前言
前段时间学习了深度学习入门课程斯坦福CS231n,巩固和理解课程的最佳方式就是完成课后代码作业。在这里记录下本人对作业的思考和解析,以供大家参考。
关于CS231n学习笔记翻译,强烈推荐 知乎专栏 。
配合笔记看视频,再完成作业,非常有效。
一、准备工作
1、作业Assignment1下载
链接:https://pan.baidu.com/s/11tDkdBRy5ndwkKue_1vMuw 密码:jvgk
2、环境Anaconda2安装
完成作业需要Python以及许多相关科学计算环境,建议大家直接安装Anaconda2,能够快速方便的配置,直接上手写作业。代码是Python 2.x版本,故选择Anaconda2-4.2.0-Windows-x86_64.exe,否则可能出现某些函数不兼容,造成不必要的麻烦。
3、数据集CIFAR-10下载
简单介绍一下,CIFAR-10是一个只有10类图片的彩色图片数据库,共6w张32*32像素大小的图片,其中5w张是训练图片,1w张是测试图片,所有图片均已标记。
选择CIFAR10 python version下载,解压后将数据放在cs231n/datasets目录下。
4、验证WebSockets是否可用
作业在Ipython Notebook内完成,需要支持WebSockets。打开下述网页,若显示“WORK FOR YOU”,表示可用。
测试链接:http://websocketstest.com/
出现以上字样即可。
若无法使用,需要翻墙。网上可用VPN有很多,大家可以搜索一下。
至此准备工作全部完成,可以开始写作业啦。
二、作业解析
1、连接服务器
如图所示,打开cmd,在[your path]/assignment1/目录下输入Ipython notebook 或者 Jupyter notebook,回车。
更多Ipython notebook教程可参考 这里 。
注意:一定要在目录下打开,否则后面会出现找不到包的情况。
退出的话输入ctrl+c,等待关闭即可。
成功连上服务器后,你将在浏览器下看到如下界面:
这次的作业就需要在knn.ipynb下完成。
2、运行代码
选中代码块,ctrl+enter或shift+enter运行,区别是shift+enter会自动跳转到下一部分。页面右上角的圆圈表示运行状态,空心表示结束,实心表示正在运行。
下面对各部分代码进行补充和运行。
kNN分类器包括两个部分:
- 在训练的时候,分类器载入数据并仅仅记住他们(不做其他处理)
- 在测试的时候,分类器依靠对测试图片和训练图片做对比,并且选出k个最相似的标签
- 另外,k是依靠交叉验证确定的
在这个作业中,我们将实现这些步骤并且理解基本的图像分类过程,理解交叉验证,并且学会写高效的向量化代码。
3、代码解析
3.1 一些基本的初始化
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20# Run some setup code for this notebook. importrandom importnumpy as np fromcs231n.data_utils import load_CIFAR10 importmatplotlib.pyplot as plt from__future__ import print_function #This is a bit of magic to make matplotlib figures appear inline in the notebook #rather than in a new window. %matplotlibinline plt.rcParams['figure.figsize']= (10.0, 8.0) # set default size of plots plt.rcParams['image.interpolation']= 'nearest' plt.rcParams['image.cmap']= 'gray' #Some more magic so that the notebook will reload external python modules; #see http://stackoverflow.com/questions/1907993/autoreload-of-modules-in-ipython %load_extautoreload %autoreload2
一些初始化代码,载入必要的包,保证图像输出在网页中而不新建窗口。
3.2 载入数据
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9# Load the raw CIFAR-10 data. cifar10_dir= 'cs231n/datasets/cifar-10-batches-py' X_train,y_train, X_test, y_test = load_CIFAR10(cifar10_dir) #As a sanity check, we print out the size of the training and test data. print('Trainingdata shape: ', X_train.shape) print('Traininglabels shape: ', y_train.shape) print('Testdata shape: ', X_test.shape) print('Testlabels shape: ', y_test.shape)
载入CIFAR-10数据。输出数据格式:
由于是彩图3通道,故大小为32*32*3.
