块元素逐点积

块元素逐点积

方法1

使用

np.einsum

-

np.einsum('ijkl,ilm->ijkm',m0,m1)

涉及的步骤：

• 保持输入的第一个轴对齐。

• 在减少总和中使最后一个轴相

  m0

  对于第二个轴丢失

  m1

  。

• 让其余的轴以外积方式从元素展开

  m0

  并

  m1

  展开 /扩展。

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方法＃2

如果您正在寻找性能并且求和轴的长度较小，那么最好使用单循环并使用

matrix-multiplication

with

np.tensordot

，例如-

s0,s1,s2,s3 = m0.shapes4 = m1.shape[-1]r = np.empty((s0,s1,s2,s4))for i in range(s0):    r[i] = np.tensordot(m0[i],m1[i],axes=([2],[0]))

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方法＃3

现在，

np.dot

可以将其有效地用于2D输入，以进一步提高性能。因此，有了它，修改后的版本虽然更长一些，但希望性能最好的版本是-

s0,s1,s2,s3 = m0.shapes4 = m1.shape[-1]m0.shape = s0,s1*s2,s3   # Get m0 as 3D for temporary usager = np.empty((s0,s1*s2,s4))for i in range(s0):    r[i] = m0[i].dot(m1[i])r.shape = s0,s1,s2,s4m0.shape = s0,s1,s2,s3  # Put m0 back to 4D

运行时测试

功能定义-

def original_app(m0, m1):    s0,s1,s2,s3 = m0.shape    s4 = m1.shape[-1]    r = np.empty((s0,s1,s2,s4))    for i in range(s0):        for j in range(s1): r[i, j] = np.dot(m0[i, j], m1[i])    return rdef einsum_app(m0, m1):    return np.einsum('ijkl,ilm->ijkm',m0,m1)def tensordot_app(m0, m1):    s0,s1,s2,s3 = m0.shape    s4 = m1.shape[-1]    r = np.empty((s0,s1,s2,s4))    for i in range(s0):        r[i] = np.tensordot(m0[i],m1[i],axes=([2],[0]))    return rdef dot_app(m0, m1):    s0,s1,s2,s3 = m0.shape    s4 = m1.shape[-1]    m0.shape = s0,s1*s2,s3   # Get m0 as 3D for temporary usage    r = np.empty((s0,s1*s2,s4))    for i in range(s0):        r[i] = m0[i].dot(m1[i])    r.shape = s0,s1,s2,s4    m0.shape = s0,s1,s2,s3  # Put m0 back to 4D    return r

时间和验证-

In [291]: # Inputs     ...: m0 = np.random.rand(50,30,20,20)     ...: m1 = np.random.rand(50,20,20)     ...:In [292]: out1 = original_app(m0, m1)     ...: out2 = einsum_app(m0, m1)     ...: out3 = tensordot_app(m0, m1)     ...: out4 = dot_app(m0, m1)     ...:      ...: print np.allclose(out1, out2)     ...: print np.allclose(out1, out3)     ...: print np.allclose(out1, out4)     ...: TrueTrueTrueIn [293]: %timeit original_app(m0, m1)     ...: %timeit einsum_app(m0, m1)     ...: %timeit tensordot_app(m0, m1)     ...: %timeit dot_app(m0, m1)     ...: 100 loops, best of 3: 10.3 ms per loop10 loops, best of 3: 31.3 ms per loop100 loops, best of 3: 5.12 ms per loop100 loops, best of 3: 4.06 ms per loop