cs231n作业：Assignment1-KNN
note:
曼哈顿距离依赖于坐标系统的选择（向量中的元素可能都有实际的意义）
d 1 ( I 1 , I 2 ) = ∑ p ∣ I 1 p − I 2 p ∣ d_{1}(I_{1}, I_{2}) = \sum_{p}|I_{1}^{p}-I_{2}^{p}| d1(I1,I2)=∑p∣I1p−I2p∣
欧式距离对距离的排序不会受到坐标系统的影响
d 2 ( I 1 , I 2 ) = ∑ p ( I 1 p − I 2 p ) 2 d_{2}(I_{1}, I_{2}) =\sqrt{ \sum_{p}(I_{1}^{p}-I_{2}^{p})^{2}} d2(I1,I2)=∑p(I1p−I2p)2
超参数：我们在一个算法中的设置而不是通过算法学习得到（具体问题具体设置）
容易导致过拟合的做法：
（1）选择在测试集准确率最高的，可能导致过拟合
（2）将数据分成测试集和训练集，选择在测试集上准确率最高的分类器，这组参数可能只适合于这个测试集数据
（3）Better.将数据分成训练集，验证集，测试集。通过验证集选出超参数，用测试集评估。
交叉验证可能效果更好，但对于深度学习来说，成本太高所以不常用
KNN的弊端：
测试时间长实时性不好，
向量化的距离函数不太适合表示视觉之间的相似度，可能不适用于图像很难分清差别
纬度灾难，一旦维度上升，为了有好的训练效果 让数据分布密集（避免最近邻的点距离很远），导致训练数据可能指数倍的增长。
作业代码
from builtins import range
from builtins import object
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 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, nor use np.linalg.norm(). #
#####################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
dists[i][j] = np.sqrt(np.sum(np.square(X[i]-self.X_train[j])))
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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, :]. #
# Do not use np.linalg.norm(). #
#######################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
dists[i] = np.sqrt(np.sum(np.square(self.X_train - X[i]),axis=1))
#广播机制，np会自动补全
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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, #
# nor use np.linalg.norm(). #
# #
# HINT: Try to formulate the l2 distance using matrix multiplication #
# and two broadcast sums. #
#########################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
x_1 = np.sum(np.multiply(X,X),axis=1,keepdims=True).reshape(num_test,1)
y_1 = np.sum(np.multiply(self.X_train,self.X_train),axis=1,keepdims=True).reshape(1,num_train)
xy_1 = -2 * np.dot(X,self.X_train.T)
dists = np.sqrt(x_1 + y_1 + xy_1)
# 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) # python广播机制
# dists = np.add(dists, sq2)
# dists = np.sqrt(dists)
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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. #
#########################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
closest_y = self.y_train[np.argsort(dists[i])]
closest_y = closest_y[:k]
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
#########################################################################
# 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. #
#########################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
y_pred[i] = np.argmax(np.bincount(closest_y))
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
return y_pred
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. #
################################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
#print(X_train.shape)
x_train_folds = np.array_split(X_train,num_folds)
y_train_folds = np.array_split(y_train,num_folds)
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# 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 = {}
classifier = KNearestNeighbor()
for i in range(len(k_choices)):
l = []
for j in range(num_folds):
testx = np.array(x_train_folds[j])
testy = np.array(y_train_folds[j])
#交叉验证让中间一个部分当评估集
trainx = x_train_folds[:j]+x_train_folds[j+1:]
trainy = y_train_folds[:j]+y_train_folds[j+1:]
trainx = np.array([y for x in trainx for y in x])
trainy = np.array([y for x in trainy for y in x])
#展平后作为一个整的数据集用来训练
classifier.train(trainx, trainy)
dists = classifier.compute_distances_no_loops(testx)
y_test_pred = classifier.predict_labels(dists, k=k_choices[i])
y_test_num = np.sum(y_test_pred == testy)
l.append(y_test_num / len(testy))
id = k_choices[i]
k_to_accuracies[id] = l
################################################################################
# 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. #
################################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
pass
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
# 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))
numpy中keepdims的理解
np.split()与np.array_split()函数 | Numpy | Python
numpy 统计数组每一行出现次数最多的数字,np.argmax,np.bincount,np.argsort
CS231n课程学习笔记（一）——KNN的实现
cs231n作业：Assignment1-KNN
CS231n：作业1——KNN