# Python: faster local maximum in 2nd matrix

Given: R is an mxn float matrix

Output: O is an mxn matrix, where O [i, j] = R [i, j] if (i, j) is local max and O [i, j] = 0 otherwise. The local maximum is defined as the maximum element in a 3x3 block centered at i, j.

What's a faster way to accomplish this operation in python using numpy and scipy.

m,n = R.shape
for i in range(m):
for j in range(n):
R[i,j]  *= (1 if R[min(0,i-1):max(m, i+2), min(0,j-1):max(n,j+2)].max() == R[i,j] else 0)

+3

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In [28]: from scipy.ndimage import maximum_filter

Here's a sample R

:

In [29]: R
Out[29]:
array([[3, 3, 0, 0, 3],
[0, 0, 2, 1, 3],
[0, 1, 1, 1, 2],
[3, 2, 1, 2, 0],
[2, 2, 1, 2, 1]])

Get the most out of 3x3 windows:

In [30]: mx = maximum_filter(R, size=3)

In [31]: mx
Out[31]:
array([[3, 3, 3, 3, 3],
[3, 3, 3, 3, 3],
[3, 3, 2, 3, 3],
[3, 3, 2, 2, 2],
[3, 3, 2, 2, 2]])

Compare mx

with R

; this is a boolean matrix:

In [32]: mx == R
Out[32]:
array([[ True,  True, False, False,  True],
[False, False, False, False,  True],
[False, False, False, False, False],
[ True, False, False,  True, False],
[False, False, False,  True, False]], dtype=bool)

Use np.where

to create O

:

In [33]: O = np.where(mx == R, R, 0)

In [34]: O
Out[34]:
array([[3, 3, 0, 0, 3],
[0, 0, 0, 0, 3],
[0, 0, 0, 0, 0],
[3, 0, 0, 2, 0],
[0, 0, 0, 2, 0]])

+9

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