(),cv2.RETR\_LIST,cv2.CHAIN\_APPROX\_SIMPLE)\[1\]

cnts = sorted(cnts,key=cv2.contourArea,reverse=1)\[:17\] \# 提取面积最大的17个特征，就是外框特征，剔除具体的数字和花色特征

cnts = sort\_contours(cnts, method="left-to-right")\[0\]\#排序，从左到右，从上到下 这次排序主要是看距离最左边点的距离来排序，这样就可以找出每个框的特征是什么了

cv2.drawContours(template,cnts,-1,(0,0,255),3)

cv\_show('template',template)

draw\_img = template.copy()

digits\_and\_type = \{\} \# 存放数字模板和花色模板

for (i,c) in enumerate(cnts):

x,y,w,h = cv2.boundingRect(c)

\# cv2.rectangle(draw\_img,(x,y),(x+w,y+h),(0,255,0),2)

\# cv\_show("draw\_img",draw\_img)

roi = ref\[y:y+h,x:x+w\]

digits\_and\_type\[i\] = roi

'''

第三步 进行模板匹配

'''

OutPut = \[\]

scores = \[\]

save\_img2 = cv2.resize(save\_img2,(60,90))

save\_img1 = cv2.resize(save\_img1,(60,90))

\#save\_img1 = cv2.threshold(save\_img1,0,255,

\# cv2.THRESH\_BINARY | cv2.THRESH\_OTSU)\[1\]

\#save\_img2 = cv2.threshold(save\_img2,0,255,

\# cv2.THRESH\_BINARY | cv2.THRESH\_OTSU)\[1\]

for (digit,digitROI) in digits\_and\_type.items():

\# 模板匹配

result = cv2.matchTemplate(save\_img1,digitROI,cv2.TM\_CCOEFF)

(\_,score,\_,\_) = cv2.minMaxLoc(result)

scores.append(score)

OutPut.append(np.argmax(scores)+1)

scores = \[\] \# 清空

for (digit,digitROI) in digits\_and\_type.items():

\# 模板匹配

result = cv2.matchTemplate(save\_img2,digitROI,cv2.TM\_CCOEFF)

(\_,score,\_,\_) = cv2.minMaxLoc(result)

scores.append(score)

OutPut.append(np.argmax(scores)+1)

dict = \{1:"A",2:"2",3:"3",4:"4",5:"5",6:"6",7:"7",8:"8",9:"9",10:"10",11:"J",12:"Q",13:"K",14:"红桃",15:"方块",16:"梅花",17:"黑桃"\}

print("the result:")

print("当前检测卡牌的数字为" + dict\[OutPut\[0\]\] +" 当前检测卡牌的花色为" + dict\[OutPut\[1\]\])

python均分纸牌_Opencv学习笔记（3）---纸牌数字识别练习实践项目

发表评论取消回复

还没有评论，来说两句吧...

相关阅读

相关 python均分纸牌_Opencv学习笔记（3）---纸牌数字识别练习实践项目

相关 P1031 均分纸牌（洛谷训练场）题解

相关 P1031 [NOIP2002 提高组] 均分纸牌

相关 c语言发纸牌实验报告,纸牌游戏实验报告.docx

相关基于深度学习yolov4的扑克牌识别、纸牌识别

相关 Vijos P1123均分纸牌

相关洛谷P1031 均分纸牌

相关 NOIP 2002 均分纸牌(贪心||模拟)

相关 P1031 均分纸牌

相关 P1031 均分纸牌——————贪心