Converting numpy arrays of arrays into one whole numpy array

Question

I want to turn my array of array into just a single array. From something like :

array([ array([[0, 0, 0, ..., 1, 0, 0],
       [0, 1, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 2, 0, 0],
       ..., 
       array([[0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 8, 0, 2],
       ..., 
       [0, 0, 0, ..., 0, 0, 0],
       [1, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 1, 0, 0]], dtype=uint8)], dtype=object)

which has size (10,) to just the 3D numpy array which is of size (10,518, 32)

array([[[0, 0, 0, ..., 0, 0, 0],
        [0, 0, 0, ..., 0, 0, 0],
        [0, 0, 0, ..., 0, 0, 0],
        ..., 
        [0, 0, 0, ..., 0, 0, 0],
        [0, 0, 0, ..., 0, 0, 0],
        [0, 0, 0, ..., 0, 0, 0]]], dtype=uint8)

I've tried converting everything into a list then do np.asarray and also tried defining everything as the same dtype=uint8 but I couldn't get it into the 3D form.

This looks like a case where you should fix the problem upstream. Why do you even have arrays of arrays? This should have been a single 3D array from the start. — user2357112
– user2357112, Commented Feb 4, 2016 at 4:40
I agree with the first comment, but you could also do a list comprehension with the np.array.tolist(). Something like np.array(arr.tolist() for arr in my_arrays) — A Small Shell Script
– A Small Shell Script, Commented Feb 4, 2016 at 5:18
Could you please post an example that we can run, say with a (2,4,3) array of arrays? If I type what I think is a small version of your example, I don't get a shape (10,). — tom10
– tom10, Commented Feb 4, 2016 at 5:48

hpaulj · Accepted Answer · 2016-02-04 07:49:20Z

23

np.concatenate should do the trick:

Make an object array of arrays:

In [23]: arr=np.empty((4,),dtype=object)
In [24]: for i in range(4):arr[i]=np.ones((2,2),int)*i
In [25]: arr
Out[25]: 
array([array([[0, 0],
       [0, 0]]), array([[1, 1],
       [1, 1]]),
       array([[2, 2],
       [2, 2]]), array([[3, 3],
       [3, 3]])], dtype=object)

In [28]: np.concatenate(arr)
Out[28]: 
array([[0, 0],
       [0, 0],
       [1, 1],
       [1, 1],
       [2, 2],
       [2, 2],
       [3, 3],
       [3, 3]])

Or with a reshape:

In [26]: np.concatenate(arr).reshape(4,2,2)
Out[26]: 
array([[[0, 0],
        [0, 0]],

       [[1, 1],
        [1, 1]],

       [[2, 2],
        [2, 2]],

       [[3, 3],
        [3, 3]]])
In [27]: _.shape
Out[27]: (4, 2, 2)

concatenate effectively treats its input as a list of arrays. So it works regardless of whether this is an object array, a list, or 3d array.

This can't be done simply with a reshape. arr is an array of pointers - pointing to arrays located elsewhere in memory. To get a single 3d array, all of the pieces will have to be copied into one buffer. That's what concatenate does - it creates a large empty file, and copies each array, but it does it in compiled code.

np.array does not change it:

In [37]: np.array(arr).shape
Out[37]: (4,)

but treating arr as a list of arrays does work (but is slower than the concatenate version - array analyses its inputs more).

In [38]: np.array([x for x in arr]).shape
Out[38]: (4, 2, 2)

edited Feb 4, 2016 at 7:49

answered Feb 4, 2016 at 6:07

hpaulj

233k14 gold badges260 silver badges392 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

norok2 Over a year ago

The [x for x in arr] should be replaced by arr.tolist(). This would lead to faster code than np.concatenate().

hpaulj Over a year ago

@norok2, I don't see a consistent difference in timings.

norok2 · Accepted Answer · 2020-09-28 09:05:17Z

Perhaps late to the party, but I believe the most efficient approach is:

np.array(arr.tolist())

