python - Pythonic way to create a numpy array from a list of numpy arrays

Question

I generate a list of one dimensional numpy arrays in a loop and later convert this list to a 2d numpy array. I would've preallocated a 2d numpy array if i knew the number of items ahead of time, but I don't, therefore I put everything in a list.

The mock up is below:

>>> list_of_arrays = map(lambda x: x*ones(2), range(5))
>>> list_of_arrays
[array([ 0.,  0.]), array([ 1.,  1.]), array([ 2.,  2.]), array([ 3.,  3.]), array([ 4.,  4.])]
>>> arr = array(list_of_arrays)
>>> arr
array([[ 0.,  0.],
       [ 1.,  1.],
       [ 2.,  2.],
       [ 3.,  3.],
       [ 4.,  4.]])

(My question is the following:)

Is there a better way (performancewise) to go about the task of collecting sequential numerical data (in my case numpy arrays) than putting them in a list and then making a numpy.array out of it (I am creating a new obj and copying the data)? Is there an "expandable" matrix data structure available in a well tested module? A typical size of my 2d matrix would be between 100x10 and 5000x10 floats (EDIT:) In this example i'm using map, but in my actual application I have a for loop question from:https://stackoverflow.com/questions/2106287/pythonic-way-to-create-a-numpy-array-from-a-list-of-numpy-arrays

Answer 1

Convenient way, using numpy.concatenate. I believe it's also faster, than @unutbu's answer:

In [32]: import numpy as np 

In [33]: list_of_arrays = list(map(lambda x: x * np.ones(2), range(5)))

In [34]: list_of_arrays
Out[34]: 
[array([ 0.,  0.]),
 array([ 1.,  1.]),
 array([ 2.,  2.]),
 array([ 3.,  3.]),
 array([ 4.,  4.])]

In [37]: shape = list(list_of_arrays[0].shape)

In [38]: shape
Out[38]: [2]

In [39]: shape[:0] = [len(list_of_arrays)]

In [40]: shape
Out[40]: [5, 2]

In [41]: arr = np.concatenate(list_of_arrays).reshape(shape)

In [42]: arr
Out[42]: 
array([[ 0.,  0.],
       [ 1.,  1.],
       [ 2.,  2.],
       [ 3.,  3.],
       [ 4.,  4.]])