一、说明参考代码
再上一篇简单了解了手写数字训练，https://blog.csdn.net/bashendixie5/article/details/109754138，可以应用训练结果到数独识别，并使用python的sudoku进行求解。
参考代码如下
在使用epochs=20的训练结果，识别效果很差(无法求解)，
在使用epochs=50的训练结果的时候有个别识别的不对(无法求解)，
最后使用epochs=100的训练结果的时候求解正确。
# 导入相关的包
from imutils.perspective import four_point_transform
from skimage.segmentation import clear_border
import numpy as np
import imutils
import cv2
from tensorflow.keras.preprocessing.image import img_to_array
from tensorflow.keras.models import load_model
# pip install py-sudoku 并且把包内的sudoku.py文件copy到当前程序的同级目录
from sudoku import Sudoku
import tensorflow as tf
# 处理图片
def find_puzzle(image, debug=False):
# 灰度化并高斯模糊
gray = cv2.cvtColor(image, cv2.COLOR_BGR2GRAY)
blurred = cv2.GaussianBlur(gray, (7, 7), 3)
# 自适应二值化
thresh = cv2.adaptiveThreshold(blurred, 255, cv2.ADAPTIVE_THRESH_GAUSSIAN_C, cv2.THRESH_BINARY, 11, 2)
thresh = cv2.bitwise_not(thresh)
# check to see if we are visualizing each step of the image
# processing pipeline (in this case, thresholding)
if debug:
cv2.imshow("Puzzle Thresh", thresh)
cv2.waitKey(0)
# 寻找轮廓并从大到小排序
cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = imutils.grab_contours(cnts)
cnts = sorted(cnts, key=cv2.contourArea, reverse=True)
# initialize a contour that corresponds to the puzzle outline
puzzleCnt = None
# loop over the contours
for c in cnts:
# 多边形拟合
peri = cv2.arcLength(c, True)
approx = cv2.approxPolyDP(c, 0.02 * peri, True)
# if our approximated contour has four points, then we can
# assume we have found the outline of the puzzle
if len(approx) == 4:
puzzleCnt = approx
break
# 如果没有找到适合的，则抛出异常
if puzzleCnt is None:
raise Exception(("Could not find Sudoku puzzle outline. "
"Try debugging your thresholding and contour steps."))
# check to see if we are visualizing the outline of the detected
# Sudoku puzzle
if debug:
# draw the contour of the puzzle on the image and then display
# it to our screen for visualization/debugging purposes
output = image.copy()
cv2.drawContours(output, [puzzleCnt], -1, (0, 255, 0), 2)
cv2.imshow("Puzzle Outline", output)
cv2.waitKey(0)
# 对原始图像和灰度图像都应用四点透视变换，以获得自上而下的拼图鸟瞰图
puzzle = four_point_transform(image, puzzleCnt.reshape(4, 2))
warped = four_point_transform(gray, puzzleCnt.reshape(4, 2))
# check to see if we are visualizing the perspective transform
if debug:
# show the output warped image (again, for debugging purposes)
cv2.imshow("Puzzle Transform", puzzle)
cv2.waitKey(0)
# return a 2-tuple of puzzle in both RGB and grayscale
return (puzzle, warped)
# 提取数字
def extract_digit(cell, debug=False):
# apply automatic thresholding to the cell and then clear any
# connected borders that touch the border of the cell
thresh = cv2.threshold(cell, 0, 255, cv2.THRESH_BINARY_INV | cv2.THRESH_OTSU)[1]
thresh = clear_border(thresh)
# check to see if we are visualizing the cell thresholding step
if debug:
cv2.imshow("Cell Thresh", thresh)
cv2.waitKey(0)
# find contours in the thresholded cell
cnts = cv2.findContours(thresh.copy(), cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
cnts = imutils.grab_contours(cnts)
# if no contours were found than this is an empty cell
if len(cnts) == 0:
return None
# otherwise, find the largest contour in the cell and create a
# mask for the contour
c = max(cnts, key=cv2.contourArea)
mask = np.zeros(thresh.shape, dtype="uint8")
cv2.drawContours(mask, [c], -1, 255, -1)
# compute the percentage of masked pixels relative to the total
# area of the image
(h, w) = thresh.shape
percentFilled = cv2.countNonZero(mask) / float(w * h)
# if less than 3% of the mask is filled then we are looking at
# noise and can safely ignore the contour
if percentFilled < 0.03:
return None
# apply the mask to the thresholded cell
digit = cv2.bitwise_and(thresh, thresh, mask=mask)
# check to see if we should visualize the masking step
if debug:
cv2.imshow("Digit", digit)
cv2.waitKey(0)
# return the digit to the calling function
return digit
# 加载训练模型并识别
def loadmodelandocr():
# 从本地磁盘加载手写识别的模型
model = load_model('C:\\Users\\xiaomao\\Desktop\\mnist.h5')
# 加载数独图片
image = cv2.imread('C:\\Users\\xiaomao\\Desktop\\sudoku.jpg')
# 改变尺寸
image = imutils.resize(image, width=600)
# 调用方法获取仿射变换后的图片
(puzzleImage, warped) = find_puzzle(image, False)
# 初始化9*9的数独board
board = np.zeros((9, 9), dtype="int")
# a Sudoku puzzle is a 9x9 grid (81 individual cells), so we can
# infer the location of each cell by dividing the warped image
# into a 9x9 grid
stepX = warped.shape[1] // 9
stepY = warped.shape[0] // 9
# initialize a list to store the (x, y)-coordinates of each cell
# location
cellLocs = []
na = 0;
# loop over the grid locations
for y in range(0, 9):
# initialize the current list of cell locations
row = []
for x in range(0, 9):
# compute the starting and ending (x, y)-coordinates of the
# current cell
startX = x * stepX
startY = y * stepY
endX = (x + 1) * stepX
endY = (y + 1) * stepY
# add the (x, y)-coordinates to our cell locations list
row.append((startX, startY, endX, endY))
# crop the cell from the warped transform image and then
# extract the digit from the cell
cell = warped[startY:endY, startX:endX]
digit = extract_digit(cell, False)
# verify that the digit is not empty
if digit is not None:
# 改变图片大小
roi = cv2.resize(digit, (28, 28))
roi = tf.reshape(roi, (1, 28 * 28))
# 预测
pred = model.predict(roi).argmax(axis=1)[0]
board[y, x] = pred
# add the row to our cell locations
cellLocs.append(row)
# 构建一个数独board
puzzle = Sudoku(3, 3, board=board.tolist())
puzzle.show()
# 数独求解
solution = puzzle.solve()
solution.show_full()
# 遍历位置绘制识别出来的数字和数独答案
for (cellRow, boardRow) in zip(cellLocs, solution.board):
# loop over individual cell in the row
for (box, digit) in zip(cellRow, boardRow):
# unpack the cell coordinates
startX, startY, endX, endY = box
# compute the coordinates of where the digit will be drawn
# on the output puzzle image
textX = int((endX - startX) * 0.33)
textY = int((endY - startY) * -0.2)
textX += startX
textY += endY
# draw the result digit on the Sudoku puzzle image
cv2.putText(puzzleImage, str(digit), (textX, textY), cv2.FONT_HERSHEY_SIMPLEX, 0.9, (0, 255, 255), 2)
# show the output image
cv2.imshow("Sudoku Result", puzzleImage)
cv2.waitKey(0)
loadmodelandocr()
二、效果如下
