Einsum计算C距离非常慢

Question

我正在通过np.einsum计算马哈拉诺比斯距离：

np.einsum('nj,jk,nk->n', delta, VI, delta)

其中，VI是协方差矩阵的逆矩阵，是783 x 783，而增量是6000 x 783。这行代码需要在我的2016 Macbook Pro上执行10秒。我怎样才能使速度更快？

我必须计算这条线200k至300​​k次。向量化可能不是一种选择，因为每个类的VI都不相同。

Answer 1

不需要Einsum，可以使用点和元素乘积，也可以使用总和：

VI = np.random.rand(783, 783)
delta = np.random.rand(6000, 783)

%timeit np.einsum('nj,jk,nk->n', delta, VI, delta)
# 7.05 s ± 89.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit np.sum((delta @ VI) * delta, axis=-1)
# 90 ms ± 4.72 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

out_1 = np.einsum('nj,jk,nk->n', delta, VI, delta)
out_2 = np.sum((delta @ VI) * delta, axis=-1)
np.allclose(out_1, out_2)
# True

我是怎么到达的？

nj,jk->nk是一个点积：

tmp_1 = np.einsum('nj,jk->nk', delta, VI)
tmp_2 = delta @ VI
np.allclose(tmp_1, tmp_2)  # True

nk,nk->nk是元素产品：

tmp_3 = np.einsum('nk,nk->nk', tmp_1, delta)
tmp_4 = tmp_2 * delta
np.allclose(tmp_3, tmp_4)  # True

和nk->n是最后一个轴上的总和：

tmp_5 = np.einsum('nk->n', tmp_3)
tmp_6 = np.sum(tmp_4, axis=-1)
np.allclose(tmp_5, tmp_6)  # True

矢量化VI

您会注意到，沿第一个轴矢量化VI将正常工作：

# Vectorized `VI`
nd_VI = np.random.rand(3, 783, 783)
# Unvectorized `VI`, for comparison
VI = nd_VI[0, :]
delta = np.random.rand(6000, 783)

out = np.sum((delta @ VI) * delta, axis=-1)
out.shape
# (6000,)

nd_out = np.sum((delta @ nd_VI) * delta, axis=-1)
nd_out.shape
# (3, 6000)

# Result of vectorized and unvectorized `IV` are the same
np.allclose(out, nd_out[0, :])
# True

逐元素矢量化VI和delta

与向量化VI和delta相同，只需在VI和delta的开头添加相同数量的元素

# Vectorized `VI`
nd_VI = np.random.rand(3, 783, 783)
# Unvectorized `VI`, for comparison
VI = nd_VI[0, ...]
# Vectorized `delta`
nd_delta = np.random.rand(3, 6000, 783)
# Unvectorized `delta`, for comparison
delta = nd_delta[0, ...]

out = np.sum((delta @ VI) * delta, axis=-1)
out.shape
# (6000,)

nd_out = np.sum((nd_delta @ nd_VI) * nd_delta, axis=-1)
nd_out.shape
# (3, 6000)

# Result of vectorized and unvectorized `IV` are the same
np.allclose(out, nd_out[0, ...])
# True

分别向量化VI和delta

或者，如果要计算VI中每个元素与delta中每个可能元素的马氏距离，可以使用广播：

# Vectorized `VI`, note the extra empty dimension (where `delta` has 3)
nd_VI = np.random.rand(4, 1, 783, 783)
# Unvectorized `VI`, for comparison
VI = nd_VI[0, 0, ...]
# Vectorized `delta`, note the extra empty dimension (where `VI` has 4)
nd_delta = np.random.rand(1, 3, 6000, 783)
# Unvectorized `delta`, for comparison
delta = nd_delta[0, 0, ...]

out = np.sum((delta @ VI) * delta, axis=-1)
out.shape
# (6000,)

nd_out = np.sum((nd_delta @ nd_VI) * nd_delta, axis=-1)
nd_out.shape
# (4, 3, 6000)

# Result of vectorized and unvectorized `IV` are the same
np.allclose(out, nd_out[0, 0, ...])
# True