Numpy element-wise dot产品

Question

是否有一种优雅,numpy的方式来应用点积元素？或者如何将以下代码翻译成更好的版本？

m0 # shape (5, 3, 2, 2)
m1 # shape (5,    2, 2)
r = np.empty((5, 3, 2, 2))
for i in range(5):
    for j in range(3):
        r[i, j] = np.dot(m0[i, j], m1[i])

提前致谢!

Answer 1

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.

• 在总和减少中丢失最后一个轴m0与第二个轴m1.

• 让其他轴从m0和m1 散开的 /在外部产品的方式与按元素乘法扩大.

方法#2

如果您正在寻找性能和具有较小长度和还原的轴,你是关闭一个循环更好,使用matrix-multiplication与np.tensordot,像这样-

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]))

方法#3

现在,np.dot可以有效地用于2D输入以进一步提高性能.所以,有了它,修改后的版本,虽然有点长,但希望性能最高的版本是 -

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

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 r

def 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 r        

def 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)
     ...: 
True
True
True

In [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 loop
10 loops, best of 3: 31.3 ms per loop
100 loops, best of 3: 5.12 ms per loop
100 loops, best of 3: 4.06 ms per loop