使用矩阵乘法用numpy计算L2距离

Question

I'm trying to do it by myself the assignments from Stanford CS231n 2017 CNN course .

I'm trying to compute L2 distance using only matrix multiplication and sum broadcasting with Numpy. L2 distance is:

And I think I can do it if I use this formula:

The following code shows three methods to compute L2 distance. If I compare the output from the method compute_distances_two_loops with the output from method compute_distances_one_loop , both are equals. But I compare the output from the method compute_distances_two_loops with the output from the method compute_distances_no_loops , where I have implemented the L2 distance using only matrix multiplication and sum broadcasting, they are different.

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

You can find a full working testable code here .

Do you know what am I doing wrong in compute_distances_no_loops , or wherever?

UPDATE:

The code that throws the error message is:

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')

And the error message:

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

Answer 1

Here is how you can compute pairwise distances between rows of X and Y without creating any 3-dimensional matrices:

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

It's a late response, but I've solved it in a different way and wanted to post it. When I was solving this, I was unaware of the numpy's column-row vector subtractions from the matrix. As it turns out, we can subtract nx1 or 1xm vector from nxm and when we do it, subtracts from every row-column vector. If one works with a library that doesn't support this kind of behavior, he/she can use mine. For this situation, I've worked it out the math, and the result is the following one:

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

The downside of this approach is it uses more memory.

Answer 3

I think that you are looking for the pairwise distance.

There is an amazing trick to do that in a single line. You have to cleverly play with the boradcasting:

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

I believe the issue comes from inconsistent array shapes.

 #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)

Answer 5

This is my solution for function compute_distances_no_loops() that the OP asked for. I don't use the sqrt() function for performance reason:

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