在3D

Question

更快的矩陣運算更換順序積和在我目前的theano腳本的瓶頸是下面的代碼：

import numpy as np 

axis = 0 
prob = np.random.random((1, 1000, 50)) 
cases = np.random.random((1000, 1000, 50)) 

start = time.time() 
for i in xrange(1000): 
    result = (cases * prob).sum(axis=1-axis, keepdims=True) 
print '3D naive method took {} seconds'.format(time.time() - start) 
print result.shape 
print 

我曾在2D情況下看到以點產品替代的elementwise +總和了我5倍加速。在這種情況下是否有任何矩陣操作可以幫助我？

編輯：

Divakar給了我一個版本的基礎上einsum。但是，我的意圖是將其移植到theano和einsum不支持theano。因此，歡迎使用theano。

Answer 1

我們可以用np.einsum -

result = np.einsum('ijk,ijk->ik', prob, cases)[:,None,:] 

另外一個與np.matmul -

result = np.matmul(prob.transpose(2,0,1), cases.T).T 

運行測試 -

In [70]: axis = 0 
    ...: prob = np.random.random((1, 1000, 50)) 
    ...: cases = np.random.random((1000, 1000, 50)) 
    ...: 

In [71]: out1 = (cases * prob).sum(axis=1-axis, keepdims=True) 

In [72]: out2 = np.einsum('ijk,ijk->ik', prob, cases)[:,None,:] 

In [73]: out3 = np.matmul(prob.transpose(2,0,1), cases.T).T 

In [74]: np.allclose(out1, out2) 
Out[74]: True 

In [75]: np.allclose(out1, out3) 
Out[75]: True 

In [76]: %timeit (cases * prob).sum(axis=1-axis, keepdims=True) 
10 loops, best of 3: 101 ms per loop 

In [77]: %timeit np.einsum('ijk,ijk->ik', prob, cases)[:,None,:] 
10 loops, best of 3: 44.1 ms per loop 

In [78]: %timeit np.matmul(prob.transpose(2,0,1), cases.T).T 
10 loops, best of 3: 44 ms per loop