numpy数组的固定大小的子矩阵的索引 [英] Indices of fixed size sub-matrices of numpy array

问题描述

我正在实现一种算法，该算法要求我查看(严格二维)numpy数组内的非重叠连续子矩阵.例如12点乘12点

I am implementing an algorithm which requires me to look at non-overlapping consecutive submatrices within a (strictly two dimensional) numpy array. eg, for the 12 by 12

>>> a = np.random.randint(20, size=(12, 12)); a
array([[ 4,  0, 12, 14,  3,  8, 14, 12, 11, 18,  6,  6],
       [15, 13,  2, 18, 15, 15, 16,  2,  9, 16,  6,  4],
       [18, 18,  3,  8,  1, 15, 14, 13, 13, 13,  7,  0],
       [ 1,  9,  3,  6,  0,  4,  3, 15,  0,  9, 11, 12],
       [ 5, 15,  5,  6,  4,  4, 18, 13, 10, 17, 11,  8],
       [13, 17,  8, 15, 17, 12,  7,  1, 13, 15,  0, 18],
       [ 2,  1, 11, 12,  3, 16, 11,  9, 10, 15,  4, 16],
       [19, 11, 10,  7, 10, 19,  7, 13, 11,  9, 17,  8],
       [14, 14, 17,  0,  0,  0, 11,  1, 10, 14,  2,  7],
       [ 6, 15,  6,  7, 15, 19,  2,  4,  6, 16,  0,  3],
       [ 5, 10,  7,  5,  0,  8,  5,  8,  9, 14,  4,  3],
       [17,  2,  0,  3, 15, 10, 14,  1,  0,  7, 16,  2]])

并查看3x3子矩阵，我希望第一个3x3子矩阵位于左上角:

and looking at 3x3 submatrices, I would want the first 3x3 submatrix to be from the upper left corner:

>>> a[0:3, 0:3]
array([[ 4,  0, 12],
       [15, 13,  2],
       [18, 18,  3]])

下一步由a[0:3, 3:6]给定，依此类推.每一行或每一列中的最后一组这样的索引是否从数组的末尾开始都没关系-numpy仅仅给出切片中存在的部分的行为就足够了.

The next along to be given by a[0:3, 3:6] and so on. It doesn't matter if the last such set of indices in each row or column runs off the end of the array - numpy's behavior of simply giving the portion within the slice that exists is sufficient.

我想要一种以编程方式为任意大小的矩阵和子矩阵生成这些切片索引的方法.我目前有这个:

I want a way to generate these slice indices programatically for arbitrarily sized matrices and submatrices. I currently have this:

size = 3
x_max = a.shape[0]
xcoords = range(0, x_max, size)
xcoords = zip(xcoords, xcoords[1:])

并类似地生成y_coords，因此一系列索引由itertools.product(xcoords, ycoords)给出.

and similarly to generate y_coords, so that the series of indices is given by itertools.product(xcoords, ycoords).

我的问题是:是否有更直接的方法可以执行此操作，可能使用

My question is: is there a more direct way to do this, perhaps using numpy.mgrid or some other numpy technique?

推荐答案

获取索引

这是获取特定size x size块的快速方法:

base = np.arange(size) # Just the base set of indexes
row = 1                # Which block you want
col = 0                
block = a[base[:, np.newaxis] + row * size, base + col * size]

如果您愿意，可以建立类似于xcoords的矩阵，例如:

If you wanted you could build up matrices similar to your xcoords like:

y, x = np.mgrid[0:a.shape[0]/size, 0:a.shape[1]/size]
y_coords = y[..., np.newaxis] * size + base
x_coords = x[..., np.newaxis] * size + base

然后您可以访问如下所示的块:

Then you could access a block like this:

block = a[y_coords[row, col][:, np.newaxis], x_coords[row, col]]

---

直接获取积木

如果您只想获取块(而不是块条目的索引)，则可以使用

---

Getting the blocks directly

If you just want to get the blocks (and not the indexes of the block entries), I'd use np.split (twice):

blocks = map(lambda x : np.split(x, a.shape[1]/size, 1), # Split the columns
                        np.split(a, a.shape[0]/size, 0)) # Split the rows

然后您将获得一个size x size个块的二维列表:

then you have a 2D list of size x size blocks:

>>> blocks[0][0]
array([[ 4,  0, 12],
       [15, 13,  2],
       [18, 18,  3]])

>>> blocks[1][0]
array([[ 1,  9,  3],
       [ 5, 15,  5],
       [13, 17,  8]])

然后可以将其设为一个numpy数组，并使用与上面相同的索引样式:

You could then make this a numpy array and use the same indexing style as above:

>>> blocks = np.array(blocks)
>>> blocks.shape
(4, 4, 3, 3)

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