动态地在numpy中构造一个特殊矩阵 [英] Constructing a special matrix in numpy dynamically

问题描述

所以我的目标是，给定矩阵s的大小，我试图创建一个如下所示的矩阵，但大小为sxs:

So my objective is the following, given the size of the matrix s, I am attempting to create a matrix that looks like the following, but for size sxs:

[1 1 0]
[1 1 1]
[0 1 1]

对于4x4的大小，它看起来类似于以下内容:

For the size of 4x4, it would look something like the following:

[1 1 0 0]
[1 1 1 0]
[0 1 1 1]
[0 0 1 1]

因此，您可以观察到一个模式:有s-1个重叠的迷你2x2个矩阵.

Therefore you can observe a pattern: there are s-1 number of overlapping mini 2x2 ones matrices.

我当时正在考虑创建一个2x2个1矩阵，然后使用像B[:-1,:-1] = ones_matrix这样的动态引用(用于循环吗?)，其中B是大小为sxs的零矩阵.但是我不确定如何在此处合并for循环，因为如果我们说一个4x4矩阵，那么我们将必须以三种方式引用B:B[:-1,:-1] = ones_matrix, B[1:-1,1:-1] = ones_matrix, B[2:,2:]=ones_matrix.而且我想不出一种方法来动态处理n大小的零矩阵.也许还有另一种方法可以做到这一点?

I was thinking of creating a 2x2 ones matrix and then by using a dynamic referencing (for loop?) like B[:-1,:-1] = ones_matrix, where B is the zeros matrix of size sxs. But I am not sure how to incorporate a for loop here, because if we take say a 4x4 matrix, then we would have to reference B in three ways like so: B[:-1,:-1] = ones_matrix, B[1:-1,1:-1] = ones_matrix, B[2:,2:]=ones_matrix. And I can't figure out a way to do that dynamically for n-sized zeros matrix. Is there perhaps another way to do this?

推荐答案

方法#1:与其一堆2x2矩阵相比，将它看成三个对角线1并结合起来比较容易:

Method #1: Instead of a bunch of 2x2 matrices, it might be easier to look at it as three diagonals of 1 and combine those:

>>> s = 3
>>> np.diag([1]*s,0) + np.diag([1]*(s-1),-1) + np.diag([1]*(s-1), 1)
array([[1, 1, 0],
       [1, 1, 1],
       [0, 1, 1]])
>>> s = 4
>>> np.diag([1]*s,0) + np.diag([1]*(s-1),-1) + np.diag([1]*(s-1), 1)
array([[1, 1, 0, 0],
       [1, 1, 1, 0],
       [0, 1, 1, 1],
       [0, 0, 1, 1]])

方法2 :(受Divankar的回答启发)，我们可以根据距中心的距离进行思考:

Method #2: (inspired by Divankar's answer), we can think in terms of distance from the centre:

>>> s = 4
>>> i,j = np.indices((s,s))
>>> (abs(i-j) <= 1).astype(int)
array([[1, 1, 0, 0],
       [1, 1, 1, 0],
       [0, 1, 1, 1],
       [0, 0, 1, 1]])

方法3:我们可以利用tril或triu进行一些算术:

Method #3: we could take advantage of tril or triu and do some arithmetic:

>>> m = np.tril(np.ones((s,s)),1)
>>> m * m.T
array([[ 1.,  1.,  0.,  0.,  0.],
       [ 1.,  1.,  1.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  0.],
       [ 0.,  0.,  1.,  1.,  1.],
       [ 0.,  0.,  0.,  1.,  1.]])
>>> m = np.tril(np.ones((s,s)),2)
>>> m * m.T
array([[ 1.,  1.,  1.,  0.,  0.],
       [ 1.,  1.,  1.,  1.,  0.],
       [ 1.,  1.,  1.,  1.,  1.],
       [ 0.,  1.,  1.,  1.,  1.],
       [ 0.,  0.,  1.,  1.,  1.]])

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