NumPy append vs Python append

Question

In Python I can append to an empty array like:

>>> a = []
>>> a.append([1,2,3])
>>> a.append([1,2,3])
>>> a
[[1, 2, 3], [1, 2, 3]]

How can I do the same in NumPy? np.append flattens the array, unfortunately (and I need to have an empty array at the beginning).

Answer 1

OP intended to start with empty array. So, here's one approach using NumPy

In [2]: a = np.empty((0,3), int)

In [3]: a
Out[3]: array([], shape=(0L, 3L), dtype=int32)

In [4]: a = np.append(a, [[1,2,3]], axis=0)

In [5]: a
Out[5]: array([[1, 2, 3]])

In [6]: a = np.append(a, [[1,2,3]], axis=0)

In [7]: a
Out[7]:
array([[1, 2, 3],
       [1, 2, 3]])

BUT, if you're appending in a large number of loops. It's faster to append list first and convert to array than appending NumPy arrays.

In [8]: %%timeit
   ...: list_a = []
   ...: for _ in xrange(10000):
   ...:     list_a.append([1, 2, 3])
   ...: list_a = np.asarray(list_a)
   ...:
100 loops, best of 3: 5.95 ms per loop

In [9]: %%timeit
   ....: arr_a = np.empty((0, 3), int)
   ....: for _ in xrange(10000):
   ....:     arr_a = np.append(arr_a, np.array([[1,2,3]]), 0)
   ....:
10 loops, best of 3: 110 ms per loop

Answer 2

I think you're looking for vstack:

>>> import numpy as np
>>> a = [1, 2, 3]
>>> b = [1, 2, 3]
>>> np.vstack((a, b))
array([[1, 2, 3],
       [1, 2, 3]])