Porter Vectorizing Sliding Windows

Question

I have the following numpy arrays:

arr_1 = [[1,2],[3,4],[5,6]]   # 3 X 2 
arr_2 = [[0.5,0.6],[0.7,0.8],[0.9,1.0],[1.1,1.2],[1.3,1.4]]  # 5 X 2

arr_1is explicitly an array 3 X 2, whereas it arr_2is an array 5 X 2.

Now, without a loop, I want to incrementally multiply arr_1 and arr_2 in order to apply the sliding window method (window size 3) to arr_2.

Example:

Multiplication 1:  np.multiply(arr_1,arr_2[:3,:])

Multiplication 2: np.multiply(arr_1,arr_2[1:4,:])

Multiplication 3: np.multiply(arr_1,arr_2[2:5,:])

I want to do this in the form of a matrix multiplication form, to make it faster than my current solution having the form:

for i in (2):
   np.multiply(arr_1,arr_2[i:i+3,:])  

So, if the number of lines in arr_2 is large (of the order of tens of thousands), this solution does not really scale very well.

Any help would be greatly appreciated.

Answer 1

NumPy broadcasting, . arr_2 , 3D, 2D arr_1, broadcasting .

, -

W = arr_1.shape[0] # Window size
idx = np.arange(arr_2.shape[0]-W+1)[:,None] + np.arange(W)
out = arr_1*arr_2[idx]

-

In [143]: # Input arrays
     ...: arr_1 = np.random.rand(3,2)
     ...: arr_2 = np.random.rand(10000,2)
     ...: 
     ...: def org_app(arr_1,arr_2):
     ...:     W = arr_1.shape[0] # Window size
     ...:     L = arr_2.shape[0]-W+1
     ...:     out = np.empty((L,W,arr_1.shape[1]))
     ...:     for i in range(L):
     ...:        out[i] = np.multiply(arr_1,arr_2[i:i+W,:])
     ...:     return out
     ...: 
     ...: def vectorized_app(arr_1,arr_2):
     ...:     W = arr_1.shape[0] # Window size
     ...:     idx = np.arange(arr_2.shape[0]-W+1)[:,None] + np.arange(W)
     ...:     return arr_1*arr_2[idx]
     ...: 

In [144]: np.allclose(org_app(arr_1,arr_2),vectorized_app(arr_1,arr_2))
Out[144]: True

In [145]: %timeit org_app(arr_1,arr_2)
10 loops, best of 3: 47.3 ms per loop

In [146]: %timeit vectorized_app(arr_1,arr_2)
1000 loops, best of 3: 1.21 ms per loop

Answer 2

as_strided Divakar.

In [281]: %%timeit 
     ...: out=np.empty((L,W,arr1.shape[1]))
     ...: for i in range(L):
     ...:    out[i]=np.multiply(arr1,arr2[i:i+W,:])
     ...: 
10 loops, best of 3: 48.9 ms per loop
In [282]: %%timeit
     ...: idx=np.arange(L)[:,None]+np.arange(W)
     ...: out=arr1*arr2[idx]
     ...: 
100 loops, best of 3: 2.18 ms per loop
In [283]: %%timeit
     ...: arr3=as_strided(arr2, shape=(L,W,2), strides=(16,16,8))
     ...: out=arr1*arr3
     ...: 
1000 loops, best of 3: 805 µs per loop

Numpy .