8

What is the easiest way to expand a given NumPy array over an extra dimension?

For example, suppose I have

>>> np.arange(4)
array([0, 1, 2, 3])
>>> _.shape
(4,)
>>> expand(np.arange(4), 0, 6)
array([[0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3]])
>>> _.shape
(6, 4)

or this one, a bit more complicated:

>>> np.eye(2)
array([[ 1.,  0.],
       [ 0.,  1.]])
>>> _.shape
(2, 2)
>>> expand(np.eye(2), 0, 3)
array([[[ 1.,  0.],
        [ 0.,  1.]],

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

       [[ 1.,  0.],
        [ 0.,  1.]]])
>>> _.shape
(3, 2, 2)
Improve this question
2
  • have you tried pandas ? Commented Jul 10, 2013 at 21:13
  • No I haven't, would probably be overkill for what I want. Commented Jul 10, 2013 at 21:28

2 Answers 2

Reset to default
6

I would recommend np.tile.

>>> a=np.arange(4)
>>> a
array([0, 1, 2, 3])
>>> np.tile(a,(6,1))
array([[0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3]])

>>> b= np.eye(2)
>>> b
array([[ 1.,  0.],
       [ 0.,  1.]])
>>> np.tile(b,(3,1,1))
array([[[ 1.,  0.],
        [ 0.,  1.]],

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

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

Expanding in many dimensions is pretty easy also:

>>> np.tile(b,(2,2,2))
array([[[ 1.,  0.,  1.,  0.],
        [ 0.,  1.,  0.,  1.],
        [ 1.,  0.,  1.,  0.],
        [ 0.,  1.,  0.,  1.]],

       [[ 1.,  0.,  1.,  0.],
        [ 0.,  1.,  0.,  1.],
        [ 1.,  0.,  1.,  0.],
        [ 0.,  1.,  0.,  1.]]])
Improve this answer
Sign up to request clarification or add additional context in comments.

2 Comments

"Construct an array by repeating A the number of times given by reps." Yeah, exactly, thanks!
If i understand correctly, this creates a new array that uselessly takes up a lot of memory, while a simple view of the existing array would suffice...
2

I think modifying the strides of the array makes it easy to write expand:

def expand(arr, axis, length):
    new_shape = list(arr.shape)
    new_shape.insert(axis, length)
    new_strides = list(arr.strides)
    new_strides.insert(axis, 0)
    return np.lib.stride_tricks.as_strided(arr, new_shape, new_strides)

The function returns a view of the original array, that doesn't take extra memory.

The stride corresponding to the new axis is 0, so that no matter the index for that axis values remain the same, essentially giving you the desired behaviour.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.