Create an array from n copies of subarray in python

arrays, numpy, python

Solution

In [34]: x = np.arange(4)

In [35]: np.tile(x,(3,1))
Out[35]: 
array([[0, 1, 2, 3],
       [0, 1, 2, 3],
       [0, 1, 2, 3]])

But be careful -- you may be able to use broadcasting instead of duplicating the same row over and over.

For example, suppose you have some array of shape (3,4):

In [40]: y = np.arange(12).reshape(3,4)

In [41]: y
Out[41]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

And here is your `x`:

In [42]: x = np.arange(4)

In [43]: x
Out[43]: array([0, 1, 2, 3])

You can add `x` (whose shape is `(4,)`) with `y` (whose shape is `(3,4)`), and NumPy will automatically "broadcast" `x` to shape (3,4):

In [44]: x + y
Out[44]: 
array([[ 0,  2,  4,  6],
       [ 4,  6,  8, 10],
       [ 8, 10, 12, 14]])

Compare the result with

In [45]: np.tile(x,(3,1)) + y
Out[45]: 
array([[ 0,  2,  4,  6],
       [ 4,  6,  8, 10],
       [ 8, 10, 12, 14]])

As you can see, there is no need to `tile` `x` first. In fact, by not tiling `x`, you save memory.

Problem

Is there a more efficient way (or at least pythonic) to stack n copies of a subarray in order to create a new array? ``` import numpy as np x = np.arange(4) for i in range(n-1): x = hstack((x,arange(4))) ``` Thanks,

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