как бы рассчитать L2 или евклидово расстояние между рядами разных 2D векторов

Question

Мне дали 2 разных 2D массива, и меня попросили вычислить расстояние L2 между строками массива x и строками в массиве y. Форма массива x - (M, D), а форма массива y - (N, D). Конечный массив ответов должен иметь форму (M, N).

Я не очень хорош в питоне. Я действительно просто делаю случайные вещи и смотрю, что происходит.

import numpy as np

def compute_distances(x, y):
    """ Write a function that computes the L2 distance between each row 
        in `x` and `y`.

        Parameters
        ----------
        x : numpy.ndarray
            x.shape must be (M, D)
            Each row of `x` is a flattened vector representing the pixel 
            values of a single image. Thus `x` represents
            M images, each one described by a length-D vector.

        y : numpy.ndarray
            y.shape must be (N, D)
            Each row of `y` is a flattened vector representing the pixel 
            values of a single image. Thus `y` represents
            N images, each one described by a length-D vector.

        Returns
        -------
        distances : numpy.ndarray
            distances.shape = (M, N)
            distances[i, j] = the L2 distance between x[i] and y[j]
    """
    # student code goes here

    M = x.shape[0]
    N = y.shape[0]
    dists = np.array (M, N)
    dists[i, j] = np.sqrt(np.sum(np.square(x.shape[0] - y.shape[0]), axis = 0))

    pass

Answer 1

Мне нравится этот, так как он работает для 1, 2 и 3D массивов

def e_dist(a, b, metric='euclidean'):
    """Distance calculation for 1D, 2D and 3D points using einsum

    preprocessing :
        use `_view_`, `_new_view_` or `_reshape_` with structured/recarrays

    Parameters
    ----------
    a, b : array like
        Inputs, list, tuple, array in 1, 2 or 3D form
    metric : string
        euclidean ('e', 'eu'...), sqeuclidean ('s', 'sq'...),

    Notes
    -----
    mini e_dist for 2d points array and a single point

    >>> def e_2d(a, p):
            diff = a - p[np.newaxis, :]  # a and p are ndarrays
            return np.sqrt(np.einsum('ij,ij->i', diff, diff))

    See Also
    --------
    cartesian_dist : function
        Produces pairs of x,y coordinates and the distance, without duplicates.
    """
    a = np.asarray(a)
    b = np.atleast_2d(b)
    a_dim = a.ndim
    b_dim = b.ndim
    if a_dim == 1:
        a = a.reshape(1, 1, a.shape[0])
    if a_dim >= 2:
        a = a.reshape(np.prod(a.shape[:-1]), 1, a.shape[-1])
    if b_dim > 2:
        b = b.reshape(np.prod(b.shape[:-1]), b.shape[-1])
    diff = a - b
    dist_arr = np.einsum('ijk,ijk->ij', diff, diff)
    if metric[:1] == 'e':
        dist_arr = np.sqrt(dist_arr)
    dist_arr = np.squeeze(dist_arr)
    return dist_arr

Выход

a = np.random.rand(3, 2) 
b = np.random.rand(5, 2)
e_dist(a, b)
array([[0.62, 0.45, 0.88, 0.7 , 0.33],
       [0.46, 0.57, 0.47, 0.25, 0.4 ],
       [0.94, 0.68, 0.16, 0.35, 0.62]])

и для ...

a = np.random.rand(2, 3, 2) 
b = np.random.rand(2, 5, 2)
e_dist(a, b)
array([[0.72, 0.39, 0.89, 0.25, 0.29, 0.41, 0.31, 0.6 , 0.8 , 0.39],
       [0.25, 0.26, 0.69, 0.45, 0.45, 0.63, 0.57, 0.39, 0.25, 0.78],
       [0.28, 0.27, 0.74, 0.47, 0.46, 0.69, 0.62, 0.45, 0.25, 0.84],
       [0.43, 0.48, 0.47, 0.52, 0.54, 0.2 , 0.17, 0.21, 0.55, 0.38],
       [0.15, 0.36, 0.54, 0.51, 0.52, 0.52, 0.48, 0.23, 0.23, 0.69],
       [0.87, 0.76, 0.78, 0.67, 0.71, 0.26, 0.28, 0.64, 0.99, 0.06]])

И для сравнения вариантов numpy и scipy для 2d

from scipy.spatial import distance_matrix
distance_matrix(a, b)
array([[0.87, 0.89, 0.91, 0.07, 0.68, 0.95, 0.89],
       [0.78, 0.52, 0.28, 0.85, 0.22, 0.65, 0.51],
       [0.75, 0.46, 0.51, 1.08, 0.5 , 0.57, 0.45]])

e_dist(a, b)
array([[0.87, 0.89, 0.91, 0.07, 0.68, 0.95, 0.89],
       [0.78, 0.52, 0.28, 0.85, 0.22, 0.65, 0.51],
       [0.75, 0.46, 0.51, 1.08, 0.5 , 0.57, 0.45]])

Таким образом, существует множество вариантов, в зависимости от размера массивов, с которыми вы работаете.