knn
作者：互联网
写在前面
以下的代码只是学习中的随笔，如有侵权请联系删除
K_Nearest_Neighbor
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):
# 求 欧拉距离的双重循环函数
"""
:param X: 为测试集，以当前的代码（适配knn实现时）而言 500 * 3072
:return:
"""
"""
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.dot(X[i], self.X_train[j])) it is Euclid dis, not dot
dists[i][j] = np.sqrt(np.sum(np.square(X[i] - self.X_train[j])))
"""
训练集中第j行减去 测试集中第i行 ，得到一个1*3072的小矩阵， 然后对每个元素进行平方再求和 然后开方
得到 i 与 j的距离
"""
# *****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(X[i] - self.X_train), axis=1))
"""
self.X_train - X[i, :] 训练集矩阵中每行都减去 测试集的第i行 （临时）
np.square(self.X_train - X[i, :]) 返回 每个元素都平方后的矩阵
np.sum(np.square(self.X_train - X[i, :]), axis=1) 对每行进行求和, 最后得到 一个 1 * 5000 的矩阵
！ axis = 0 沿着 纵轴对矩阵进行操作， axis = 1 沿着 横轴对 矩阵进行操作， 如果不指定 则会对所有元素进行操作
np.sqrt(np.sum(np.square(self.X_train - X[i, :]), axis=1)) ： 数组中每个元素都进行开方
双循环和单循环所完成的任务是一样的
"""
# *****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_square = np.sum(X * X, axis=1, keepdims=True)
Xtr_square = np.sum(self.X_train * self.X_train, axis=1, keepdims=True).T
dists = np.sqrt(-2 * np.dot(X, self.X_train.T) + X_square + Xtr_square)
# *****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)*****
'''
index = np.argsort(dists[i])
for j in range(k):
label = self.y_train[index[j]]
closest_y.append(label)
'''
closest_y = self.y_train[np.argsort(dists[i])[:k]]
# *****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))
# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
return y_pred
knn.ipynb
标签：knn,num,self,np,train,dists,test 来源： https://www.cnblogs.com/wojiuyishui/p/16110468.html