cs231n assignment1_Q1_KNN Classifier

今天开始将进行学习cs231n课程并完成相关的作业，在此记录。

配置环境
首先在做作业之前，需要配置作业相关的环境才能进行，我的步骤如下：

1. 下载并安装anaconda：https://www.anaconda.com/download/ ，这是一个Python的科学包，它集成了很多机器学习、深度学习要用的相关库，也自带了一个Python。
2. 安装Jupyter notebook，直接可以在命令行安装，pip install jupyter
3. 下载数据集CIFAR10，网址：http://www.cs.toronto.edu/~kriz/cifar.html
4. 在有Pycharm的前提下，在Pycharm上使用Jupyter notebook进行编写。
   Jupyter这个工具可以及时在工作界面上看到运行的结果。

做作业

题目：

k-最近邻（kNN）练习

完成并将完成的工作表（包括其输出和工作表之外的任何支持代码）与您的作业提交一起提交。 有关详细信息，请参阅课程网站上的作业页面。

kNN分类器包括两个阶段：

• 在训练期间，分类器获取训练数据并简单地记住它
• 在测试期间，kNN通过与所有训练图像进行比较并且转移k个最相似训练示例的标签来对每个测试图像进行分类
• k的值是交叉验证的

在本练习中，您将实现这些步骤并理解基本的图像分类管道，交叉验证，并获得编写高效矢量化代码的熟练程度。

步骤

1. 到官网下载作业包，网址：https://github.com/cs231n/cs231n.github.io/blob/master/assignments/2017/assignment1.md
   然后解压。
2. 进入pycharm，新建项目到之前步骤1解压的目录下，将下载的CIFAR10压缩包解压到cs231n\datasets下。
3. 在pycharm中打开assignment1/knn.ipynb进行编写。

下面贴上代码

knn.ipynb

import random
import numpy as np
from cs231n.data_utils import load_CIFAR10
import matplotlib.pyplot as plt


# This is a bit of magic to make matplotlib figures appear inline in the notebook
# rather than in a new window.
%matplotlib inline 

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_ext autoreload
%autoreload 2

# 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('Training data shape: ', X_train.shape)   
print('Training labels shape: ', y_train.shape)
print('Test data shape: ', X_test.shape)
print('Test labels shape: ', y_test.shape)

Training data shape: (50000, 32, 32, 3)
Training labels shape: (50000,)
Test data shape: (10000, 32, 32, 3)
Test labels shape: (10000,)

# Visualize some examples from the 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

#############
#
# enumerate函数示例：
# seasons = ['Spring', 'Summer', 'Fall', 'Winter']
# >>> list(enumerate(seasons))
# [(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]
#
#############

for y, cls in enumerate(classes):  
    idxs = np.flatnonzero(y_train == y) #此方法返回了训练集中有相同标签的索引值
    idxs = np.random.choice(idxs, samples_per_class, replace=False) #产生一个随机采样
    for i, idx in enumerate(idxs):
        plt_idx = i * num_classes + y + 1
        plt.subplot(samples_per_class, num_classes, plt_idx) #行数，列数，每行的第几个图像
        plt.imshow(X_train[idx].astype('uint8'))
        plt.axis('off')
        if i == 0:
            plt.title(cls)
plt.show()

# Subsample the data for more efficient code execution in this exercise
#为了后面减少计算量，随机采样数据集，训练集采样5000张，测试集采样500张
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]

# Reshape the image data into rows
#将32*32*3的图片reshape成一行
X_train = np.reshape(X_train, (X_train.shape[0], -1))  #逗号表达式， -1表示不确定。
X_test = np.reshape(X_test, (X_test.shape[0], -1))
print(X_train.shape, X_test.shape)

from cs231n.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) #仅存储训练样本数据

# Open cs231n/classifiers/k_nearest_neighbor.py and implement
# compute_distances_two_loops.

# Test your implementation:
dists = classifier.compute_distances_two_loops(X_test) #用两层循环计算L2距离
print(dists.shape)

(500, 5000)

# 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()

# 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
#print(y_test_pred.shape,y_test.shape)
num_correct = np.sum(y_test_pred == y_test) #分类正确的次数
accuracy = float(num_correct) / num_test
print('Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy))

Got 137 / 500 correct => accuracy: 0.274000

y_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))

Got 139 / 500 correct => accuracy: 0.278000

# 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)

# 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')

Difference was: 0.000000
Good! The distance matrices are the same

# 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')

Difference was: 0.000000
Good! The distance matrices are the same

# 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

Two loop version took 55.624931 seconds
One loop version took 103.568292 seconds
No loop version took 0.680607 seconds

num_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.                                #
################################################################################

