CS231n assignment1 KNN
时间：2019/3/25
内容：assignment1-KNN
参考网上大佬的代码。还是计算机视觉小白，有错误欢迎指正~
https://blog.csdn.net/donghai_yu/article/details/79344539
https://blog.csdn.net/qq_26414307/article/details/79318555
KNN部分
完整代码
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)
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((X[i,:]-self.X_train[j,:])**2))
#####################################################################
# 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 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(self.X_train-X[i,:]),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]
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.multiply(np.dot(X,self.X_train.T),-2)
sq1 = np.sum(np.square(X),axis=1,keepdims = True)
sq2 = np.sum(np.square(self.X_train),axis=1)
dists = np.add(dists,sq1)
dists = np.add(dists,sq2)
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]
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. #
#########################################################################
y_indicies=np.argsort(dists[i,:],axis=0)
#########################################################################
# 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. #
#########################################################################
closest_y=self.y_train[y_indicies[: k]]
y_pred[i]=np.argmax(np.bincount(closest_y))
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred
代码理解
习题中使用了三种方法来计算L2，首先是用两个循环实现：
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] # 读取X的第一维长度，X的第一维即num_test（见上面注释）
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. #
#####################################################################
# [i,:]数组切片，是取第1维中下标为i的元素的所有数据，也就是第i行全部元素（从0开始），此处表示第i个样本点
# KNN算法需要计算测试集的点与训练集所有点的距离，也就是训练集与测试集两两点求距离
# L2是欧氏距离（所有点差的平方之和，再开根号）
dists[i,j]=np.sqrt(np.sum((X[i,:]-self.X_train[j,:])**2))
#####################################################################
# END OF YOUR CODE #
#####################################################################
return dists
然后用一个循环实现：
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
"""
# 以下三行与两个循环的一样
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, :]. #
#######################################################################
# axis=1是按行相加
# numpy的broadcast
dists[i,:]=np.sqrt(np.sum(np.square(self.X_train-X[i,:]),axis=1))
#######################################################################
# END OF YOUR CODE #
#######################################################################
return dists
不用循环实现：
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. #
#########################################################################
# 思想：(a-b)^2 = a^2 + b^2 -2ab
dists = np.multiply(np.dot(X,self.X_train.T),-2)
#-2*X*X_train.T ==> [num_test,num_train]
sq1 = np.sum(np.square(X),axis=1,keepdims = True)
#np.square为对元素进行平方
#keepdims=True:如果不加，结果为列向量；如果加，则结果为[num_test,1](矩阵)
sq2 = np.sum(np.square(self.X_train),axis=1)
#结果为[num_train,1]
dists = np.add(dists,sq1)
#矩阵加也有broadcast!!
dists = np.add(dists,sq2)
#矩阵加一维向量，将向量转为行向量，broadcast!!
dists = np.sqrt(dists)
#########################################################################
# END OF YOUR CODE #
#########################################################################
return dists
补充关于nump.square的用法：
对于矩阵做平方，结果矩阵和输入的shape相同，结果中的每一个元素是原来位置元素的平方
numpy库中的reshape(-1,1)中-1被理解为unspecified value，意思是未指定为给定的。因此reshape(-1,1)变成1列，同理reshape(1,-1)会变成1行
可以参考： https://blog.csdn.net/wld914674505/article/details/80460042
预测部分:
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] # 表示dist的行数
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. #
#########################################################################
# argsort返回数组中值（此处是距离）从小到大的索引值
# [i,:]表示第i行的所有元素
# 主要y_indicies中都是一些索引值
# 由于dists形状是(num_test,num_train)，因此每行最前面的是训练集中距离该测试点最近的
y_indicies=np.argsort(dists[i,:],axis=0)
#########################################################################
# 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. #
#########################################################################
# [: k]表示从开始到k
# argmax返回最大值（此处为次数最大值）对应的索引值；具体分类也是用具体数字表示的所以不会有问题
closest_y=self.y_train[y_indicies[: k]] # 距离最近的训练集前k个点的label
y_pred[i]=np.argmax(np.bincount(closest_y))
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred
补充：np.bincount的用法
交叉验证部分
完整代码
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
y_train_ = y_train.reshape(-1, 1)
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. #
################################################################################
# Your code
for k in k_choices:
accuracies = []
for i in range(num_folds):
X_test_cv = X_train_folds[i]
X_train_cv = np.vstack(X_train_folds[:i] + X_train_folds[i+1:])
y_test_cv = y_train_folds[i]
y_train_cv = np.hstack(y_train_folds[:i]+y_train_folds[i+1:])
classifier.train(X_train_cv, y_train_cv)
dists_cv = classifier.compute_distances_no_loops(X_test_cv)
y_test_pred = classifier.predict_labels(dists_cv, k)
num_correct = np.sum(y_test_pred == y_test_cv)
accuracies.append(float(num_correct) * num_folds / num_training)
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))
代码理解
y_train_ = y_train.reshape(-1, 1)
X_train_folds = np.array_split(X_train, num_folds)
y_train_folds = np.array_split(y_train, num_folds)
for k in k_choices:
accuracies = []
for i in range(num_folds):
X_test_cv = X_train_folds[i]
# 具体这样拼接的原因要根据data的形状分析
X_train_cv = np.vstack(X_train_folds[:i] + X_train_folds[i+1:])
y_test_cv = y_train_folds[i]
y_train_cv = np.hstack(y_train_folds[:i]+y_train_folds[i+1:])
classifier.train(X_train_cv, y_train_cv)
dists_cv = classifier.compute_distances_no_loops(X_test_cv)
y_test_pred = classifier.predict_labels(dists_cv, k)
num_correct = np.sum(y_test_pred == y_test_cv)
accuracies.append(float(num_correct) * num_folds / num_training)
k_to_accuracies[k] = accuracies