使用矩阵乘法通过 numpy 计算 L2 距离

Question

我正在尝试自己完成斯坦福 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：

我想如果我使用这个公式我可以做到：

以下代码显示了计算 L2 距离的三种方法。如果我将 method 的输出compute_distances_two_loops与 method 的输出进行比较compute_distances_one_loop，两者相等。但我将该方法的输出compute_distances_two_loops与该方法的输出进行比较compute_distances_no_loops，其中我仅使用矩阵乘法和求和广播实现了 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

您可以在这里找到完整的工作可测试代码。

你知道我在compute_distances_no_loops哪里做错了什么吗？

更新：

抛出错误消息的代码是：

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')

以及错误消息：

Difference was: 372100.327569
Uh-oh! The distance matrices are different

Answer 1

以下是如何在不创建任何 3 维矩阵的情况下计算 X 行和 Y 行之间的成对距离：

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)