使用矩阵乘法用 numpy 计算 L2 距离

Question

我正在尝试自己完成Stanford CS231n 2017 CNN course 的作业。

我正在尝试仅使用矩阵乘法和 Numpy 广播求和来计算 L2 距离。 L2距离为：

如果我使用这个公式，我想我可以做到：

以下代码显示了计算 L2 距离的三种方法。如果我将compute_distances_two_loops 方法的输出与compute_distances_one_loop 方法的输出进行比较，两者都是相等的。但我将compute_distances_two_loops 方法的输出与compute_distances_no_loops 方法的输出进行比较，其中我只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

您可以找到完整的工作可测试代码here。

你知道我在compute_distances_no_loops 做错了什么吗？

更新：

抛出错误信息的代码是：

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')

还有错误信息：

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

Answer 1

以下是在不创建任何 3 维矩阵的情况下计算 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)

Answer 2

这是一个迟到的回应，但我已经以不同的方式解决了它并想发布它。当我解决这个问题时，我不知道从矩阵中减去 numpy 的列行向量。事实证明，我们可以从 nxm 中减去 nx1 或 1xm 向量，当我们这样做时，从每个行列向量中减去。如果有人使用不支持这种行为的库，他/她可以使用我的。对于这种情况，我算了一下，结果如下：

sum_x_train=np.sum(self.X_train**2,axis=1, keepdims=True)
sum_x_test=np.sum(X**2,axis=1, keepdims=True)
sum_2x_tr_te=np.dot(self.X_train,X.T)*2
sum_x_train=np.dot(sum_x_train,np.ones((1,X.shape[0])))
sum_x_test=np.dot(sum_x_test,np.ones((1,self.X_train.shape[0])))
dists=np.sqrt(sum_x_test.T+sum_x_train-sum_2x_tr_te).T

这种方法的缺点是它使用更多的内存。

Answer 3

我认为您正在寻找成对距离。

有一个惊人的技巧可以在一行中做到这一点。你必须巧妙地玩转播：

X_test = np.expand_dims(X, 0) # shape: [1, num_tests, D]
X_train = np.expand_dims(self.X_train, 1) # shape: [num_train, 1, D]
dists = np.square(X_train - X_test) # Thanks to broadcast [num_train, num_tests, D]
dists = np.sqrt(np.sum(dists, axis=-1)) # [num_train, num_tests]

Answer 4

这是我对 OP 要求的函数 compute_distances_no_loops() 的解决方案。出于性能原因，我不使用sqrt() 函数：

def compute_distances_no_loops(self, X):
    num_test = X.shape[0]
    num_train = self.X_train.shape[0]
    dists = np.zeros((num_test, num_train))
    #---------------
    # Get square of X and X_train
    X_sq = np.sum(X**2, axis=1, keepdims=True) 
    Xtrain_sq = np.sum(self.X_train**2, axis=1, keepdims=True)
    # Calculate (squared) dists as (X_train - X)**2 = X_train**2 - 2*X_train*X + X**2
    dists = -2*X.dot(self.X_train.T) + X_sq + Xtrain_sq.T
    #---------------  
    return dists

Answer 5

我认为问题来自不一致的数组形状。

#a^2 matrix (500, 1) 
alpha = np.sum(np.square(X), axis=1)
alpha = alpha.reshape(len(alpha), 1)
print(alpha.shape)
#b^2 matrix (1, 5000) 
beta  = np.sum(np.square(self.X_train.T), axis=0)
beta = beta.reshape(1, len(beta))
print(beta.shape)
#ab matrix (500, 5000)
alphabeta = np.dot(X, self.X_train.T)
print(alphabeta.shape)

dists = np.sqrt(-2 * alphabeta + alpha + beta)