How to convert list of numpy arrays into single numpy array?
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 3x5 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—even along the axis of concatenation.
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.
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
Enjoy,
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.
Timing:
|------------|----------------|-------------------|
| | 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.
Code:
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)
l.append(new_np)
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}')