How to convert list of numpy arrays into single numpy array?


Suppose I have ;

LIST = [[array([1, 2, 3, 4, 5]), array([1, 2, 3, 4, 5],[1,2,3,4,5])] # inner lists are numpy arrays

I try to convert;

array([[1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5])

I am solving it by iteration on vstack right now but it is really slow for especially large LIST

What do you suggest for the best efficient way?

Asked By: erogol



In general you can concatenate a whole sequence of arrays along any axis:

numpy.concatenate( LIST, axis=0 )

but you do have to worry about the shape and dimensionality of each array in the list (for a 2-dimensional 3×5 output, you need to ensure that they are all 2-dimensional n-by-5 arrays already). If you want to concatenate 1-dimensional arrays as the rows of a 2-dimensional output, you need to expand their dimensionality.

As Jorge’s answer points out, there is also the function stack, introduced in numpy 1.10:

numpy.stack( LIST, axis=0 )

This takes the complementary approach: it creates a new view of each input array and adds an extra dimension (in this case, on the left, so each n-element 1D array becomes a 1-by-n 2D array) before concatenating. It will only work if all the input arrays have the same shape.

vstack (or equivalently row_stack) is often an easier-to-use solution because it will take a sequence of 1- and/or 2-dimensional arrays and expand the dimensionality automatically where necessary and only where necessary, before concatenating the whole list together. Where a new dimension is required, it is added on the left. Again, you can concatenate a whole list at once without needing to iterate:

numpy.vstack( LIST )

This flexible behavior is also exhibited by the syntactic shortcut numpy.r_[ array1, ...., arrayN ] (note the square brackets). This is good for concatenating a few explicitly-named arrays but is no good for your situation because this syntax will not accept a sequence of arrays, like your LIST.

There is also an analogous function column_stack and shortcut c_[...], for horizontal (column-wise) stacking, as well as an almost-analogous function hstack—although for some reason the latter is less flexible (it is stricter about input arrays’ dimensionality, and tries to concatenate 1-D arrays end-to-end instead of treating them as columns).

Finally, in the specific case of vertical stacking of 1-D arrays, the following also works:

numpy.array( LIST )

…because arrays can be constructed out of a sequence of other arrays, adding a new dimension to the beginning.

Answered By: jez

Starting in NumPy version 1.10, we have the method stack. It can stack arrays of any dimension (all equal):

# List of arrays.
L = [np.random.randn(5,4,2,5,1,2) for i in range(10)]

# Stack them using axis=0.
M = np.stack(L)
M.shape # == (10,5,4,2,5,1,2)
np.all(M == L) # == True

M = np.stack(L, axis=1)
M.shape # == (5,10,4,2,5,1,2)
np.all(M == L) # == False (Don't Panic)

# This are all true    
np.all(M[:,0,:] == L[0]) # == True
all(np.all(M[:,i,:] == L[i]) for i in range(10)) # == True


Answered By: Jorge E. Cardona

I checked some of the methods for speed performance and find that there is no difference!
The only difference is that using some methods you must carefully check dimension.


|            | shape (10000)  |  shape (1,10000)  |
| np.concat  |    0.18280     |      0.17960      |
|  np.stack  |    0.21501     |      0.16465      |
| np.vstack  |    0.21501     |      0.17181      |
|  np.array  |    0.21656     |      0.16833      |

As you can see I tried 2 experiments – using np.random.rand(10000) and np.random.rand(1, 10000)
And if we use 2d arrays than np.stack and np.array create additional dimension – result.shape is (1,10000,10000) and (10000,1,10000) so they need additional actions to avoid this.


from time import perf_counter
from tqdm import tqdm_notebook
import numpy as np
l = []
for i in tqdm_notebook(range(10000)):
    new_np = np.random.rand(10000)

start = perf_counter()
stack = np.stack(l, axis=0 )
print(f'np.stack: {perf_counter() - start:.5f}')

start = perf_counter()
vstack = np.vstack(l)
print(f'np.vstack: {perf_counter() - start:.5f}')

start = perf_counter()
wrap = np.array(l)
print(f'np.array: {perf_counter() - start:.5f}')

start = perf_counter()
l = [el.reshape(1,-1) for el in l]
conc = np.concatenate(l, axis=0 )
print(f'np.concatenate: {perf_counter() - start:.5f}')
Answered By: Mikhail_Sam

Other solution is to use the asarray function:

numpy.asarray( LIST )

Answered By: Dadou

I found a much more robust numpy function reshape.

Problem of stack and vstack is, that it fails for an empty list.

>>> LIST = [np.array([1, 2, 3, 4, 5]), np.array([1, 2, 3, 4, 5]),np.array([1,2,3,4,5])]
>>> s = np.vstack(LIST)
>>> s.shape
(3, 5)
>>> s = np.vstack([])
ValueError: need at least one array to concatenate

The alternative is reshape

>>> s = np.reshape(LIST, (len(LIST),5))
>>> s.shape
(3, 5)
>>> LIST = []
>>> s = np.reshape(LIST, (len(LIST),5))
>>> s.shape

Drawback is, you need to know the length/shape of your inner array

Answered By: svebert
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