在2D numpy数组的每个滚动窗口中获得最大值

Question

我有一个2D numpy数组,我希望获得每个2d滚动窗口中包含的最大值,该值从左到右,从上到下,每次滚动一行或一列.最天真的方法是遍历所有滚动窗口并获得此滚动窗口中包含的所有值的最大值.我在下面写下了这个方法:

import numpy as np
shape=(1050,300)
window_size=(120,60)
a = np.arange(shape[1]*shape[0]).reshape(shape[1],shape[0])
max_Map=np.full((shape[1]-window_size[1]+1,shape[0]-window_size[0]+1),0,dtype='uint32')

for i in range(shape[1]-window_size[1]+1):
    for j in range(shape[0]-window_size[0]+1):
        window_max=np.max(a[i:i+window_size[1],j:j+window_size[0]])
        max_Map[i][j]=window_max

但这非常低效,因为每次滑动之间只有2行(或2列)发生了变化,但我的代码没有考虑2个连续滚动窗口之间的任何相关性.我能想到的一个改进是每个滑动窗口(假设水平滚动)我将计算最左列的最大值和剩余列的最大值,并将2个值的最大值作为当前窗口最大值.对于下一个滚动窗口,最大值将是新添加的列和之前剩余的列的最大值...但我仍然不认为这是优化的...

我真的很感激,如果有人能指出我正确的方向,我觉得这应该是一个研究得很好的问题,但我无法在任何地方找到解决方案......提前谢谢!

Answer 1

方法 #1使用Scipy's 2D max filter-

from scipy.ndimage.filters import maximum_filter as maxf2D

# Store shapes of inputs
N,M = window_size
P,Q = a.shape

# Use 2D max filter and slice out elements not affected by boundary conditions
maxs = maxf2D(a, size=(M,N))
max_Map_Out = maxs[M//2:(M//2)+P-M+1, N//2:(N//2)+Q-N+1]

方法 #2使用Scikit's 2D sliding window views-

from skimage.util.shape import view_as_windows

N,M = window_size
max_Map_Out = view_as_windows(a, (M,N)).max(axis=(-2,-1))

关于窗口大小及其使用的注意事项：原始方法以翻转的方式对齐窗口大小，即window_size沿第二轴滑动的第一个形状参数，而第二个形状参数决定窗口如何沿第一轴滑动。对于进行滑动最大过滤的其他问题，情况可能并非如此，我们通常使用第一个形状参数作为数组的第一个轴2D，类似地使用第二个形状参数。因此，要解决这些情况，只需使用 :M,N = window_size并按原样使用其余代码。

运行时测试

方法 -

def org_app(a, window_size):
    shape = a.shape[1], a.shape[0]
    max_Map=np.full((shape[1]-window_size[1]+1,
                     shape[0]-window_size[0]+1),0,dtype=a.dtype)
    for i in range(shape[1]-window_size[1]+1):
        for j in range(shape[0]-window_size[0]+1):
            window_max=np.max(a[i:i+window_size[1],j:j+window_size[0]])
            max_Map[i][j]=window_max
    return max_Map

def maxf2D_app(a, window_size):
    N,M = window_size
    P,Q = a.shape
    maxs = maxf2D(a, size=(M,N))
    return maxs[M//2:(M//2)+P-M+1, N//2:(N//2)+Q-N+1]

def view_window_app(a, window_size):
    N,M = window_size
    return view_as_windows(a, (M,N)).max(axis=(-2,-1))

时间安排和验证 -

In [573]: # Setup inputs
     ...: shape=(1050,300)
     ...: window_size=(120,60)
     ...: a = np.arange(shape[1]*shape[0]).reshape(shape[1],shape[0])
     ...: 

In [574]: np.allclose(org_app(a, window_size), maxf2D_app(a, window_size))
Out[574]: True

In [575]: np.allclose(org_app(a, window_size), view_window_app(a, window_size))
Out[575]: True

In [576]: %timeit org_app(a, window_size)
1 loops, best of 3: 2.11 s per loop

In [577]: %timeit view_window_app(a, window_size)
1 loops, best of 3: 1.14 s per loop

In [578]: %timeit maxf2D_app(a, window_size)
100 loops, best of 3: 3.09 ms per loop

In [579]: 2110/3.09  # Speedup using Scipy's 2D max filter over original approach
Out[579]: 682.8478964401295