X_train_folds = np.split(X_train,5,axis=0)
y_train_folds = np.split(y_train,5,axis=0)

################################################################################
#                                 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 = []
    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)
        y_test_pred = classifier.predict_labels(dists,k)
        num_correct = np.sum(y_test_pred == y_valid_cv)
        accuracy = float(num_correct) / y_valid_cv.shape[0]
        #print (accuracy)
        accuracies.append(accuracy)
    
    k_to_accuracies[k]= accuracies

################################################################################
#                                 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))

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

# 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()

# 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.
best_k = 10

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))

Got 141 / 500 correct => accuracy: 0.282000

接下来是K_nearest_neighbor.py的代码

import numpy as np
from past.builtins import xrange


class KNearestNeighbor(object):
  """ a kNN classifier with L2 distance """

  def __init__(self):
    pass

  def train(self, X, y):
 .X_train = X
    self.y_train = y
    
  def predict(self, X, k=1, num_loops=0):
oops == 0:
      dists = self.compute_distances_no_loops(X)
    elif num_loops == 1:
      dists = self.compute_distances_one_loop(X)
    elif num_loops == 2:
      dists = self.compute_distances_two_loops(X)
    else:
      raise ValueError('Invalid value %d for num_loops' % num_loops)

    return self.predict_labels(dists, k=k)

  def compute_distances_two_loops(self, X):
  t = X.shape[0] #dists矩阵行数
    num_train = self.X_train.shape[0] #dists矩阵列数
    dists = np.zeros((num_test, num_train)) #距离初始化为0
    for i in xrange(num_test):
      for j in xrange(num_train):
        #####################################################################
        # TODO:                                                             #
        # Compute the l2 distance between the ith test point and the jth    #
        # training point, and store the result in dists[i, j]. You should   #
        # not use a loop over dimension.                                    #
        #####################################################################

        dists[i][j] = np.sqrt(np.sum(np.square(X[i]-self.X_train[j]))) #两向量之间的L2距离。

        #####################################################################
        #                       END OF YOUR CODE                            #
        #####################################################################
    return dists

  def 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))#按行来执行

      #######################################################################
      #                         END OF YOUR CODE                            #
      #######################################################################
    return dists

  def 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.                                         #
    #########################################################################

    #就是将距离函数拆开，完成各个单项后再合并（x - y）^2 = x^2 - 2*x*y + y^2
    #当axis为0时,是压缩行,即将每一列的元素相加,将矩阵压缩为一行
    #当axis为1时,是压缩列,即将每一行的元素相加,将矩阵压缩为一列
    #这里的计算要特别注意维度的不同

    ab = np.dot(X, self.X_train.T)  # num_test * num_train 点积
    a2 = np.sum(np.square(X), axis=1).reshape(-1, 1)  # num_test * 1
    b2 = np.sum(np.square(self.X_train.T), axis=0).reshape(1, -1)  # 1 * num_train
    dists = -2 * ab + a2 + b2  # 不同维度计算会自动 broadcast
    dists = np.sqrt(dists)


    #########################################################################
    #                         END OF YOUR CODE                              #
    #########################################################################
    return dists

  def predict_labels(self, dists, k=1):
    """
    Given a matrix of distances between test points 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 test point 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 predicted label for the test point X[i].  
    """
    num_test = dists.shape[0] #dists矩阵行数
    #print(dists.shape[0])
    y_pred = np.zeros(num_test)
    for i in xrange(num_test):
      # A list of length k storing the labels of the k nearest neighbors to
      # the ith test point.

      #closest_y = []

      #########################################################################
      # TODO:                                                                 #
      # Use the distance matrix to find the k nearest neighbors of the ith    #
      # testing point, and use self.y_train to find the labels of these       #
      # neighbors. Store these labels in closest_y.                           #
      # Hint: Look up the function numpy.argsort.                             #
      #########################################################################

      #argsort函数返回的是数组值从小到大的索引值。

      closest_y = self.y_train[ np.argsort(dists[i])[0:k] ]

      #########################################################################
      # TODO:                                                                 #
      # Now that you have found the labels of the k nearest neighbors, you    #
      # need to find the most common label in the list closest_y of labels.   #
      # Store this label in y_pred[i]. Break ties by choosing the smaller     #
      # label.                                                                #
      #########################################################################

      #argmax函数沿给定轴返回最大元素的索引
      y_pred[i] = np.argmax(np.bincount(closest_y))  #bincount函数统计结果的“票数”

      #########################################################################
      #                           END OF YOUR CODE                            # 
      #########################################################################

    return y_pred

其中non-loop的算法原理：

本次作业重点就是在对于矩阵的处理上，掌握numpy的基本操作。