计算两个点阵列之间的成对角度矩阵

我有两个点向量，x and y， 成形(n, p) and (m, p)分别。举个例子：

x = np.array([[ 0.     , -0.16341,  0.98656],
              [-0.05937, -0.25205,  0.96589],
              [ 0.05937, -0.25205,  0.96589],
              [-0.11608, -0.33488,  0.93508],
              [ 0.     , -0.33416,  0.94252]])
y = np.array([[ 0.     , -0.36836,  0.92968],
              [-0.12103, -0.54558,  0.82928],
              [ 0.12103, -0.54558,  0.82928]])

我想计算一个(n, m)- 大小的矩阵，包含两点之间的角度，a lathis https://stackoverflow.com/questions/2827393/angles-between-two-n-dimensional-vectors-in-python问题。即，矢量化版本：

theta = np.array(
            [ np.arccos(np.dot(i, j) / (la.norm(i) * la.norm(j)))
                 for i in x for j in y ]
        ).reshape((n, m))

Note: n and m每个的数量级可以约为 10000。

有多种方法可以做到这一点：

import numpy.linalg as la
from scipy.spatial import distance as dist

# Manually
def method0(x, y):
    dotprod_mat = np.dot(x,  y.T)
    costheta = dotprod_mat / la.norm(x, axis=1)[:, np.newaxis]
    costheta /= la.norm(y, axis=1)
    return np.arccos(costheta)

# Using einsum
def method1(x, y):
    dotprod_mat = np.einsum('ij,kj->ik', x, y)
    costheta = dotprod_mat / la.norm(x, axis=1)[:, np.newaxis]
    costheta /= la.norm(y, axis=1)
    return np.arccos(costheta)

# Using scipy.spatial.cdist (one-liner)
def method2(x, y):
    costheta = 1 - dist.cdist(x, y, 'cosine')
    return np.arccos(costheta)

# Realize that your arrays `x` and `y` are already normalized, meaning you can
# optimize method1 even more
def method3(x, y):
    costheta = np.einsum('ij,kj->ik', x, y) # Directly gives costheta, since
                                            # ||x|| = ||y|| = 1
    return np.arccos(costheta)

(n, m) = (1212, 252) 的计时结果：

>>> %timeit theta = method0(x, y)
100 loops, best of 3: 11.1 ms per loop
>>> %timeit theta = method1(x, y)
100 loops, best of 3: 10.8 ms per loop
>>> %timeit theta = method2(x, y)
100 loops, best of 3: 12.3 ms per loop
>>> %timeit theta = method3(x, y)
100 loops, best of 3: 9.42 ms per loop

时间差异随着元素数量的增加而减小。对于 (n, m) = (6252, 1212)：

>>> %timeit -n10 theta = method0(x, y)
10 loops, best of 3: 365 ms per loop
>>> %timeit -n10 theta = method1(x, y)
10 loops, best of 3: 358 ms per loop
>>> %timeit -n10 theta = method2(x, y)
10 loops, best of 3: 384 ms per loop
>>> %timeit -n10 theta = method3(x, y)
10 loops, best of 3: 314 ms per loop

但是，如果您省略np.arccos步骤，即假设您可以通过costheta，并且没有need theta本身，那么：

>>> %timeit costheta = np.einsum('ij,kj->ik', x, y)
10 loops, best of 3: 61.3 ms per loop
>>> %timeit costheta = 1 - dist.cdist(x, y, 'cosine')
10 loops, best of 3: 124 ms per loop
>>> %timeit costheta = dist.cdist(x, y, 'cosine')
10 loops, best of 3: 112 ms per loop

这是针对(6252, 1212)的情况。所以实际上np.arccos占用了80%的时间。在这种情况下我发现np.einsum is much比...快dist.cdist。所以你肯定想使用einsum.

Summary:结果theta大部分相似，但是np.einsum对我来说最快，特别是当你没有额外计算规范时。尽量避免计算theta并与刚刚合作costheta.

Note:我没有提到的重要一点是浮点精度的有限性可能会导致np.arccos给予nan价值观。method[0:3]为以下价值观而努力x and y当然，这还没有得到适当的标准化。但method3给了一些nans。我通过预归一化解决了这个问题，这自然会破坏使用中的任何收益method3，除非您需要对一小组预归一化矩阵进行多次计算（无论出于何种原因）。