基于机器学习（K-means）算法的手写数字识别系统，同时配套与前端进行了交互适配，可以在网页段调用

# 基于机器学习的手写数字识别系统 > **写在前面** > > 怎么说呢， K-means做识别感觉好像准确率不太，想用CNN，但是好像这是基于机器学习的大作业，用CNN好像不妥 > > 于是就只用了K-means， 但其实我要是复习的差不多有时间的话想增加...（不太可能了） > > 第一次尝试前端与后端交互共同实现，也是学到了很多新东西，顺便把web作业也水了，开心:smile:..... > > 总结一下吧：**在 0~5 的情况下准确率较高， 6~9 准确率不堪入目** > > <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/4KK.png" alt="4" style="zoom:80%;" /> > > **下面就开始正式的汇报** ## 流程 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/image-20230526204706611.png" alt="image-20230526204706611" style="zoom:80%;" /> ## 训练数据处理 本次训练的数据集为MNIST数据集，训练初期目标为 $32\times32$ , 数据集大小为 $28\times 28$ ，所以要先放大图片 接着对图片进行二值化，对于训练的预期是输入白底黑字的图片，而MINIST的图片是黑底白字，所以对于训练集，如果灰度大于$127$，取特征为 $1$ ， 否则取 $0$ 。 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/image-20230526205914048.png" alt="image-20230526205914048" style="zoom: 67%;" /> **核心代码** ```py def readImg(Road): image = cv2.imread(Road, cv2.IMREAD_GRAYSCALE) down_width = 32 down_height = 32 down_points = (down_width, down_height) image = cv2.resize(image, down_points, interpolation=cv2.INTER_LINEAR) Info = [] for i in range(32): for j in range(32): if image[i][j] >= 127: Info.append(1) else: Info.append(0) return Info ``` ## 模型选择 考虑四个模型， 分别是 *K-means* ， *贝叶斯*， *决策树* ， *SVC* 对同样的数据进行处理准确率分别为： * K-means : 0.6324 * 贝叶斯 ： 0.4236 * 决策树 ： 0.5012 * SVC ： 0.5326 经过第一步粗略的统计，决定采用K-means算法 ```py trainFeature, trainLabel = DataPre.getTrainFeature() testFeature, Label2 = DataPre.getInputFeature("../data/Img", "../data/Text") Label2 = [int(pr) for pr in Label2] # KNN def k_means(K): model = KNeighborsClassifier(n_neighbors=K) model.fit(trainFeature, trainLabel) return model # Bayes def Bayes(): model = GaussianNB() model.fit(trainFeature, trainLabel) return model.predict(testFeature) # Tree def Tree(): dtc = DecisionTreeClassifier() dtc.fit(trainFeature, trainLabel) return dtc.predict(testFeature) def SVC_(): model = SVC(kernel="linear") model.fit(trainFeature, trainLabel) return model.predict(testFeature) ``` ## 模型优化 对于 K-means算法来说，最重要的莫过于K值的选择，如何判断最优情况，决定遍历选择 此次遍历遍历 K 从 1 ~ 9 的情况 为了简介表述以及方便观看，我们只挑选有代表性的展示 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/1.png" alt="1" style="zoom:80%;" /> 可以观察到 K = 1 的时候， 后面数字的准确度堪忧，所以我们舍弃 K = 1 的情况， K = 1， 2， 3 均为此种情况 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/4KK.png" alt="4" style="zoom:80%;" /> <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/5KK.png" alt="5" style="zoom:80%;" /> 可以观察到 K = 4， 5， 6 ~ 9 的时候较为合理，所以我们选取的 K 在这个区间段。 