如何在numpy中沿小尺寸优化点积？

Question

我有两个np.ndarray

• a是形状(13000, 8, 315000)和类型uint8的数组
• b是形状(8,)和类型float32的数组

我想将第二维度（8）上的每个片段乘以b中相应的元素，然后沿该维度求和（即沿第二轴的点积）。结果将是形状(13000, 315000)

我设计了两种方法：

• np.einsum('ijk,j->ik', a, b)：使用%timeit它给出49 s ± 12.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
• np.dot(a.transpose(0, 2, 1), b)：使用%timeit它给出1min 8s ± 3.54 s per loop (mean ± std. dev. of 7 runs, 1 loop each)

有更快的选择吗？你知道吗

补充信息

np.show_config()返回：

blas_mkl_info:
  NOT AVAILABLE
openblas_lapack_info:
    libraries = ['openblas', 'openblas']
    language = c
    library_dirs = ['/usr/local/lib']
    define_macros = [('HAVE_CBLAS', None)]
lapack_mkl_info:
  NOT AVAILABLE
openblas_info:
    libraries = ['openblas', 'openblas']
    language = c
    library_dirs = ['/usr/local/lib']
    define_macros = [('HAVE_CBLAS', None)]
blis_info:
  NOT AVAILABLE
lapack_opt_info:
    libraries = ['openblas', 'openblas']
    language = c
    library_dirs = ['/usr/local/lib']
    define_macros = [('HAVE_CBLAS', None)]
blas_opt_info:
    libraries = ['openblas', 'openblas']
    language = c
    library_dirs = ['/usr/local/lib']
    define_macros = [('HAVE_CBLAS', None)]

a.flags：

C_CONTIGUOUS : True
F_CONTIGUOUS : False
OWNDATA : True
WRITEABLE : True
ALIGNED : True
WRITEBACKIFCOPY : False
UPDATEIFCOPY : False

b.flags：

C_CONTIGUOUS : True
F_CONTIGUOUS : True
OWNDATA : True
WRITEABLE : True
ALIGNED : True
WRITEBACKIFCOPY : False
UPDATEIFCOPY : False

Answer 1

我们可以利用^{} with ^{} module处理大数据，提高内存效率，从而提高性能-

import numexpr as ne

d = {'a0':a[:,0],'b0':b[0],'a1':a[:,1],'b1':b[1],\
     'a2':a[:,2],'b2':b[2],'a3':a[:,3],'b3':b[3],\
     'a4':a[:,4],'b4':b[4],'a5':a[:,5],'b5':b[5],\
     'a6':a[:,6],'b6':b[6],'a7':a[:,7],'b7':b[7]}
eval_str = 'a0*b0 + a1*b1 + a2*b2 + a3*b3 + a4*b4 + a5*b5 + a6*b6 + a7*b7'
out = ne.evaluate(eval_str,d)

计时运行示例-

In [474]: # Setup with ~10x smaller than posted one, as my system can't handle those
     ...: np.random.seed(0)
     ...: a = np.random.randint(0,9,(1000,8,30000)).astype(np.uint8)
     ...: b = np.random.rand(8).astype(np.float32)

In [478]: %timeit np.einsum('ijk,j->ik', a, b)
1 loop, best of 3: 247 ms per loop

# einsum with optimize flag set as True
In [479]: %timeit np.einsum('ijk,j->ik', a, b, optimize=True)
1 loop, best of 3: 248 ms per loop

In [480]: d = {'a0':a[:,0],'b0':b[0],'a1':a[:,1],'b1':b[1],\
     ...:      'a2':a[:,2],'b2':b[2],'a3':a[:,3],'b3':b[3],\
     ...:      'a4':a[:,4],'b4':b[4],'a5':a[:,5],'b5':b[5],\
     ...:      'a6':a[:,6],'b6':b[6],'a7':a[:,7],'b7':b[7]}

In [481]: eval_str = 'a0*b0 + a1*b1 + a2*b2 + a3*b3 + a4*b4 + a5*b5 + a6*b6 + a7*b7'

In [482]: %timeit ne.evaluate(eval_str,d)
10 loops, best of 3: 94.3 ms per loop

~2.6x改善。你知道吗

创建评估部分的更好（不易出错）和通用方法如下-

d = {'a'+str(i):a[:,i] for i in range(8)}
d.update({'b'+str(i):b[i] for i in range(8)})
eval_str = ' + '.join(['a'+str(i)+'*'+'b'+str(i) for i in range(8)])