如何使用Numpy重复这一对2d对？

Question

I have an array that I want to repeat.

test = numpy.array([(1, 11,), (2, 22), (3, 33)])

Now

numpy.repeat(test, 2, 0)
numpy.repeat(test, 2, 1)

results in

array([[ 1, 11],
       [ 1, 11],
       [ 2, 22],
       [ 2, 22],
       [ 3, 33],
       [ 3, 33]])
array([[ 1,  1, 11, 11],
       [ 2,  2, 22, 22],
       [ 3,  3, 33, 33]]).

While

numpy.tile(test, 2)

results in

array([[ 1, 11,  1, 11],
       [ 2, 22,  2, 22],
       [ 3, 33,  3, 33]]).

How can I get this result instead?

array([[ 1, 11],
       [ 2, 22],
       [ 3, 33],
       [ 1, 11],
       [ 2, 22],
       [ 3, 33]])

Alternatively, for my use case I only use the repeated values once. To avoid the memory allocations, is there a way to have a generator of the repeated series instead somehow?

Answer 1

np.tile lets you specify repeats for each axis (as a tuple)

In [370]: np.tile(test,(2,1))
Out[370]: 
array([[ 1, 11],
       [ 2, 22],
       [ 3, 33],
       [ 1, 11],
       [ 2, 22],
       [ 3, 33]])

Answer 2

Here is one option:

In [15]: test = np.array([(1, 11,), (2, 22), (3, 33)])

In [16]: np.tile(test.T, 2).T
Out[16]: 
array([[ 1, 11],
       [ 2, 22],
       [ 3, 33],
       [ 1, 11],
       [ 2, 22],
       [ 3, 33]])