python:在二维矩阵中更快的局部最大值 [英] python: Faster local maximum in 2-d matrix

问题描述

给出:R是一个mxn浮点矩阵

Given: R is an mxn float matrix

输出:O是一个mxn矩阵，如果(i，j)是局部最大值，则O [i，j] = R [i，j]，否则O [i，j] = 0.局部最大值定义为以i，j为中心的3x3块中的最大元素.

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

使用numpy和scipy在python上执行此操作的更快方法是什么.

What's a faster way to do this operation on 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)

推荐答案

您可以使用

You can use scipy.ndimage.maximum_filter:

In [28]: from scipy.ndimage import maximum_filter

这是示例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]])

在3x3的窗口上获取最大值:

Get the maximum on 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]])

比较mx与R;这是一个布尔矩阵:

Compare mx to 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)

使用 np.where 创建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]])

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