使用矩阵乘法用numpy计算L2距离

范思法兰

我正在尝试自己完成Stanford CS231n 2017 CNN课程的作业

我正在尝试仅使用矩阵乘法和Numpy的求和广播来计算L2距离。L2距离为:

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我认为如果使用以下公式,我可以做到:

在此处输入图片说明

以下代码显示了三种计算L2距离的方法。如果我将方法compute_distances_two_loops的输出与method的输出进行比较compute_distances_one_loop,则两者相等。但是我将方法的输出与方法compute_distances_two_loops的输出进行了比较compute_distances_no_loops,在这里我仅使用矩阵乘法和求和广播来实现L2距离,它们是不同的。

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 xrange(num_test):
        for j in xrange(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))
            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.

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 xrange(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                            #
        #######################################################################
    print(dists.shape)
    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.sqrt(-2 * np.dot(X, self.X_train.T) +
                    np.sum(np.square(self.X_train), axis=1) +
                    np.sum(np.square(X), axis=1)[:, np.newaxis])
    print(dists.shape)
    #########################################################################
    #                         END OF YOUR CODE                              #
    #########################################################################
    return dists

您可以在此处找到完整的可测试代码

您知道我在compute_distances_no_loops哪里或哪里都做错了吗?

更新:

引发错误消息的代码是:

dists_two = classifier.compute_distances_no_loops(X_test)

# check that the distance matrix agrees with the one we computed before:
difference = np.linalg.norm(dists - dists_two, ord='fro')
print('Difference was: %f' % (difference, ))
if difference < 0.001:
    print('Good! The distance matrices are the same')
else:
    print('Uh-oh! The distance matrices are different')

和错误消息:

Difference was: 372100.327569
Uh-oh! The distance matrices are different
维克托里亚·玛利亚索娃(Viktoriya Malyasova)

这是在不创建任何3维矩阵的情况下如何计算X和Y行之间的成对距离的方法:

def dist(X, Y):
    sx = np.sum(X**2, axis=1, keepdims=True)
    sy = np.sum(Y**2, axis=1, keepdims=True)
    return np.sqrt(-2 * X.dot(Y.T) + sx + sy.T)

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