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).
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
For Python 3.x, use range() instead of the now-deprecated xrange().
array() conversions.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]])
array() conversions.append(), because vstack() removes the need for (and the complication of) axis=0.np.appendLet's start with an empty 2-D array:
In [8]: a = np.array([]); a = a.reshape((0, 3)); a
Out[8]: array([], shape=(0, 3), dtype=float64)
Now, let's append some rows:
In [19]: a = np.append(a, [[1, 2, 3]], axis=0 ); a
Out[19]: array([[ 1., 2., 3.]])
In [20]: a = np.append(a, [[1, 2, 3]], axis=0 ); a
Out[20]:
array([[ 1., 2., 3.],
[ 1., 2., 3.]])
np.concatenate:Again, let's start with an empty 2-D array:
In [28]: a = np.array([]); a = a.reshape((0, 3)); a
Out[28]: array([], shape=(0, 3), dtype=float64)
Now, let's concatenate some rows:
In [29]: a = np.concatenate( (a, [[1, 2, 3]]), axis=0 ); a
Out[29]: array([[ 1., 2., 3.]])
In [30]: a = np.concatenate( (a, [[1, 2, 3]]), axis=0 ); a
Out[30]:
array([[ 1., 2., 3.],
[ 1., 2., 3.]])
np.append()and at the end remove the first element/row/column. I would suggest if it possible to predefine actual array size and not to change size every time.np.array(a). Listappendis faster than arrayappend.