3.3 展示部分训练图
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18# Visualize some examples fromthe dataset. #We show a few examples of training images from each class. classes= ['plane', 'car', 'bird', 'cat', 'deer', 'dog', 'frog', 'horse', 'ship','truck'] num_classes= len(classes) samples_per_class= 7 fory, cls in enumerate(classes): idxs = np.flatnonzero(y_train == y) idxs = np.random.choice(idxs,samples_per_class, replace=False) for i, idx in enumerate(idxs): plt_idx = i * num_classes + y + 1 plt.subplot(samples_per_class,num_classes, plt_idx) plt.imshow(X_train[idx].astype('uint8')) plt.axis('off') if i == 0: plt.title(cls) plt.show()
从每一类中展示7张训练图片。结果如下:
3.4 取样数据
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10#Subsample the data for more efficient code execution in this exercise num_training= 5000 mask= list(range(num_training)) X_train= X_train[mask] y_train= y_train[mask] num_test= 500 mask= list(range(num_test)) X_test= X_test[mask] y_test= y_test[mask]
在练习中,为了更高效地执行代码,我们只取样部分数据。选取5000张测试图片,500张测试图片。
3.5 数据变形
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5#Reshape the image data into rows X_train= np.reshape(X_train, (X_train.shape[0], -1)) X_test= np.reshape(X_test, (X_test.shape[0], -1)) print(X_train.shape,X_test.shape)
将图片转化为行向量,所有图片组成二维矩阵。32*32*3=3072.
故结果如下:
1(5000L,3072L) (500L, 3072L)
3.6 载入函数
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6fromcs231n.classifiers import KNearestNeighbor #Create a kNN classifier instance. #Remember that training a kNN classifier is a noop: #the Classifier simply remembers the data and does no further processing classifier= KNearestNeighbor() classifier.train(X_train,y_train)
载入KnearestNeighbour包,创建KNearestNeighbor类的对象classifier,调用train()函数。
3.7 二重循环计算距离矩阵
现在我们开始计算代表测试图片和训练图片之间距离的矩阵。如果有Ntr张训练数据,Nte张测试数据,那么应该得到Nte*Ntr的距离矩阵,其中(i,j)表示第 i 张测试图距离第 j 张训练图片的距离。
所以,我们需要实现的cs231n/classifiers/k_nearest_neighbor.py目录下的函数compute_distances_two_loops(),使用了二重循环。
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33defcompute_distances_two_loops(self, X): """ Compute the distance between each test point in X and each trainingpoint in self.X_train using a nested loop over both the training data and the test data. Inputs: - X: A numpy array of shape (num_test, D) containing test data. Returns: - dists: A numpy array of shape (num_test, num_train) where dists[i, j] is the Euclidean distance between the ithtest point and the jth training point. """ num_test = X.shape[0] num_train = self.X_train.shape[0] dists = np.zeros((num_test, num_train)) #500*5000 for i in xrange(num_test): for j in xrange(num_train): dists[i,j] = np.sqrt(np.sum(np.square(self.X_train[j,:]- X[i,:]))) ##################################################################### # TODO: # # Compute the L2 distance between the ithtest point and the jth # # training point, and store the resultin dists[i, j]. You should # # not use a loop over dimension. # ##################################################################### #pass ##################################################################### # END OF YOUR CODE # ##################################################################### return dists
使用的是L2距离,注意 i 和 j 分别代表测试集和训练集。
3.8 测试距离矩阵
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6#Open cs231n/classifiers/k_nearest_neighbor.py and implement #compute_distances_two_loops. #Test your implementation: dists= classifier.compute_distances_two_loops(X_test) print(dists.shape)
调用刚刚写好的函数,打印距离矩阵的大小:
1(500L, 5000L)
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4#We can visualize the distance matrix: each row is a single test example and #its distances to training examples plt.imshow(dists,interpolation='none') plt.show()
将距离矩阵可视化。每一行表示测试样例距所有训练图片的距离。
如上图所示,纵坐标表示500张测试图片,横坐标表示5000张训练图片。越黑表示距离越接近,越亮表示距离越远。
3.9 预测测试图片的标签
实现预测函数predict_labels(),在cs231n/classifiers/k_nearest_neighbor.py目录下。根据3.7~3.8算出的dists距离矩阵,选出离测试图片最近的k张训练图,投票选出最可能的预测结果。若k=1,就选出距离最近的一张训练图,将该图的标签作为测试图的标签。
