获得对角线“条纹"来自 NumPy 或 PyTorch 中的矩阵 [英] Getting diagonal "stripe" from matrix in NumPy or PyTorch
问题描述
我需要得到对角线的条纹"；的矩阵.假设我有一个大小为 KxN (K>N) 的矩阵:
[[ 0 1 2][ 3 4 5][ 6 7 8][ 9 10 11]]我需要从中提取对角线条纹，在本例中，是通过截断原始条纹创建的矩阵 MxV 大小:
[[ 0 x x][ 3 4 x][×7 8][ x x 11]]所以结果矩阵为:
[[ 0 4 8][ 3 7 11]]我可以像这样定义一个 bolean 掩码:
将 numpy 导入为 npX=np.arange(12).reshape(4,3)掩码= np.asarray([[对，错，错]，[对，对，错]，[假，真，真]，[假，假，真]])>>>X数组([[ 0, 1, 2],[ 3, 4, 5],[ 6, 7, 8],[ 9, 10, 11]])>>>X.T[mask.T].reshape(3,2).T数组([[ 0, 4, 8],[ 3, 7, 11]])但我不知道如何为任意 KxN 矩阵(例如 39x9、360x96)自动生成这样的掩码
在 numpy、scipy 或 pytorch 中是否有自动执行此操作的函数?
附加问题:是否有可能获得反向条纹"?反而?即
[[ x x 2][×4 5][ 6 7 x][9 x x]]stride_tricks 解决方案:
如果 a 是一个数组，这会创建一个可写的视图，这意味着如果你觉得很喜欢，你可以做类似的事情
更新:可以以相同的精神获得左下至右上的条纹.唯一的小问题:它与原始数组的地址不同.
<预><代码>>>>def reverse_stripe(a):... a = np.asanyarray(a)... *sh, i, j = a.shape...断言 i >= j... *st, k, m = a.strides...返回 np.lib.stride_tricks.as_strided(a[..., j-1:, :], (*sh, i-j+1, j), (*st, k, m-k))...>>>a = np.arange(24).reshape(6, 4)>>>反向条纹(一)数组([[12, 9, 6, 3],[16, 13, 10, 7],[20, 17, 14, 11]])I need to get the diagonal "stripe" of a matrix. Say I have a matrix of size KxN (K>N):
[[ 0 1 2]
[ 3 4 5]
[ 6 7 8]
[ 9 10 11]]
From it I need to extract a diagonal stripe, in this case, a matrix MxV size that is created by truncating the original one:
[[ 0 x x]
[ 3 4 x]
[ x 7 8]
[ x x 11]]
So the result matrix is:
[[ 0 4 8]
[ 3 7 11]]
I could define a bolean mask like so:
import numpy as np
X=np.arange(12).reshape(4,3)
mask=np.asarray([
[ True, False, False],
[ True, True, False],
[ False, True, True],
[ False, False, True]
])
>>> X
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 6, 7, 8],
[ 9, 10, 11]])
>>> X.T[mask.T].reshape(3,2).T
array([[ 0, 4, 8],
[ 3, 7, 11]])
But I don't see how such a mask could be automatically generated to an arbitrary KxN matrix (e.g. 39x9, 360x96)
Is there a function that does this automatically either in numpy, scipy or pytorch?
Additional question: is it possible to get a "reverse stripe" instead? i.e.
[[ x x 2]
[ x 4 5]
[ 6 7 x]
[ 9 x x]]
stride_tricks do the trick:
>>> import numpy as np
>>>
>>> def stripe(a):
... a = np.asanyarray(a)
... *sh, i, j = a.shape
... assert i >= j
... *st, k, m = a.strides
... return np.lib.stride_tricks.as_strided(a, (*sh, i-j+1, j), (*st, k, k+m))
...
>>> a = np.arange(24).reshape(6, 4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15],
[16, 17, 18, 19],
[20, 21, 22, 23]])
>>> stripe(a)
array([[ 0, 5, 10, 15],
[ 4, 9, 14, 19],
[ 8, 13, 18, 23]])
If a is an array this creates a writable view, meaning that if you feel so inclined you can do things like
>>> stripe(a)[...] *= 10
>>> a
array([[ 0, 1, 2, 3],
[ 40, 50, 6, 7],
[ 80, 90, 100, 11],
[ 12, 130, 140, 150],
[ 16, 17, 180, 190],
[ 20, 21, 22, 230]])
UPDATE: bottom-left to top-right stripes can be obtained in the same spirit. Only minor complication: It is not based at the same address as the original array.
>>> def reverse_stripe(a):
... a = np.asanyarray(a)
... *sh, i, j = a.shape
... assert i >= j
... *st, k, m = a.strides
... return np.lib.stride_tricks.as_strided(a[..., j-1:, :], (*sh, i-j+1, j), (*st, k, m-k))
...
>>> a = np.arange(24).reshape(6, 4)
>>> reverse_stripe(a)
array([[12, 9, 6, 3],
[16, 13, 10, 7],
[20, 17, 14, 11]])
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