通过更大的矩阵迭代子矩阵 [英] Iterate in submatrices through a bigger matrix

问题描述

我正在寻找一种方法，使用for循环可以迭代更大的矩阵，在该矩阵中，每次迭代将输出大小(行，列，深度)(6、3、3)的子矩阵.

I looking for a way in which I using a for loop can iterate through a bigger matrix, in which each iteration will output a sub matrix of size (row, col, depth) (6,3,3).

我的大矩阵存储为numpy矩阵，并且也可能希望每次迭代都这样输出.

My big matrix is stored as numpy matrix, and would possible also like the each iteration to be outputted as such.

>>> import numpy as np
>>> a = np.random.rand(6*3,3*3,3)
>>> print a.shape
(18, 9, 3)
>>> print a
>>> b

变量b应该包含矩阵a的所有大小为(6,3,3)的子矩阵. 每个子矩阵都不应与之前的子矩阵重叠.

The variable b should contain all the sub matrixes of size (6,3,3) from matrix a. Each submatrix should not overlap with the prior.

推荐答案

方法1

我假设我们正在寻找non-overlapping/distinct块.因此，我们可以使用 Scikit-image's view_as_blocks 实用程序-

I am assuming we are looking for non-overlapping/distinct blocks. As such we could use Scikit-image's view_as_blocks utility -

from skimage.util.shape import view_as_blocks

BSZ = (6,3,3)
out = view_as_blocks(a,BSZ).reshape((-1,)+ (BSZ))

样品运行-

In [279]: a = np.random.rand(6*3,3*3,3)

In [280]: out = view_as_blocks(a,BSZ).reshape((-1,)+ (BSZ))

In [281]: out.shape
Out[281]: (9, 6, 3, 3)

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方法2

仅使用本机NumPy工具(如reshaping和transpose)，这是一种方法-

Using just native NumPy tools like reshaping and transpose, here's one way -

m,n,r = a.shape
split_shp = m//BSZ[0], BSZ[0], n//BSZ[1], BSZ[1], r//BSZ[2], BSZ[2]
out = a.reshape(split_shp).transpose(0,2,4,1,3,5).reshape((-1,)+ (BSZ))

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