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45def predict_labels(self, dists, k=1): """ Given a matrix of distances between testpoints and training points, predict a label for each test point. Inputs: - dists: A numpy array of shape (num_test,num_train) where dists[i, j] gives the distance betwen the ith testpoint and the jth training point. Returns: - y: A numpy array of shape (num_test,)containing predicted labels for the test data, where y[i] is the predictedlabel for the test point X[i]. """ num_test = dists.shape[0] y_pred = np.zeros(num_test) #500*1 for i in xrange(num_test): # A list of length k storing the labelsof the k nearest neighbors to # the ith test point. closest_y = [] ######################################################################### # TODO: # # Use the distance matrix to find the knearest neighbors of the ith # # testing point, and use self.y_train tofind the labels of these # # neighbors. Store these labels inclosest_y. # # Hint: Look up the functionnumpy.argsort. # ######################################################################### closest_y = np.argsort(dists[i,:]) # i'm socool #pass ######################################################################### # TODO: # # Now that you have found the labels ofthe k nearest neighbors, you # # need to find the most common label inthe list closest_y of labels. # # Store this label in y_pred[i]. Breakties by choosing the smaller # # label. # ######################################################################### y_pred[i] =np.argmax(np.bincount(self.y_train[closest_y[:k]])) #pass ######################################################################### # END OF YOURCODE # ######################################################################### return y_pred
对所有的测试图片遍历,将dists[]矩阵按行排序(由大到小),索引放于closet_y向量中。对cloest_y中前k个向量进行计数np.bincount(),最终用np.argmax()得到票数最多的下标,作为最终的标签y_pred[i]。
举个例子:
假如cloest_y向量中排名前五的数字分别为[1,1,1,3,2],那么np.bincount()将会返回索引在该数组内出现的次数:
array([0, 3, 1, 1])
因为[1,1,1,3,2]中最大数字为3,故bincount()结果有4个数字,索引值为0~3.数组表示0出现0次,1出现3次,2出现1次,3出现1次。这时候取最大值下标np.argmax(),正好得到索引值1,就是我们希望的结果。
3.10 运行预测代码
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7# Now implement the function predict_labels and run the code below: # We use k = 1 (which is Nearest Neighbor). y_test_pred = classifier.predict_labels(dists, k=1) # Compute and print the fraction of correctly predicted examples num_correct = np.sum(y_test_pred == y_test) accuracy = float(num_correct) / num_test print('Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))
结果如下:
1Got 137/500 correct => accuracy: 0.274000
3.11 测试k=5的情况
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5y_test_pred = classifier.predict_labels(dists, k=5) num_correct = np.sum(y_test_pred == y_test) accuracy = float(num_correct) / num_test print('Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))
结果如下:
1Got 139/500 correct => accuracy: 0.278000
比起刚才的27.4%,准确度有了些微的提升。
3.12 半向量化代码
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23def compute_distances_one_loop(self, X): """ Compute the distance between each test point in X and each training point in self.X_train using a single loop over the test data. Input / Output: Same as compute_distances_two_loops """ num_test = X.shape[0] num_train = self.X_train.shape[0] dists = np.zeros((num_test, num_train)) for i in xrange(num_test): ####################################################################### # TODO: # # Compute the L2 distance between the ith test point and all training # # points, and store the result in dists[i, :]. # ####################################################################### dists[i,:] = np.sqrt(np.sum(np.square(self.X_train - X[i,:]),axis = 1)) #pass ####################################################################### # END OF YOUR CODE # ####################################################################### return dist
和双重循环的区别在于,单层循环只遍历了测试图片,对训练图片的遍历采用了向量化代码完成。axis=1表示沿着水平方向累加。这是因为要对某张测试图片计算他距离每张训练图片的距离。
3.13 测试半向量化代码
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18# Now lets speed up distance matrix computation by using partial vectorization # with one loop. Implement the function compute_distances_one_loop and run the # code below: dists_one = classifier.compute_distances_one_loop(X_test) print(dists_one.shape) # To ensure that our vectorized implementation is correct, we make sure that it # agrees with the naive implementation. There are many ways to decide whether # two matrices are similar; one of the simplest is the Frobenius norm. In case # you haven't seen it before, the Frobenius norm of two matrices is the square # root of the squared sum of differences of all elements; in other words, reshape # the matrices into vectors and compute the Euclidean distance between them. difference = np.linalg.norm(dists - dists_one, ord='fro') print('Difference was: %f' % (difference, )) if difference < 0.001: print('Good! The distance matrices are the same') else: print('Uh-oh! The distance matrices are different')

证明我们的半向量化代码是正确的。
3.14 向量化代码
这一部分是最核心、最难以理解,也是最高效的代码。要求不能包含任何循环,依靠numpy提供的广播机制来计算矩阵。
