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 ]])
我想把它翻转成这样:
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 ]])
也就是说,我想取所有的正值,并在正值内反转,留下后面的零。这不是fliplr
所做的:
>>> 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 ]])
有什么建议吗?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)
。每个(4,4)
块看起来像上面的例子(在200、20个不同的条目中具有不同的数字)。
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 ]])
同样的方法将推广到任何由多个下三角子矩阵组成的任意N维阵列:
# 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 ]])
对于上三角矩阵,您可以简单地从c
中减去r
,例如:
r, c = np.triu_indices_from(B.T)
B.T[r, c] = B.T[c - r, c]
这里有一种2D
数组情况的方法-
mask = np.tril(np.ones((4,4),dtype=bool))
out = np.zeros_like(B)
out[mask] = B[:,::-1][mask[:,::-1]]
您可以使用相同的2D
掩码将其扩展到3D
阵列情况,masking
使用它的最后两个轴,如so-
out = np.zeros_like(B)
out[:,mask] = B[:,:,::-1][:,mask[:,::-1]]
类似地,对于4D
阵列情况,如so-
out = np.zeros_like(B)
out[:,:,mask] = B[:,:,:,::-1][:,:,mask[:,::-1]]
可以看出,我们将掩模过程保持在(4,4)
的最后两个轴,并且解决方案基本保持不变。
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 ]])