python中不同形状的数组的元素乘法

Question

说我有两个数组a和b,

  a.shape = (5,2,3)
  b.shape = (2,3)

然后c = a * b会给我一个数组c的形状(5,2,3)用c[i,j,k] = a[i,j,k]*b[j,k].

现在的情况是,

  a.shape = (5,2,3)
  b.shape = (2,3,8)

我想c有一个形状(5,2,3,8)与c[i,j,k,l] = a[i,j,k]*b[j,k,l].

如何有效地做到这一点？我a和b实际上相当大.

Answer 1

这应该工作:

a[..., numpy.newaxis] * b[numpy.newaxis, ...]

用法:

In : a = numpy.random.randn(5,2,3)

In : b = numpy.random.randn(2,3,8)

In : c = a[..., numpy.newaxis]*b[numpy.newaxis, ...]

In : c.shape
Out: (5, 2, 3, 8)

参考:数组广播在numpy

编辑:更新了参考URL

Answer 2

我认为以下应该有效:

import numpy as np
a = np.random.normal(size=(5,2,3))
b = np.random.normal(size=(2,3,8))
c = np.einsum('ijk,jkl->ijkl',a,b)

和:

In [5]: c.shape
Out[5]: (5, 2, 3, 8)

In [6]: a[0,0,1]*b[0,1,2]
Out[6]: -0.041308376453821738

In [7]: c[0,0,1,2]
Out[7]: -0.041308376453821738

np.einsum 可能有点棘手,但对于这些索引问题非常强大:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.einsum.html

另请注意,这需要numpy> = v1.6.0

我不确定你的特定问题的效率,但如果它的性能不如所需,那么一定要考虑使用Cython和显式for循环,并且可能使用它来并行化 prange

UPDATE

In [18]: %timeit np.einsum('ijk,jkl->ijkl',a,b)
100000 loops, best of 3: 4.78 us per loop

In [19]: %timeit a[..., np.newaxis]*b[np.newaxis, ...]
100000 loops, best of 3: 12.2 us per loop


In [20]: a = np.random.normal(size=(50,20,30))
In [21]: b = np.random.normal(size=(20,30,80))

In [22]: %timeit np.einsum('ijk,jkl->ijkl',a,b)
100 loops, best of 3: 16.6 ms per loop

In [23]: %timeit a[..., np.newaxis]*b[np.newaxis, ...]
100 loops, best of 3: 16.6 ms per loop

In [2]: a = np.random.normal(size=(500,20,30))
In [3]: b = np.random.normal(size=(20,30,800))

In [4]: %timeit np.einsum('ijk,jkl->ijkl',a,b)
1 loops, best of 3: 3.31 s per loop

In [5]: %timeit a[..., np.newaxis]*b[np.newaxis, ...]
1 loops, best of 3: 2.6 s per loop