问题
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_1
is clearly a 3 X 2
array, whereas arr_2
is a 5 X 2
array.
Now without looping, I want to element-wise multiply arr_1 and arr_2 so that I apply a sliding window technique (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 some sort of a matrix multiplication form to make it faster than my current solution which is of the form:
for i in (2):
np.multiply(arr_1,arr_2[i:i+3,:])
So if the number of rows in arr_2 are large (of the order of tens of thousands), this solution doesn't really scale very well.
Any help would be much appreciated.
回答1:
We can use NumPy broadcasting to create those sliding windowed indices in a vectorized manner. Then, we can simply index into arr_2
with those to create a 3D
array and perform element-wise multiplication with 2D
array arr_1
, which in turn will bring on broadcasting
again.
So, we would have a vectorized implementation like so -
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]
Runtime test and verify results -
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
回答2:
This is a nice case to test the speed of as_strided
and Divakar's broadcasting.
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
Create Numpy array without enumerating array for more of a comparison of these methods.
来源:https://stackoverflow.com/questions/39232790/numpy-vectorization-of-sliding-window-operation