To give some idea of how it would work:

import numpy as np


N, M, K = 4, 3, 2
arr = np.empty((N,), dtype=object)
for i in range(N):
    arr[i] = np.full((M, K), i)


print(arr)
# [array([[0, 0],
#        [0, 0],
#        [0, 0]])
#  array([[1, 1],
#        [1, 1],
#        [1, 1]])
#  array([[2, 2],
#        [2, 2],
#        [2, 2]])
#  array([[3, 3],
#        [3, 3],
#        [3, 3]])]


new_arr = np.array(arr.tolist())
print(new_arr)
# [[[0 0]
#   [0 0]
#   [0 0]]

#  [[1 1]
#   [1 1]
#   [1 1]]

#  [[2 2]
#   [2 2]
#   [2 2]]

#  [[3 3]
#   [3 3]
#   [3 3]]]

...and the timings:

%timeit np.array(arr.tolist())
# 100000 loops, best of 3: 2.48 µs per loop
%timeit np.concatenate(arr).reshape(N, M, K)
# 100000 loops, best of 3: 3.28 µs per loop
%timeit np.array([x for x in arr])
# 100000 loops, best of 3: 3.32 µs per loop

Ricardo Decal · Accepted Answer · 2020-04-30 00:35:37Z

I had the same issue extracting a column from a Pandas DataFrame containing an array in each row:

joined["ground truth"].values
# outputs
array([array([0, 0, 0, 0, 0, 0, 0, 0]), array([0, 0, 0, 0, 0, 0, 0, 0]),
       array([0, 0, 0, 0, 0, 0, 0, 0]), ...,
       array([0, 0, 0, 0, 0, 0, 0, 0]), array([0, 0, 0, 0, 0, 0, 0, 0]),
       array([0, 0, 0, 0, 0, 0, 0, 0])], dtype=object)

np.concatenate didn't help because it merged the arrays into a flat array (same as np.hstack). Instead, I needed to vertically stack them with np.vstack:

array([[0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       ...,
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0],
       [0, 0, 0, ..., 0, 0, 0]])

Keith Brodie · Accepted Answer · 2016-02-04 05:16:07Z

One way is to allocate the target array and copy the objects in as a loop.

import numpy as np

x = np.array([ np.array([[0, 0, 0, 1, 0, 0],
                         [0, 1, 0, 0, 0, 0],
                         [0, 0, 3, 7, 0, 0],
                         [0, 0, 0, 2, 0, 0]], dtype=np.uint8),
               np.array([[0, 0, 0, 0, 0, 0],
                         [0, 0, 0, 0, 0, 0],
                         [0, 0, 4, 8, 0, 0],
                         [0, 0, 0, 8, 0, 2]], dtype=np.uint8),
               np.array([[0, 0, 0, 0, 0, 0],
                         [1, 0, 0, 0, 0, 0],
                         [0, 0, 5, 9, 0, 0],
                         [0, 0, 0, 1, 0, 0]], dtype=np.uint8)], dtype=object)

print len(x)

print x[0].shape

y=np.zeros([len(x),x[0].shape[0],x[0].shape[1]],dtype=np.uint8)

print y.shape

for i in range(len(x)):
    y[i,:,:] = x[i]

print y

If I understand what you're asking this is the desired result:

3
(4L, 6L)
(3L, 4L, 6L)
[[[0 0 0 1 0 0]
  [0 1 0 0 0 0]
  [0 0 3 7 0 0]
  [0 0 0 2 0 0]]

 [[0 0 0 0 0 0]
  [0 0 0 0 0 0]
  [0 0 4 8 0 0]
  [0 0 0 8 0 2]]

 [[0 0 0 0 0 0]
  [1 0 0 0 0 0]
  [0 0 5 9 0 0]
  [0 0 0 1 0 0]]]

Collectives™ on Stack Overflow

Converting numpy arrays of arrays into one whole numpy array

4 Answers 4

2 Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related