python 实现 pairwise_l2_distances

问题:
$$X={\left[\begin{array}{c}—x_1—\\ ⋮\\ —x_M—\end{array}\right]}_{M\times C},Y={\left[\begin{array}{c}—y_1—\\ ⋮\\ —y_N—\end{array}\right]}_{N\times C}$$
其中 $x_i,y_i\in R^C$.
要求$X,Y$每一行之间的距离, 即$S={\left[S_{ij}\right]}_{M\times N}={\left[||x_i-y_j|{|}_2^2\right]}_{M\times N}$.

---

易知
$$S_{ij}=||x_i-y_j|{|}_2^2=(x_i-y_j{)}^{\top }(x_i-y_j)=x_i^{\top }{x}_i-2x_i^{\top }{y}_j+y_j^{\top }{y}_j$$

def pairwise_l2_distances(X, Y):
    """
    A fast, vectorized way to compute pairwise l2 distances between rows in `X`
    and `Y`.
    Notes
    -----
    An entry of the pairwise Euclidean distance matrix for two vectors is
    .. math::
        d[i, j]  &=  \sqrt{(x_i - y_i) @ (x_i - y_i)} \\\\
                 &=  \sqrt{sum (x_i - y_j)^2} \\\\
                 &=  \sqrt{sum (x_i)^2 - 2 x_i y_j + (y_j)^2}
    The code below computes the the third line using numpy broadcasting
    fanciness to avoid any for loops.
    Parameters
    ----------
    X : :py:class:`ndarray <numpy.ndarray>` of shape `(N, C)`
        Collection of `N` input vectors
    Y : :py:class:`ndarray <numpy.ndarray>` of shape `(M, C)`
        Collection of `M` input vectors. If None, assume `Y` = `X`. Default is
        None.
    Returns
    -------
    dists : :py:class:`ndarray <numpy.ndarray>` of shape `(N, M)`
        Pairwise distance matrix. Entry (i, j) contains the `L2` distance between
        :math:`x_i` and :math:`y_j`.
    """
    D = -2 * X @ Y.T + np.sum(Y ** 2, axis=1) + np.sum(X ** 2, axis=1)[:, np.newaxis]
    D[D < 0] = 0  # clip any value less than 0 (a result of numerical imprecision)
    return np.sqrt(D)