Reputation: 103
I have a 3D tensor A x B x C
. For each matrix B x C
, I want to extract the leading diagonal.
Is there a vectorized way of doing this in numpy or pytorch instead of looping over A
?
Upvotes: 2
Views: 1022
Reputation: 16404
Use numpy.diagonal()
:
np.diagonal(a, axis1=1, axis2=2)
Example:
In [10]: a = np.arange(3*4*5).reshape(3,4,5)
In [11]: a
Out[11]:
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, 24],
[25, 26, 27, 28, 29],
[30, 31, 32, 33, 34],
[35, 36, 37, 38, 39]],
[[40, 41, 42, 43, 44],
[45, 46, 47, 48, 49],
[50, 51, 52, 53, 54],
[55, 56, 57, 58, 59]]])
In [12]: np.diagonal(a, axis1=1, axis2=2)
Out[12]:
array([[ 0, 6, 12, 18],
[20, 26, 32, 38],
[40, 46, 52, 58]])
Upvotes: 3
Reputation: 221584
Assuming that the leading diagonal for a generic non-squared (BxC) slice starts off from the top-left corner, we can reshape and slice -
a.reshape(a.shape[0],-1)[:,::a.shape[-1]+1]
Sample run -
In [193]: np.random.seed(0)
In [194]: a = np.random.randint(11,99,(3,4,5))
In [195]: a
Out[195]:
array([[[55, 58, 75, 78, 78],
[20, 94, 32, 47, 98],
[81, 23, 69, 76, 50],
[98, 57, 92, 48, 36]],
[[88, 83, 20, 31, 91],
[80, 90, 58, 75, 93],
[60, 40, 30, 30, 25],
[50, 43, 76, 20, 68]],
[[43, 42, 85, 34, 46],
[86, 66, 39, 45, 11],
[11, 47, 64, 16, 49],
[28, 90, 15, 53, 69]]])
In [196]: a.reshape(a.shape[0],-1)[:,::a.shape[-1]+1]
Out[196]:
array([[55, 94, 69, 48],
[88, 90, 30, 20],
[43, 66, 64, 53]])
Upvotes: 1