NumPy之计算两个矩阵的成对平方欧氏距离

问题描述

设 \({X_{m \times k}} = \left[ {\vec x_1^T;\vec x_2^T; \cdots ;\vec x_m^T} \right]\) (; 表示纵向连接) 和 \({Y_{n \times k}} = \left[ {\vec y_1^T;\vec y_2^T; \cdots ;\vec y_n^T} \right]\), 计算矩阵 \({X_{m \times k}}\) 中每一个行向量和矩阵 \({Y_{n \times k}}\) 中每一个行向量的平方欧氏距离 (pairwise squared Euclidean distance), 即计算:

\(\left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec x}_1} - {{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec x}_1} - {{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_1} - {{\vec y}_n}} \right\|_2^2} \\
{\left\| {{{\vec x}_2} - {{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec x}_2} - {{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_2} - {{\vec y}_n}} \right\|_2^2} \\
\vdots & \vdots & \ddots & \vdots \\
{\left\| {{{\vec x}_m} - {{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec x}_m} - {{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_m} - {{\vec y}_n}} \right\|_2^2}
\end{array}} \right]\) (这是一个 \(m \times n\) 矩阵).

这个计算在度量学习, 图像检索, 行人重识别等算法的性能评估中有着广泛的应用.

公式推导

在 NumPy 中直接利用上述原式来计算两个矩阵的成对平方欧氏距离, 要显式地使用二重循环, 而在 Python 中循环的效率是相当低下的. 如果想提高计算效率, 最好是利用 NumPy 的特性将原式转化为数组/矩阵运算. 下面就尝试进行这种转化.

先将原式展开为:

\(\left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec x}_1}} \right\|_2^2}&{\left\| {{{\vec x}_1}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_1}} \right\|_2^2} \\
{\left\| {{{\vec x}_2}} \right\|_2^2}&{\left\| {{{\vec x}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_2}} \right\|_2^2} \\
\vdots & \vdots & \ddots & \vdots \\
{\left\| {{{\vec x}_m}} \right\|_2^2}&{\left\| {{{\vec x}_m}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_m}} \right\|_2^2}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec y}_n}} \right\|_2^2} \\
{\left\| {{{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec y}_n}} \right\|_2^2} \\
\vdots & \vdots & \ddots & \vdots \\
{\left\| {{{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec y}_n}} \right\|_2^2}
\end{array}} \right] - 2\left[ {\begin{array}{*{20}{c}}
{\left\langle {{{\vec x}_1},{{\vec y}_1}} \right\rangle }&{\left\langle {{{\vec x}_1},{{\vec y}_2}} \right\rangle }& \cdots &{\left\langle {{{\vec x}_1},{{\vec y}_n}} \right\rangle } \\
{\left\langle {{{\vec x}_2},{{\vec y}_1}} \right\rangle }&{\left\langle {{{\vec x}_2},{{\vec y}_2}} \right\rangle }& \cdots &{\left\langle {{{\vec x}_2},{{\vec y}_n}} \right\rangle } \\
\vdots & \vdots & \ddots & \vdots \\
{\left\langle {{{\vec x}_m},{{\vec y}_1}} \right\rangle }&{\left\langle {{{\vec x}_m},{{\vec y}_2}} \right\rangle }& \cdots &{\left\langle {{{\vec x}_m},{{\vec y}_n}} \right\rangle }
\end{array}} \right]\)

下面逐项地化简或转化为数组/矩阵运算的形式:

\(\left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec x}_1}} \right\|_2^2}&{\left\| {{{\vec x}_1}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_1}} \right\|_2^2} \\
{\left\| {{{\vec x}_2}} \right\|_2^2}&{\left\| {{{\vec x}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_2}} \right\|_2^2} \\
\vdots & \vdots & \ddots & \vdots \\
{\left\| {{{\vec x}_m}} \right\|_2^2}&{\left\| {{{\vec x}_m}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_m}} \right\|_2^2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec x}_1}} \right\|_2^2} \\
{\left\| {{{\vec x}_2}} \right\|_2^2} \\
\vdots \\
{\left\| {{{\vec x}_m}} \right\|_2^2}
\end{array}} \right]\vec 1_n^T = \left( {\left( {X \circ X} \right){{\vec 1}_k}} \right)\vec 1_n^T = \left( {X \circ X} \right){\vec 1_k}\vec 1_n^T\)

式中, \(\circ\) 表示按元素积 (element-wise product), 又称为 Hadamard 积; \({\vec 1_k}\) 表示维的全1向量 (all-ones vector), 余者类推. 上式中 \({\vec 1_k}\) 的作用是计算 \(X \circ X\) 每行元素的和, 返回一个列向量; \(\vec 1_n^T\) 的作用类似于 NumPy 中的广播机制, 在这里是将一个列向量扩展为一个矩阵, 矩阵的每一列都是相同的.

