Identity matrix of 3 Dimensional matrix-CodePudding

Identity = array([[[1., 0., 0., 0.],
    [0., 1., 0., 0.],
    [0., 0., 1., 0.],
    [0., 0., 0., 1.]],

   [[1., 0., 0., 0.],
    [0., 1., 0., 0.],
    [0., 0., 1., 0.],
    [0., 0., 0., 1.]],

   [[1., 0., 0., 0.],
    [0., 1., 0., 0.],
    [0., 0., 1., 0.],
    [0., 0., 0., 1.]]])

There is a task which wants me to create identity matrix in 3D.

I have made assumptions that the above is the identity matrix in 3D with the shape (3,4,4).

I have seen other variations of identity of 3D matrix which I didn't understand. Check this What's the best way to create a "3D identity matrix" in Numpy? for reference.

If I am right in my above assumption of identity matrix. Please assist me to construct the same with numpy.

CodePudding user response：

You can use np.identity() to generate an identity matrix and then use np.broadcast_to() to add the third dimension:

import numpy as np
n = 4
print(np.broadcast_to(np.identity(n), (3, n, n)))

CodePudding user response：

You can also use np.tile:

n = 4
k = 3
out = np.tile(np.identity(n), (k,1)).reshape(k,n,n)

Output:

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

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

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