KNN 图像分类python实现

KNN思路相对比较简单，主要包括训练过程和分类过程。在训练过程上，只需要将训练集存储起来就可以。在分类过程中，将测试集和训练集中的每一张图片去比较，选取差别最小的那张图片就可以了。
数据集使用的是CIFAR-10，下载链接http://www.cs.toronto.edu/~kriz/cifar.html

学习心得：如果数据集多，就把训练集分成两部分，一小部分作为验证集（假的测试集），剩下的都为训练集（一般来说是70%-90%，具体多少取决于需要调整的超参数的多少，如果超参数多，验证集占比就更大一点）。验证集的好处是用来调节超参数，为什么不直接用测试集呢？这样容易过拟合，泛化能力差。如果数据集不多，就使用交叉验证的方法来调节参数。但是交叉验证的代价比较高，如果可以支付的起计算代价，使用交叉验证当然是更好的。而且K折交叉验证，K越大越好，但是代价也更高。
kNN缺点：
1. 分类器必须存储所有的训练集数据，用来和未来的测试数据集比较。
2. 要将测试集和所有的训练集进行比较，因此代价很高。
由于图像的维度一般都很高，所以一般不使用KNN，因为计算距离的代价很高。作为新手熟悉kNN的过程，实现kNN还是很好的。

具体实现过程如下：

import numpy as npclass kNearestNeighbor:    def __init__(self):        pass    def train(self, X, y):        self.Xtr = X        self.ytr = y    def predict(self, X, k=1):        num_test = X.shape[0]        Ypred = np.zeros(num_test, dtype = self.ytr.dtype)        for i in range(num_test):            distances = np.sum(np.abs(self.Xtr - X[i,:]), axis = 1)            closest_y = y_train[np.argsort(distances)[:k]]            u, indices = np.unique(closest_y, return_inverse=True)            Ypred[i] = u[np.argmax(np.bincount(indices))]        return Ypred

load_CIFAR_batch()和load_CIFAR10()是用来加载CIFAR-10数据集的

import pickledef load_CIFAR_batch(filename):    """ load single batch of cifar """    with open(filename, 'rb') as f:        datadict = pickle.load(f, encoding='latin1')        X = datadict['data']        Y = datadict['labels']        X = X.reshape(10000, 3, 32, 32).transpose(0,2,3,1).astype("float")        Y = np.array(Y)        return X, Y

import osdef load_CIFAR10(ROOT):    """ load all of cifar """    xs = []    ys = []    for b in range(1,6):        f = os.path.join(ROOT, 'data_batch_%d' %(b))        X, Y = load_CIFAR_batch(f)        xs.append(X)        ys.append(Y)    Xtr = np.concatenate(xs) #使变成行向量    Ytr = np.concatenate(ys)    del X,Y    Xte, Yte = load_CIFAR_batch(os.path.join(ROOT, 'test_batch'))    return Xtr, Ytr, Xte, Yte

Xtr, Ytr, Xte, Yte = load_CIFAR10('cifar10')Xtr_rows = Xtr.reshape(Xtr.shape[0], 32 * 32 * 3)Xte_rows = Xte.reshape(Xte.shape[0], 32 * 32 * 3)

#由于数据集稍微有点大，在电脑上跑的很慢，所以取训练集5000个，测试集500个num_training = 5000num_test = 500x_train = Xtr_rows[:num_training, :]y_train = Ytr[:num_training]x_test = Xte_rows[:num_test, :]y_test = Yte[:num_test]

knn = kNearestNeighbor()knn.train(x_train, y_train)y_predict = knn.predict(x_test, k=7)acc = np.mean(y_predict == y_test)print('accuracy : %f' %(acc))

accuracy : 0.302000

#k值取什么最后的效果会更好呢？可以使用交叉验证的方法，这里使用的是5折交叉验证num_folds = 5k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100]x_train_folds = np.array_split(x_train, num_folds)y_train_folds = np.array_split(y_train, num_folds)k_to_accuracies = {}for k_val in k_choices:    print('k = ' + str(k_val))    k_to_accuracies[k_val] = []    for i in range(num_folds):        x_train_cycle = np.concatenate([f for j,f in enumerate (x_train_folds) if j!=i])        y_train_cycle = np.concatenate([f for j,f in enumerate (y_train_folds) if j!=i])        x_val_cycle = x_train_folds[i]        y_val_cycle = y_train_folds[i]        knn = kNearestNeighbor()        knn.train(x_train_cycle, y_train_cycle)        y_val_pred = knn.predict(x_val_cycle, k_val)        num_correct = np.sum(y_val_cycle == y_val_pred)        k_to_accuracies[k_val].append(float(num_correct) / float(len(y_val_cycle)))

k = 1k = 3k = 5k = 8k = 10k = 12k = 15k = 20k = 50k = 100

for k in sorted(k_to_accuracies):    for accuracy in k_to_accuracies[k]:        print('k = %d, accuracy = %f' % (int(k), accuracy))

k = 1, accuracy = 0.098000k = 1, accuracy = 0.148000k = 1, accuracy = 0.205000k = 1, accuracy = 0.233000k = 1, accuracy = 0.308000k = 3, accuracy = 0.089000k = 3, accuracy = 0.142000k = 3, accuracy = 0.215000k = 3, accuracy = 0.251000k = 3, accuracy = 0.296000k = 5, accuracy = 0.096000k = 5, accuracy = 0.176000k = 5, accuracy = 0.240000k = 5, accuracy = 0.284000k = 5, accuracy = 0.309000k = 8, accuracy = 0.100000k = 8, accuracy = 0.175000k = 8, accuracy = 0.263000k = 8, accuracy = 0.289000k = 8, accuracy = 0.310000k = 10, accuracy = 0.099000k = 10, accuracy = 0.174000k = 10, accuracy = 0.264000k = 10, accuracy = 0.318000k = 10, accuracy = 0.313000k = 12, accuracy = 0.100000k = 12, accuracy = 0.192000k = 12, accuracy = 0.261000k = 12, accuracy = 0.316000k = 12, accuracy = 0.318000k = 15, accuracy = 0.087000k = 15, accuracy = 0.197000k = 15, accuracy = 0.255000k = 15, accuracy = 0.322000k = 15, accuracy = 0.321000k = 20, accuracy = 0.089000k = 20, accuracy = 0.225000k = 20, accuracy = 0.270000k = 20, accuracy = 0.319000k = 20, accuracy = 0.306000k = 50, accuracy = 0.079000k = 50, accuracy = 0.248000k = 50, accuracy = 0.278000k = 50, accuracy = 0.287000k = 50, accuracy = 0.293000k = 100, accuracy = 0.075000k = 100, accuracy = 0.246000k = 100, accuracy = 0.275000k = 100, accuracy = 0.284000k = 100, accuracy = 0.277000

可视化交叉验证的结果

import matplotlib.pyplot as pltplt.rcParams['figure.figsize'] = (10.0, 8.0)plt.rcParams['image.interpolation'] = 'nearest'plt.rcParams['image.cmap'] = 'gray'

for k in k_choices:    accuracies = k_to_accuracies[k]    plt.scatter([k] * len(accuracies), accuracies)accuracies_mean = np.array([np.mean(v) for k,v in sorted(k_to_accuracies.items())])accuracies_std = np.array([np.std(v) for k,v in sorted(k_to_accuracies.items())])plt.errorbar(k_choices, accuracies_mean, yerr=accuracies_std)plt.title('Cross-validation on k')plt.xlabel('k')plt.ylabel('Cross-validation accuracy')plt.show()

结果如图：