\(\left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec y}_n}} \right\|_2^2} \\
{\left\| {{{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec y}_n}} \right\|_2^2} \\
\vdots & \vdots & \ddots & \vdots \\
{\left\| {{{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec y}_n}} \right\|_2^2}
\end{array}} \right] = {\vec 1_m}{\left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec y}_1}} \right\|_2^2} \\
{\left\| {{{\vec y}_2}} \right\|_2^2} \\
\vdots \\
{\left\| {{{\vec y}_n}} \right\|_2^2}
\end{array}} \right]^T} = {\vec 1_m}{\left( {\left( {Y \circ Y} \right){{\vec 1}_k}} \right)^T} = {\vec 1_m}\vec 1_k^T{\left( {Y \circ Y} \right)^T}\)

\(\left[ {\begin{array}{*{20}{c}}
{\left\langle {{{\vec x}_1},{{\vec y}_1}} \right\rangle }&{\left\langle {{{\vec x}_1},{{\vec y}_2}} \right\rangle }& \cdots &{\left\langle {{{\vec x}_1},{{\vec y}_n}} \right\rangle } \\
{\left\langle {{{\vec x}_2},{{\vec y}_1}} \right\rangle }&{\left\langle {{{\vec x}_2},{{\vec y}_2}} \right\rangle }& \cdots &{\left\langle {{{\vec x}_2},{{\vec y}_n}} \right\rangle } \\
\vdots & \vdots & \ddots & \vdots \\
{\left\langle {{{\vec x}_m},{{\vec y}_1}} \right\rangle }&{\left\langle {{{\vec x}_m},{{\vec y}_2}} \right\rangle }& \cdots &{\left\langle {{{\vec x}_m},{{\vec y}_n}} \right\rangle }
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\vec x_1^T} \\
{\vec x_2^T} \\
\vdots \\
{\vec x_m^T}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{{\vec y}_1}}&{{{\vec y}_2}}& \cdots &{{{\vec y}_n}}
\end{array}} \right] = X{Y^T}\)

所以:

\(\left[ {\begin{array}{*{20}{c}}
{\left\| {{{\vec x}_1} - {{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec x}_1} - {{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_1} - {{\vec y}_n}} \right\|_2^2} \\
{\left\| {{{\vec x}_2} - {{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec x}_2} - {{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_2} - {{\vec y}_n}} \right\|_2^2} \\
\vdots & \vdots & \ddots & \vdots \\
{\left\| {{{\vec x}_m} - {{\vec y}_1}} \right\|_2^2}&{\left\| {{{\vec x}_m} - {{\vec y}_2}} \right\|_2^2}& \cdots &{\left\| {{{\vec x}_m} - {{\vec y}_n}} \right\|_2^2}
\end{array}} \right] = \left( {X \circ X} \right){\vec 1_k}\vec 1_n^T + {\vec 1_m}\vec 1_k^T{\left( {Y \circ Y} \right)^T} - 2X{Y^T}\)

上述转化式中出现了 \(X{Y^T}\) (矩阵乘) , 矩阵乘在 NumPy 等在很多库中都有高效的实现, 对代码的优化是有好处的.

代码实现

sklearn 中已经包含了用 NumPy 实现的计算 "两个矩阵的成对平方欧氏距离" 的函数 (sklearn.metrics.euclidean_distances), 它利用的就是上面的转化公式. 这里, 我们利用上面的转化公式并借鉴 sklearn, 用 NumPy 重新实现一个轻量级且易于理解的版本:

import numpy as np

def euclidean_distances(x, y, squared=True):
    """Compute pairwise (squared) Euclidean distances.
    """
    assert isinstance(x, np.ndarray) and x.ndim == 2
    assert isinstance(y, np.ndarray) and y.ndim == 2
    assert x.shape[1] == y.shape[1]

    x_square = np.sum(x*x, axis=1, keepdims=True)
    if x is y:
        y_square = x_square.T
    else:
        y_square = np.sum(y*y, axis=1, keepdims=True).T
    distances = np.dot(x, y.T)
    # use inplace operation to accelerate
    distances *= -2
    distances += x_square
    distances += y_square
    # result maybe less than 0 due to floating point rounding errors.
    np.maximum(distances, 0, distances)
    if x is y:
        # Ensure that distances between vectors and themselves are set to 0.0.
        # This may not be the case due to floating point rounding errors.
        distances.flat[::distances.shape[0] + 1] = 0.0
    if not squared:
        np.sqrt(distances, distances)
    return distances

如果想进一步加速, 可以将

x_square = np.sum(x*x, axis=1, keepdims=True)

替换为

x_square = np.expand_dims(np.einsum('ij,ij->i', x, x), axis=1)

将

y_square = np.sum(y*y, axis=1, keepdims=True).T

替换为

y_square = np.expand_dims(np.einsum('ij,ij->i', y, y), axis=0)

使用 np.einsum 的好处是不会产生一个和 x 或 y 同样形状的临时数组 (x*x 或 y*y 会产生一个和 x 或 y 同样形状的临时数组).

PyTorch 中也包含了计算 "两个矩阵的成对平方欧氏距离" 的函数, 不过它利用了如下的转化公式, 感兴趣的朋友可以自己用 NumPy 实现一下.

\(\begin{aligned}
\left( {X \circ X} \right){{\vec 1}_k}\vec 1_n^T + {{\vec 1}_m}\vec 1_k^T{\left( {Y \circ Y} \right)^T} - 2X{Y^T} &= \left[ {\begin{array}{*{20}{c}}
{ - 2X}&{\left( {X \circ X} \right){{\vec 1}_k}}&{{{\vec 1}_m}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{Y^T}} \\
{\vec 1_n^T} \\
{{{\left( {Y \circ Y} \right)}^T}}
\end{array}} \right] \\
&= \left[ {\begin{array}{*{20}{c}}
{ - 2X}&{\left( {X \circ X} \right){{\vec 1}_k}}&{{{\vec 1}_m}}
\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}
Y&{{{\vec 1}_n}}&{Y \circ Y}
\end{array}} \right]^T} \\
\end{aligned}\)

参考

• sklearn.metrics.euclidean_distances
• _euclidean_dist in PyTorch

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