首先考虑两个需要计算的矩阵。一个包含测试图片的矩阵X,大小是500*3072,表示有500张测试图,每一行代表图片的像素情况;另一个是包含训练图片的矩阵X_train,有5000张图片,故大小为5000*3072。需要做的就是将测试矩阵的每一行和训练矩阵的每一行做L2距离计算,结果存在dists矩阵(500*5000)中。
考虑L2距离的计算公式:
对求和符号内部公式展开得:x^2 + y ^2 – 2*x*y。
所谓广播机制,是指两个矩阵在每一维上维度相等或者其中一个矩阵的维度是1的情况下,较小的矩阵将自动扩展为较大矩阵同样的大小。举例如下:
矩阵a = [[1,2,3]
[1,2,3]
[1,2,3]]
b = [1 1 1]
a矩阵维度为3*3,b矩阵维度为1*3.由于a和b在列上维度相同,b矩阵在行上维度为1。故a = a+b的结果将为:
a = [[2,3,4]
[2,3,4]
[2,3,4]]
相当于将b矩阵按行广播,变为[[1,1,1][1,1,1][1,1,1]]了。
更多广播机制的内容参见 这里 。
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32def compute_distances_no_loops(self, X): """ Compute the distance between each test point in X and each training point in self.X_train using no explicit loops. Input / Output: Same as compute_distances_two_loops """ num_test = X.shape[0] num_train = self.X_train.shape[0] dists = np.zeros((num_test, num_train)) ######################################################################### # TODO: # # Compute the L2 distance between all test points and all training # # points without using any explicit loops, and store the result in # # dists. # # # # You should implement this function using only basic array operations; # # in particular you should not use functions from scipy. # # # # HINT: Try to formulate the l2 distance using matrix multiplication # # and two broadcast sums. # ######################################################################### dists += np.sum(self.X_train ** 2, axis=1).reshape(1, num_train) #1*5000,第一次广播 dists += np.sum(X ** 2, axis=1).reshape(num_test,1) #500*1,第二次广播 dists -= 2 * np.dot(X, self.X_train.T) #500*5000 dists = np.sqrt(dists) #pass ######################################################################### # END OF YOUR CODE # ######################################################################### return dists
3.15 测试向量化代码
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12# Now implement the fully vectorized version inside compute_distances_no_loops # and run the code dists_two = classifier.compute_distances_no_loops(X_test) # check that the distance matrix agrees with the one we computed before: difference = np.linalg.norm(dists - dists_two, ord='fro') print('Difference was: %f' % (difference, )) if difference < 0.001: print('Good! The distance matrices are the same') else: print('Uh-oh! The distance matrices are different')
运行结果:
3.16 比较各函数效果
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22# Let's compare how fast the implementations are def time_function(f, *args): """ Call a function f with args and return the time (in seconds) that it took to execute. """ import time tic = time.time() f(*args) toc = time.time() return toc - tic two_loop_time = time_function(classifier.compute_distances_two_loops, X_test) print('Two loop version took %f seconds' % two_loop_time) one_loop_time = time_function(classifier.compute_distances_one_loop, X_test) print('One loop version took %f seconds' % one_loop_time) no_loop_time = time_function(classifier.compute_distances_no_loops, X_test) print('No loop version took %f seconds' % no_loop_time) # you should see significantly faster performance with the fully vectorized implementation
我们在3.14实现了向量化函数,在3.12实现了半向量化函数,在3.7实现了无向量化函数。对他们用时分别测试。结果如下:
3.17 交叉验证
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62num_folds = 5 k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100] X_train_folds = [] y_train_folds = [] ################################################################################ # TODO: # # Split up the training data into folds. After splitting, X_train_folds and # # y_train_folds should each be lists of length num_folds, where # # y_train_folds[i] is the label vector for the points in X_train_folds[i]. # # Hint: Look up the numpy array_split function. # ################################################################################ # split self.X_train to 5 folds avg_size = int(X_train.shape[0] / num_folds) # will abandon the rest if not divided evenly. for i in range(num_folds): X_train_folds.append(X_train[i * avg_size : (i+1) * avg_size]) y_train_folds.append(y_train[i * avg_size : (i+1) * avg_size]) pass ################################################################################ # END OF YOUR CODE # ################################################################################ # A dictionary holding the accuracies for different values of k that we find # when running cross-validation. After running cross-validation, # k_to_accuracies[k] should be a list of length num_folds giving the different # accuracy values that we