各种情况的运行日志如下 ： > read data cost 59.17566 second > > the model which K is 1 cost 101.19315 second > > * A model with K of 1 has an accuracy of **0.39235** > > the model which K is 2 cost 103.17654 second > > * A model with K of 2 has an accuracy of **0.35766** > > the model which K is 3 cost 104.81046 second > > * A model with K of 3 has an accuracy of **0.42246** > > the model which K is 4 cost 113.85034 second > > * A model with K of 4 has an accuracy of **0.62091** > > the model which K is 5 cost 115.52863 second > > * A model with K of 5 has an accuracy of **0.62909** > > the model which K is 6 cost 119.29570 second > > * A model with K of 6 has an accuracy of **0.63032** > > the model which K is 7 cost 108.92201 second > > * A model with K of 7 has an accuracy of **0.63046** > > the model which K is 8 cost 105.33319 second > > * A model with K of 8 has an accuracy of **0.62913** > > the model which K is 9 cost 106.15953 second > > * A model with K of 9 has an accuracy of **0.63036** 虽然 K = 9 的准确度更高， 但考虑到泛化误差，依然以 K = 4 作为最后的最优选择 **注意到所有 K 值的情况下在 $0$ , $1$ 上预测数量较大， 在 $4$ , $5$ 的范围内暴跌， 所以尝试控制数据集来减缓** 在初始的时候，各个数据量的分布情况如下： > **[6372. 7307. 6405. 6602. 6259. 5785. 6345. 6767. 6298. 6422.]** 于是我们分别减缓 $0$ , $1$ 的数据量来控制预测的个数以及质量 我们选取了两个典型分布分析 分布一： > **[4892. 5226. 6405. 6602. 6259. 5785. 6345. 6767. 6298. 6422.]** > > A model with K of 4 has an accuracy of **0.59888** <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/4-%E3%80%904892%EF%BC%8C5226%E3%80%91.png" alt="4-【4892，5226】" style="zoom:80%;" /> 分布二 ： > **[4685. 3830. 6405. 6602. 6259. 5785. 6345. 6767. 6298. 6422.]** > > A model with K of 4 has an accuracy of **0.59802** <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/4-%E3%80%901685%EF%BC%8C3830%E3%80%91.png" alt="4-【1685，3830】" style="zoom:80%;" /> 经过分析可知，分布二对情况四的预测情况更加下跌，而分布一在对 $8$ 的准确率下跌，其他情况类似，于是保持原数据不动 ## 用户输入图像优化 在最开始的测试情况下，因为使用的平板书写，所以痕迹较粗，转到拍照上传后发现出现痕迹过细无法识别情况，于是添加边缘检测以及特征优化功能 > 测试情况 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/image-20230526213647380.png" alt="image-20230526213647380" style="zoom:80%;" /> > 实际情况 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/Snipaste_2023-05-25_19-11-16.png" alt="Snipaste_2023-05-25_19-11-16" style="zoom: 67%;" /> **于是决定对边缘进行识别并优化** 于是打算采取 $Canny$ 算法去进行边缘检测并加在图像上 ```py Use = cv2.Canny(image, 10,10) for i in range(32): for j in range(32): if Use[i][j] == 255: Info[i * 32 + j] = 1 img[i][j] = 255 ``` > 实际效果如下 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/Snipaste_2023-05-25_20-56-12.png" alt="Snipaste_2023-05-25_20-56-12" style="zoom:60%;" /> 观察到仍有不完全存在，于是进行轮廓识别并绘制 ```py _, image = cv2.threshold(image, 127, 255, cv2.THRESH_BINARY) _, contours, _ = cv2.findContours(image, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE) cv2.drawContours(image, contours,-1, (0, 255, 0), 2) ``` > 实际效果 <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/Snipaste_2023-05-25_20-54-58.png" alt="Snipaste_2023-05-25_20-54-58" style="zoom:80%;" /> 观察到边框为白色，进行去除 ```py for i in range(32): for j in range(32): if i == 1 or i == 0 or i == 31 or i == 30: Info[i*32 + j] = 0 img[i][j] = 0 if j == 1 or j == 0 or j == 31 or j == 30: Info[i*32 + j] = 0 img[i][j] = 0 ``` <img src="https://gitee.com/qq3109778990/remem_pic/raw/master/img/Snipaste_2023-05-25_20-57-09.png" alt="Snipaste_2023-05-25_20-57-09" style="zoom:80%;" /> 至此，优化完成