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186 linhas
8.1 KiB
186 linhas
8.1 KiB
from builtins import range
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from builtins import object
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import numpy as np
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from past.builtins import xrange
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class KNearestNeighbor(object):
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""" a kNN classifier with L2 distance """
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def __init__(self):
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pass
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def train(self, X, y):
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"""
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Train the classifier. For k-nearest neighbors this is just
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memorizing the training data.
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Inputs:
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- X: A numpy array of shape (num_train, D) containing the training data
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consisting of num_train samples each of dimension D.
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- y: A numpy array of shape (N,) containing the training labels, where
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y[i] is the label for X[i].
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"""
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self.X_train = X
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self.y_train = y
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def predict(self, X, k=1, num_loops=0):
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"""
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Predict labels for test data using this classifier.
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Inputs:
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- X: A numpy array of shape (num_test, D) containing test data consisting
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of num_test samples each of dimension D.
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- k: The number of nearest neighbors that vote for the predicted labels.
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- num_loops: Determines which implementation to use to compute distances
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between training points and testing points.
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Returns:
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- y: A numpy array of shape (num_test,) containing predicted labels for the
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test data, where y[i] is the predicted label for the test point X[i].
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"""
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if num_loops == 0:
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dists = self.compute_distances_no_loops(X)
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elif num_loops == 1:
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dists = self.compute_distances_one_loop(X)
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elif num_loops == 2:
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dists = self.compute_distances_two_loops(X)
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else:
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raise ValueError('Invalid value %d for num_loops' % num_loops)
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return self.predict_labels(dists, k=k)
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def compute_distances_two_loops(self, X):
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"""
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Compute the distance between each test point in X and each training point
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in self.X_train using a nested loop over both the training data and the
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test data.
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Inputs:
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- X: A numpy array of shape (num_test, D) containing test data.
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Returns:
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- dists: A numpy array of shape (num_test, num_train) where dists[i, j]
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is the Euclidean distance between the ith test point and the jth training
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point.
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"""
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num_test = X.shape[0]
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num_train = self.X_train.shape[0]
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dists = np.zeros((num_test, num_train))
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for i in range(num_test):
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for j in range(num_train):
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#####################################################################
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# TODO:
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#计算第i个测试点与第j个训练点之间的l2距离，并将结果存储在dists[i，j]中。
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#你不应使用循环和np.linalg.norm()函数。
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#####################################################################
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# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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pass
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# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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return dists
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def compute_distances_one_loop(self, X):
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"""
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Compute the distance between each test point in X and each training point
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in self.X_train using a single loop over the test data.
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Input / Output: Same as compute_distances_two_loops
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"""
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num_test = X.shape[0]
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num_train = self.X_train.shape[0]
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dists = np.zeros((num_test, num_train))
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for i in range(num_test):
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#######################################################################
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# TODO:
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#计算第i个测试点与所有训练点之间的l2距离，并将结果存储在dists[i，：]中。
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#不要使用np.linalg.norm()。
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#######################################################################
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# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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# 注意np.sum中要加上维度axis=1才能得出正确的结果
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# 关于axis的介绍
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# https://zhuanlan.zhihu.com/p/30960190
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# 以及np.sum的介绍
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# https://docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html
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# self.X_train (5000,3072) X[i] (1,3072) (self.X_train - X[i]) (5000,3072)
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pass
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# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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return dists
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def compute_distances_no_loops(self, X):
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"""
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Compute the distance between each test point in X and each training point
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in self.X_train using no explicit loops.
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Input / Output: Same as compute_distances_two_loops
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"""
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num_test = X.shape[0]
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num_train = self.X_train.shape[0]
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dists = np.zeros((num_test, num_train))
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#########################################################################
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# TODO:
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#在不使用任何显式循环的情况下，计算所有测试点和所有训练点之间的l2距离，
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#并将结果存储在dists中。
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#您应该仅使用基本的数组操作来实现此功能。
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#不可以使用scipy中的函数以及函数np.linalg.norm()。
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#
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#提示：尝试使用矩阵乘法和广播总和来计算l2距离。
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#########################################################################
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# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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# (x-y)^2 = x^2 + y^2 - 2xy
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# reshape是为了让两个矩阵有个维度为1，这样子便可进行广播
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pass
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# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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return dists
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def predict_labels(self, dists, k=1):
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"""
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Given a matrix of distances between test points and training points,
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predict a label for each test point.
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Inputs:
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- dists: A numpy array of shape (num_test, num_train) where dists[i, j]
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gives the distance betwen the ith test point and the jth training point.
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Returns:
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- y: A numpy array of shape (num_test,) containing predicted labels for the
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test data, where y[i] is the predicted label for the test point X[i].
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"""
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num_test = dists.shape[0]
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y_pred = np.zeros(num_test)
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for i in range(num_test):
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# A list of length k storing the labels of the k nearest neighbors to
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# the ith test point.
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closest_y = []
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#########################################################################
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# TODO:
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#使用距离矩阵查找第i个测试点的k个最近邻居，
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#并使用self.y_train查找这些邻居的标签。
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#将这些标签存储在closest_y中。
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#
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#提示：查阅函数numpy.argsort。
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#########################################################################
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# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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# numpy.argsort 返回排序好的数列的索引
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pass
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# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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#########################################################################
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# TODO:
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#
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#现在，你已经找到了k个最近邻的标签，接着需要在closest_y中找到最可能的标签。 #将此标签存储在y_pred [i]中。如果有两个标签可能性一样的话选择索引更小的那个。
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#########################################################################
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# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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y_pred[i] = np.bincount(closest_y).argmax()
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# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
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return y_pred
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