Einsum 计算 c 距离非常慢
Einsum very slow to compute c distance
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)
其中协方差矩阵的逆矩阵 VI 为 783 x 783，delta 为 6000 x 783。这条线在我的 2016 Macbook Pro 上需要 10 秒才能执行。我怎样才能让它更快？
我必须计算这条线 20 万到 30 万次。矢量化可能不是一个选项，因为每个 class.
的 VI 不同
不需要 Einsum，您可以使用点和元素乘积，然后求和：
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
我是怎么得出这个结论的？
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 是元素积：
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
向量化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
矢量化 VI 和 delta 元素
与矢量化 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
独立向量化 VI 和 delta
或者，如果你想计算VI中每个元素与delta中每个可能元素的马氏距离，你可以使用广播：
# 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
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)
其中协方差矩阵的逆矩阵 VI 为 783 x 783，delta 为 6000 x 783。这条线在我的 2016 Macbook Pro 上需要 10 秒才能执行。我怎样才能让它更快？
我必须计算这条线 20 万到 30 万次。矢量化可能不是一个选项，因为每个 class.
的 VI 不同不需要 Einsum，您可以使用点和元素乘积，然后求和：
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
我是怎么得出这个结论的？
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 是元素积：
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
向量化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
矢量化 VI 和 delta 元素
与矢量化 VI 和 delta 相同，只需在 VI 和 delta
# 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
独立向量化 VI 和 delta
或者，如果你想计算VI中每个元素与delta中每个可能元素的马氏距离，你可以使用广播：
# 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