found when using that value of k. k_to_accuracies = {} ################################################################################ # TODO: # # Perform k-fold cross validation to find the best value of k. For each # # possible value of k, run the k-nearest-neighbor algorithm num_folds times, # # where in each case you use all but one of the folds as training data and the # # last fold as a validation set. Store the accuracies for all fold and all # # values of k in the k_to_accuracies dictionary. # ################################################################################ for k in k_choices: accuracies = [] print(k) for i in range(num_folds): X_train_cv = np.vstack(X_train_folds[0:i] + X_train_folds[i+1:]) y_train_cv = np.hstack(y_train_folds[0:i] + y_train_folds[i+1:]) X_valid_cv = X_train_folds[i] y_valid_cv = y_train_folds[i] classifier.train(X_train_cv, y_train_cv) dists = classifier.compute_distances_no_loops(X_valid_cv) accuracy = float(np.sum(classifier.predict_labels(dists, k) == y_valid_cv)) / y_valid_cv.shape[0] accuracies.append(accuracy) k_to_accuracies[k] = accuracies pass ################################################################################ # END OF YOUR CODE # ################################################################################ # Print out the computed accuracies for k in sorted(k_to_accuracies): for accuracy in k_to_accuracies[k]: print('k = %d, accuracy = %f' % (k, accuracy))
这部分代码不难理解,主要将训练代码分为5部分,其中一部分作为验证集,来不断改变k值,寻找最优解。其中np.vstack()表示沿着竖直方向将矩阵堆叠,np.hstack()表示沿水平方向堆叠矩阵。运行结果:
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611 3 5 8 10 12 15 20 50 100 k = 1, accuracy = 0.263000 k = 1, accuracy = 0.257000 k = 1, accuracy = 0.264000 k = 1, accuracy = 0.278000 k = 1, accuracy = 0.266000 k = 3, accuracy = 0.239000 k = 3, accuracy = 0.249000 k = 3, accuracy = 0.240000 k = 3, accuracy = 0.266000 k = 3, accuracy = 0.254000 k = 5, accuracy = 0.248000 k = 5, accuracy = 0.266000 k = 5, accuracy = 0.280000 k = 5, accuracy = 0.292000 k = 5, accuracy = 0.280000 k = 8, accuracy = 0.262000 k = 8, accuracy = 0.282000 k = 8, accuracy = 0.273000 k = 8, accuracy = 0.290000 k = 8, accuracy = 0.273000 k = 10, accuracy = 0.265000 k = 10, accuracy = 0.296000 k = 10, accuracy = 0.276000 k = 10, accuracy = 0.284000 k = 10, accuracy = 0.280000 k = 12, accuracy = 0.260000 k = 12, accuracy = 0.295000 k = 12, accuracy = 0.279000 k = 12, accuracy = 0.283000 k = 12, accuracy = 0.280000 k = 15, accuracy = 0.252000 k = 15, accuracy = 0.289000 k = 15, accuracy = 0.278000 k = 15, accuracy = 0.282000 k = 15, accuracy = 0.274000 k = 20, accuracy = 0.270000 k = 20, accuracy = 0.279000 k = 20, accuracy = 0.279000 k = 20, accuracy = 0.282000 k = 20, accuracy = 0.285000 k = 50, accuracy = 0.271000 k = 50, accuracy = 0.288000 k = 50, accuracy = 0.278000 k = 50, accuracy = 0.269000 k = 50, accuracy = 0.266000 k = 100, accuracy = 0.256000 k = 100, accuracy = 0.270000 k = 100, accuracy = 0.263000 k = 100, accuracy = 0.256000 k = 100, accuracy = 0.263000
3.18 结果可视化
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14# plot the raw observations for k in k_choices: accuracies = k_to_accuracies[k] plt.scatter([k] * len(accuracies), accuracies) # plot the trend line with error bars that correspond to standard deviation 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()

横坐标表示不同的k值选择,纵坐标表示交叉验证的准确率。
3.19 验证k值
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21# Based on the cross-validation results above, choose the best value for k, # retrain the classifier using all the training data, and test it on the test # data. You should be able to get above 28% accuracy on the test data. temp = 0 for k in k_choices: accuracies = k_to_accuracies[k] if temp < accuracies[np.argmax(accuracies)]: temp = accuracies[np.argmax(accuracies)] best_k = k print(best_k) classifier = KNearestNeighbor() classifier.train(X_train, y_train) y_test_pred = classifier.predict(X_test, k=best_k) # Compute and display the accuracy num_correct = np.sum(y_test_pred == y_test) accuracy = float(num_correct) / num_test print('Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))
我们找到最佳k值,利用测试集验证。结果如下:
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310 Got 141 / 500 correct => accuracy: 0.282000
需要指出的是,训练集、验证集和测试集是不同的三个集合。训练集是训练用的数据,在其中分裂出一部分作为验证集,用来参数调优;记住千万不能利用测试集来调优,它应该是最后用来检验模型能力的标准。
总结
可以看到,KNN模型用作图像分类任务是没有优势的,训练很简单(保存数据),测试的时候很耗费时间和计算资源(一一比对计算)。即使是最好的情况,识别率也不足30%。我们用这个模型来熟悉图像分类的大致流程,训练我们的向量化思维。
最后
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