（cs231n）assignment1_knn

k-Nearest Neighbor (kNN) exercise

Complete and hand in this completed worksheet (including its outputs and any supporting code outside of the worksheet) with your assignment submission. For more details see the assignments page on the course website.

The kNN classifier consists of two stages:

• During training, the classifier takes the training data and simply remembers it
• During testing, kNN classifies every test image by comparing to all training images and transfering the labels of the k most similar training examples
• The value of k is cross-validated

In this exercise you will implement these steps and understand the basic Image Classification pipeline, cross-validation, and gain proficiency in writing efficient, vectorized code.

本节是K近邻算法的实例，该算法需要计算测试数据与所有训练集数据集之间的距离（L1曼哈顿距离和L2欧式距离，挑选距离最小的k个值做为备选，选择k个结果中出现最多的label作为最终预测结果。主要做三个方面：
1.导入CIFAR-10数据库
2.使用L2（欧式距离）进行计算，比较代码使用双循环，单循环和不使用循环结构的运算速度
3.进行预测，根据上步得到的距离关系，挑选出k个选民，出现最多的类别即为预测结果。

载入数据集，调入函数，利用matplotlib绘制图像

from __future__ import print_function
# Run some setup code for this notebook.

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,图片像素
plt.rcParams['image.interpolation'] = 'nearest' #设置差值方式为nearest
plt.rcParams['image.cmap'] = 'gray' #设置颜色style

# 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' #cifar10的路径
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.
# Use .shape to see the dimension
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
for y, 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) #一个图中包含多个子图
        plt.imshow(X_train[idx].astype('uint8'))
        plt.axis('off')
        if i == 0:
            plt.title(cls)
plt.show()

调整数据集大小

# 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]

# 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)

(5000, 3072) (500, 3072)

使用KNN开始训练

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)

We would now like to classify the test data with the kNN classifier. Recall that we can break down this process into two steps:

1. First we must compute the distances between all test examples and all train examples.
2. Given these distances, for each test example we find the k nearest examples and have them vote for the label

Lets begin with computing the distance matrix between all training and test examples. For example, if there are Ntr training examples and Nte test examples, this stage should result in a Nte x Ntr matrix where each element (i,j) is the distance between the i-th test and j-th train example.

First, open cs231n/classifiers/k_nearest_neighbor.py and implement the function compute_distances_two_loops that uses a (very inefficient) double loop over all pairs of (test, train) examples and computes the distance matrix one element at a time.

# 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)

(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()

Inline Question #1: Notice the structured patterns in the distance matrix, where some rows or columns are visible brighter. (Note that with the default color scheme black indicates low distances while white indicates high distances.)

• What in the data is the cause behind the distinctly bright rows?
• What causes the columns?

Your Answer: fill this in.

# 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))

Got 137 / 500 correct => accuracy: 0.274000

You should expect to see approximately 27% accuracy. Now lets try out a larger k, say k = 5:

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

You should expect to see a slightly better performance than with k = 1.

验证单循环和无循环结构与双循环的精度是否一样，再比较他们的耗时

# 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
可以看到三种循环的精度是一样的

k_nearest_neighbor.py

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):
    """
    Train the classifier. For k-nearest neighbors this is just 
    memorizing the training data.

    Inputs:
    - X: A numpy array of shape (num_train, D) containing the training data
      consisting of num_train samples each of dimension D.
    - y: A numpy array of shape (N,) containing the training labels, where
         y[i] is the label for X[i].
    """
    self.X_train = X
    self.y_train = y
    
  def predict(self, X, k=1, num_loops=0):
    """
    Predict labels for test data using this classifier.

    Inputs:
    - X: A numpy array of shape (num_test, D) containing test data consisting
         of num_test samples each of dimension D.
    - k: The number of nearest neighbors that vote for the predicted labels.
    - num_loops: Determines which implementation to use to compute distances
      between training points and testing points.

    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].  
    """
    if num_loops == 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):
    """
    Compute the distance between each test point in X and each training point
    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 ith test 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))
    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.                                    #
        #####################################################################
        pass
        dists[i][j] = np.sqrt(np.sum(np.square(X[i]-self.X_train[j])))
        #####################################################################
        #                       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, :].                        #
      #######################################################################
      pass
      dists[i,:] = np.sqrt(np.sum(np.square(X[i,:] - self.X_train), axis = 1))
      #######################################################################
      #                         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] #.shape[0]读取X的第一行
    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.                                         #
    #########################################################################
    pass
    dists = np.sqrt(np.sum(self.X_train ** 2, axis =1).reshape(1, num_train) + np.sum(X ** 2, axis=1).reshape(num_test, 1) - 2 * np.dot(X, self.X_train.T))
    #########################################################################
    #                         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]
    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.  把距离最短的K个标签保存下来，self.y_train保存得是label名   #
      # Hint: Look up the function numpy.argsort. 由小到大排列                      #
      #########################################################################
      pass
      closet_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.给出了标签的K的最接近的值，在这K次的结果中，出现最多的标签就是正确标签。
      #np.bincount  指出每个标签出现的次数
      #np.argmax 挑出出现最多的标签
      #########################################################################
      pass
      y_pred[i] = np.argmax(np.bincount(closet_y))
      #########################################################################
      #                           END OF YOUR CODE                            # 
      #########################################################################

    return y_pred

比较三种代码结构的耗时

# 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 26.360300 seconds
One loop version took 50.430272 seconds
No loop version took 0.151595 seconds

的确，不使用循环结构可以有效提高运算速度

Cross-validation

We have implemented the k-Nearest Neighbor classifier but we set the value k = 5 arbitrarily. We will now determine the best value of this hyperparameter with cross-validation.

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.                                #
################################################################################
pass
X_train_folds = np.array_split(X_train, num_folds)
y_train_folds = np.array_split(y_train, num_folds)
################################################################################
#                                 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.                               #
################################################################################
pass
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
################################################################################
#                                 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))

1
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

# 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()

根据上步结果，为下方best_k赋值最好的k,(这里看到很多不同的博主赋值为不同的k，使用代码来选择应最为准确

# 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 = k_choices[ np.argmax(accuracies_mean)]

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