如何在 numpy 中使用搜索窗口并将最大值和所有其他值保留为零

Question

我有一个二维数组，例如(101 行和 100 列(。现在我想创建一个搜索窗口或块(3 行 x 3 列(，它将在数组周围移动并确定最大值，选择它并使用 python 和 numpy 将所有其他值保留为零。例如

x = ([[1,2,3,4,5,6,7,8,9,10],
[2,5,4,5,3,4,6,7,5,3],
[3,3,4,5,6,7,3,4,5,8]]

例如，使用 2x2 搜索窗口 x.some函数从左上角开始将给出结果

Result = ([0,0...
[0,5... for the 1st iteration so the the whole result should look like this 

Result = ([[0,0,0,0,0,6,0,8,0,10],
[0,5,0,5,0,0,0,0,0,0],
[3,3,0,5,0,7,0,4,0,8]]

请注意，搜索窗口的最后一行必须从 2x2 数组更改为 2x1，因为搜索窗口不重叠

您的帮助将不胜感激。 提前谢谢你

Answer 1

以下是使用skimage.util.view_as_blocks的方法：

>>> import numpy as np
>>> import skimage.util as su
>>> 
>>> def split_axis(N, n):
...     q, r = divmod(N, n)
...     left = ((np.s_[:q*n], n),) if q else ()
...     right = ((np.s_[q*n:], r),) if r else ()
...     return (*left, *right)
... 
>>> def block_max(x, block, inplace=False):
...     if not inplace:
...         x = x.copy()
...     xi, xj = x.shape
...     bi, bj = block
...     for ci, ri in split_axis(xi, bi):
...         for cj, rj in split_axis(xj, bj):
...             vab = su.view_as_blocks(x[ci, cj], (ri, rj))
...             vab[vab < vab.max(axis=(-1, -2), keepdims=True)] = 0
...     return x
... 
>>> x = ([[1,2,3,4,5,6,7,8,9,10],
...      [2,5,4,5,3,4,6,7,5,3],
...      [3,3,4,5,6,7,3,4,5,8]])
>>> 
>>> x = np.array(x)
>>> 
>>> block_max(x, (2, 2))
array([[ 0,  0,  0,  0,  0,  6,  0,  8,  0, 10],
[ 0,  5,  0,  5,  0,  0,  0,  0,  0,  0],
[ 3,  3,  0,  5,  0,  7,  0,  4,  0,  8]])

如果您没有skimage：

>>> def view_as_blocks(x, blockshape):
...     *xs, xi, xj = x.shape
...     bi, bj = blockshape
...     return np.ascontiguousarray(x).reshape(*xs, xi//bi, xj//bj, *blockshape)

您的更新问题(未经测试(：

>>> def block_max(x, block):
...     out = np.zeros_like(x)
...     xi, xj = x.shape
...     bi, bj = block
...     for ci, ri in split_axis(xi, bi):
...         for cj, rj in split_axis(xj, bj):
...             vab = su.view_as_blocks(x[ci, cj], (ri, rj))
...             oab = su.view_as_blocks(out[ci, cj], (ri, rj))
...             vmx = vab.max(axis=(-1, -2), keepdims=True)
...             vmn = vab.min(axis=(-1, -2), keepdims=True)
...             cond = vmx - vmn > 2
...             oab[cond & (vab == vmx)] == 1
...             oab[cond & (vab == vmn)] == 2
...     return out