CS231n Assignment1：KNN
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2024-03-25 10:23:12
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cs231n/classifiers/k_nearest_neighbor.py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))
"""standard answer"""
# 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.
X shape (
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)
dists2 = np.sum(np.square(X),axis=1,keepdims=True)
dists3 = np.sum(np.square(self.X_train),axis=1)
dists = np.add(dists,dists2)
dists = np.add(dists,dists3)
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] dist向量是测试集和训练集一起形成的矩阵
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. #
#########################################################################
closest_y = self.y_train[np.argsort(dists[i,:])[: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. #
#########################################################################
y_pred[i] = np.argmax(np.bincount(closest_y))
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred
knn.ipynb代码，cross validation部分：
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.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. #
################################################################################
for k in k_choices:
k_to_accuracies[k] = []
for k in k_choices:
for i in range(num_folds):
#这里出现错误，前若干的数组是[:i]不是[;,i]，这个代表着前几列
X_train_cv = np.vstack(X_train_folds[:i] + X_train_folds[i+1:]) #(4000,3072)
X_test_cv = X_train_folds[i] #(1000,3072)
y_train_cv = np.hstack(y_train_folds[:i] + y_train_folds[i+1:]) #(4000,)
y_test_cv = y_train_folds[i] #(1000,)
#这里不需要第二个classifier = KNearestNeighbor()么
classifier.train(X_train_cv,y_train_cv)
dists_cv = classifier.compute_distances_no_loops(X_test_cv) #(1000,4000)
y_test_pred_cv = classifier.predict_labels(dists_cv,k)
num_correct_cv = np.sum(y_test_pred_cv == y_test_cv)
accuracy_cv = float(num_correct_cv)/y_test_cv.shape[0]
k_to_accuracies[k].append(accuracy_cv)
################################################################################
# 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))
做作业中遇到的问题汇总：
1、numpy.sum函数：
numpy.sum(a, axis=None, dtype=None, out=None, keepdims=<no value>, initial=<no value>)
作用就是求和函数，参考文档如下所示：https://docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html
具体用到的参数为axis 和 keepdims。
参数及结果如下：
a = np.arange(12).reshape(3,4)
print(a)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]]
print(np.sum(a))
#66
print(np.sum(a,axis=0))
#[12 15 18 21]
print(np.sum(a,axis=0,keepdims=True))
#[[12 15 18 21]]
print(np.sum(a,axis=1))
# [ 6 22 38]
print(np.sum(a,axis=1,keepdims=True))
# [[ 6]
# [22]
# [38]]
2、numpy.argsort，numpy.argmax ，numpy.bincount函数:
numpy.argsort ：将矩阵a按照axis排序，并返回排序后的下标
numpy.argmax：返回数组或矩阵最大值的索引
numpy.bincount：返回数组中每个元素出现的次数，尤其适用于计算数据集的标签列（y_train）的分布
3、numpy.vstack,numpy.hstack函数：
作用是堆叠数组，vstack按行堆叠数组，hstack按列堆叠数组
代码运行结果如下：
a = np.arange(12).reshape(3,4)
print(a)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]]
print(np.vstack(a))
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]]
print(np.hstack(a))
# [ 0 1 2 3 4 5 6 7 8 9 10 11]
4、计算KNN dists数组时no loop的算法及原理：
个人理解，不一定很严谨，但是可以推论出来。
这个以后会用公式证明，简单的说下思路。
我们现在有
我们可以先推出的公式，再推出的公式,或者直接用numpy.sum对矩阵的操作获得我们想要的结果。
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