方法 #1 使用Scipy的2D最大值滤波器
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该方法使用Scipy库中的2D最大值滤波器来实现。
from scipy.ndimage.filters import maximum_filter as maxf2D
N,M = window_size
P,Q = a.shape
maxs = maxf2D(a, size=(M,N))
max_Map_Out = maxs[M//2:(M//2)+P-M+1, N//2:(N//2)+Q-N+1]
方案二 使用Scikit的2D滑动窗口视图
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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释:原始方法中,窗口大小按照相反的方式对齐,即window_size
的第一个形状参数沿着第二个轴滑动,而第二个形状参数决定窗口沿着第一个轴滑动的方式。对于其他进行滑动最大值过滤的问题,可能不是这种情况,我们通常使用第一个形状参数来操作2D
数组的第一个轴,第二个形状参数类似。因此,为了解决这些情况,只需使用:M,N = window_size
,然后使用其余的代码。
运行时间测试
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))
时间和验证 -
In [573]:
...: 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
Out[579]: 682.8478964401295
max_Map=np.full((shape[1]-window_size[1]+1,shape[0]-window_size[0]+1),..
,for i in range(shape[1]-window_size[1]+1):
和for j in range(shape[0]-window_size[0]+1):
来覆盖所有元素。 - Divakar