[英]Flip non-zero values along each row of a lower triangular numpy array
I have a lower triangular array, like B: 我有一个较低的三角形数组,如B:
B = np.array([[1,0,0,0],[.25,.75,0,0], [.1,.2,.7,0],[.2,.3,.4,.1]])
>>> B
array([[ 1. , 0. , 0. , 0. ],
[ 0.25, 0.75, 0. , 0. ],
[ 0.1 , 0.2 , 0.7 , 0. ],
[ 0.2 , 0.3 , 0.4 , 0.1 ]])
I want to flip it to look like: 我想将它翻转为:
array([[ 1. , 0. , 0. , 0. ],
[ 0.75, 0.25, 0. , 0. ],
[ 0.7 , 0.2 , 0.1 , 0. ],
[ 0.1 , 0.4 , 0.3 , 0.2 ]])
That is, I want to take all the positive values, and reverse within the positive values, leaving the trailing zeros in place. 也就是说,我想获取所有正值,并在正值内反转,留下尾随零。 This is not what
fliplr
does: 这不是
fliplr
作用:
>>> np.fliplr(B)
array([[ 0. , 0. , 0. , 1. ],
[ 0. , 0. , 0.75, 0.25],
[ 0. , 0.7 , 0.2 , 0.1 ],
[ 0.1 , 0.4 , 0.3 , 0.2 ]])
Any tips? 有小费吗? Also, the actual array I am working with would be something like
B.shape = (200,20,4,4)
instead of (4,4)
. 此外,我正在使用的实际数组将类似于
B.shape = (200,20,4,4)
而不是(4,4)
。 Each (4,4)
block looks like the above example (with different numbers across the 200, 20 different entries). 每个
(4,4)
块看起来像上面的例子(在200个不同的条目中有不同的数字)。
How about this: 这个怎么样:
# row, column indices of the lower triangle of B
r, c = np.tril_indices_from(B)
# flip the column indices by subtracting them from r, which is equal to the number
# of nonzero elements in each row minus one
B[r, c] = B[r, r - c]
print(repr(B))
# array([[ 1. , 0. , 0. , 0. ],
# [ 0.75, 0.25, 0. , 0. ],
# [ 0.7 , 0.2 , 0.1 , 0. ],
# [ 0.1 , 0.4 , 0.3 , 0.2 ]])
The same approach will generalize to any arbitrary N -dimensional array that consists of multiple lower triangular submatrices: 相同的方法将推广到由多个下三角形子矩阵组成的任意N维数组:
# creates a (200, 20, 4, 4) array consisting of tiled copies of B
B2 = np.tile(B[None, None, ...], (200, 20, 1, 1))
print(repr(B2[100, 10]))
# array([[ 1. , 0. , 0. , 0. ],
# [ 0.25, 0.75, 0. , 0. ],
# [ 0.1 , 0.2 , 0.7 , 0. ],
# [ 0.2 , 0.3 , 0.4 , 0.1 ]])
r, c = np.tril_indices_from(B2[0, 0])
B2[:, :, r, c] = B2[:, :, r, r - c]
print(repr(B2[100, 10]))
# array([[ 1. , 0. , 0. , 0. ],
# [ 0.75, 0.25, 0. , 0. ],
# [ 0.7 , 0.2 , 0.1 , 0. ],
# [ 0.1 , 0.4 , 0.3 , 0.2 ]])
For an upper triangular matrix you could simply subtract r
from c
instead, eg: 对于上三角矩阵,您可以简单地从
c
减去r
,例如:
r, c = np.triu_indices_from(B.T)
B.T[r, c] = B.T[c - r, c]
Here's one approach for a 2D
array case - 这是
2D
阵列案例的一种方法 -
mask = np.tril(np.ones((4,4),dtype=bool))
out = np.zeros_like(B)
out[mask] = B[:,::-1][mask[:,::-1]]
You can extend it to a 3D
array case using the same 2D
mask by masking
the last two axes with it, like so - 您可以使用相同的
2D
蒙版将其扩展为3D
阵列大小写,方法是用它来masking
最后两个轴,如下所示 -
out = np.zeros_like(B)
out[:,mask] = B[:,:,::-1][:,mask[:,::-1]]
.. and similarly for a 4D
array case, like so - ..和类似的
4D
阵列情况,如此 -
out = np.zeros_like(B)
out[:,:,mask] = B[:,:,:,::-1][:,:,mask[:,::-1]]
As one can see, we are keeping the masking process to the last two axes of (4,4)
and the solution basically stays the same. 可以看出,我们将掩蔽过程保持到
(4,4)
的最后两个轴,并且解决方案基本保持不变。
Sample run - 样品运行 -
In [95]: B
Out[95]:
array([[ 1. , 0. , 0. , 0. ],
[ 0.25, 0.75, 0. , 0. ],
[ 0.1 , 0.2 , 0.7 , 0. ],
[ 0.2 , 0.3 , 0.4 , 0.1 ]])
In [96]: mask = np.tril(np.ones((4,4),dtype=bool))
...: out = np.zeros_like(B)
...: out[mask] = B[:,::-1][mask[:,::-1]]
...:
In [97]: out
Out[97]:
array([[ 1. , 0. , 0. , 0. ],
[ 0.75, 0.25, 0. , 0. ],
[ 0.7 , 0.2 , 0.1 , 0. ],
[ 0.1 , 0.4 , 0.3 , 0.2 ]])
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