简要介绍KNN（K Nearest Neighbor）
Nearest Neighbor
介绍KNN以前，我们先了解一下Nearest Neighbor。Nearest Neighbor主要用于分类，以CIFAR-10数据集为例，有50000张训练图片和与之对应的标签。我们将要为测试集标上标签，我们取一张测试集中的图片，在训练集中找出与之最相似的一张图片，那么该测试集图片与该最相似的一张的图片的标签相同。但是这样也有局限性。
以改图为例，我们现在要对绿色圆形进行分类，看起来绿色圆形仿佛应该属于红色三角形，但是事实可能往往并不是这样，所以便有了KNN。
K Nearest Neighbor
如果一个样本在特征空间中的K个最相邻的样本中的大多数属于某一个类别，则该样本也属于这个类别，并具有这个类别上样本的特性。该方法在确定分类决策上只依据最邻近的一个或者几个样本的类别来决定待分样本所属的类别。 KNN方法在类别决策时，只与极少量的相邻样本有关。由于KNN方法主要靠周围有限的邻近的样本，而不是靠判别类域的方法来确定所属类别的，因此对于类域的交叉或重叠较多的待分样本集来说，KNN方法较其他方法更为适合。
KNN算法比NN算法误差要小一些，因为KNN中的K我们可以通过在训练集找到一个比较好的K，在测试集上KNN的误差比NN的误差要小。
还是以此图为例，如果用NN算法，绿色圆形本应属于红色三角形，当K=3时，也是如此。但是当K=5时，绿色圆形属于蓝色三角性，可见，绿色圆点到底属于哪一类，与具体的K有关。所以我们要看分类的对象，并训练KNN分类器，毕竟如果NN算法要好的话，也是KNN算法中K=1的特殊情况。
也就是说：K Nearest Neighbor 算法包含了 Nearest Neighbor算法
如何度量相似
L1（Manhattan）distance
L2 distance
numpy技巧
idxs = np.flatnonzero()： 返回非0元素的位置
idxs = np.random.choice()：从数组中随机抽取元素
plt.subplot()：子图绘制
np.linalg.norm()：计算范数
np.array_split()：数组划分
np.vstack(): 按垂直方向（行顺序）堆叠数组构成一个新的数组
np.hstack(): 按水平方向（列顺序）堆叠数组构成一个新的数组
np.argsort()：排序，返回索引值
np.bincount()：返回每个索引出现的次数
KNN代码实现
github：https://github.com/GIGpanda/CS231n
一共两个.py文件，knn.py和k_nearest_neighbor.py
knn.py
数据加载
加载CIFAR-10数据，并打印训练集和测试集的图片、标签尺寸。
# KNN
from __future__ import print_function
import random
import numpy as np
from cs231n.data_utils import load_CIFAR10
import matplotlib.pyplot as plt
from cs231n.classifiers import KNearestNeighbor
import time
plt.rcParams['figure.figsize'] = (10.0, 8.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'
# 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)
可视化部分数据
对于训练集中的10类数据，每类随机取出7张并可视化。
# 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) # np.flatnonzero 返回非0元素的位置
idxs = np.random.choice(idxs, samples_per_class, replace=False)
# np.random.choice 从数组中随机抽取元素
# 从数组idxs中随机抽取数字, 组成大小为samples_per_class的数组, replace=False表示不可以取相同的数字
for i, idx in enumerate(idxs):
plt_idx = i * num_classes + y + 1
# plt_idx 算出每张图片打印出来的位置 也就是同一类图片在同一列
plt.subplot(samples_per_class, num_classes, plt_idx)
# plt.subplot 参数描述子图的位置信息
# samples_per_class: 行 num_classes: 列 plt_idx: 索引值
# 索引从1开始 从左上角到右下角
plt.imshow(X_train[idx].astype('uint8'))
plt.axis('off')
# off: turn off axis line and labels
if i == 0:
plt.title(cls) # 画出每个子图的标题
plt.show()
Subsample
取出训练集和测试集中的部分数据训练KNN。
# Subsample the data for more efficient code execution in this exercise
num_training = 5000
mask = list(range(num_training))
# range(start, stop, step) 创建一个整数列表 默认从0开始
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)
Train a KNN classifier
训练K=1和K=5时，分类的准确率。
# 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)
# 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)
# 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)
# np.argsort
# 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))
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 => accuaracy: %f'%(num_correct, num_test, accuracy))
Use one loop and check
用一重循环计算L2距离，并验证。
# 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')
Use no loop and check
不用循环计算L2距离，并验证。
# 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')
比较no_loop, one_loop, two_loop的时间开销
# 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.
"""
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
交叉验证
将训练集分成5份，4份作为训练数据，1份作为测试数据。
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. #
################################################################################
# Your code
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 = {}
for k in k_choices:
k_to_accuracies[k] = []
################################################################################
# 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. #
################################################################################
# Your code
for i in k_choices:
for j in range(num_folds):
idx = j
X_train = np.vstack(X_train_folds[0:idx] + X_train_folds[idx+1:num_folds])
y_train = np.hstack(y_train_folds[0:idx] + y_train_folds[idx+1:num_folds])
X_test = X_train_folds[idx]
y_test = y_train_folds[idx]
classifier.train(X_train, y_train)
dists = classifier.compute_distances_no_loops(X_test)
y_test_pred = classifier.predict_labels(dists, i)
num_correct = np.sum(y_test_pred == y_test)
num_test = len(y_test)
accuracy = float(num_correct) / num_test
k_to_accuracies[i].append(accuracy)
################################################################################
# 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))
可视化
可视化每一个K的误差。
# 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()
选择最佳的k，训练分类器
通过可视化每一个K的误差，选择最佳的K重新训练分类器。
# 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.
# 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 = 1
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))
k_nearest_neighbor.py
初始化
import numpy as np
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)
两重循环计算距离
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 range(num_test):
for j in range(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])))
#####################################################################
# END OF YOUR CODE #
#####################################################################
return dists
一重循环计算距离
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 range(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(X[i] - self.X_train), axis=1))
#######################################################################
# END OF YOUR CODE #
#######################################################################
return dists
不用循环计算距离
我们有两个矩阵Ma, Mb，并计算这两个矩阵的L2距离。
考虑 
只要我们通过改变Ma、Mb的形状，便可以完成上述计算。
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. #
#########################################################################
# 现在我们要算(a-b)^2 = a^2 - 2*a*b - b^2
# a: X b : self.X_train
a2 = np.diag(np.dot(X, X.T))
b2 = np.diag(np.dot(self.X_train, self.X_train.T))
ab = np.dot(X, self.X_train.T)
row = ab.shape[0]
col = ab.shape[1]
a2 = np.reshape(np.repeat(a2, col), ab.shape)
b2 = np.reshape(np.repeat(b2, row), ab.T.shape)
dists = np.sqrt(a2-2*ab+b2.T)
#########################################################################
# END OF YOUR CODE #
#########################################################################
return dists
预测标签
样本的标签为相邻的K个样本中出现次数最多的那个标签。
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 range(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. #
#########################################################################
tmpdists = dists[i,: ]
idxs = np.argsort(tmpdists)
closest_y = self.y_train[idxs[: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. #
#########################################################################
count = np.bincount(closest_y)
y_pred[i] = np.argmax(count)
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred