Concatenating empty array in Numpy

Question

in Matlab I do this:

>> E = [];
>> A = [1 2 3 4 5; 10 20 30 40 50];
>> E = [E ; A]

E =

     1     2     3     4     5
    10    20    30    40    50

Now I want the same thing in Numpy but I have problems, look at this:

>>> E = array([],dtype=int)
>>> E
array([], dtype=int64)
>>> A = array([[1,2,3,4,5],[10,20,30,40,50]])

>>> E = vstack((E,A))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/numpy/core/shape_base.py", line 226, in vstack
    return _nx.concatenate(map(atleast_2d,tup),0)
ValueError: array dimensions must agree except for d_0

I have a similar situation when I do this with:

>>> E = concatenate((E,A),axis=0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: arrays must have same number of dimensions

Or:

>>> E = append([E],[A],axis=0)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/numpy/lib/function_base.py", line 3577, in append
    return concatenate((arr, values), axis=axis)
ValueError: arrays must have same number of dimensions

In MATLAB arrays are always 2D or larger. In numpy they can have 1 or even 0 dimensions. — hpaulj
– hpaulj, Commented Mar 29, 2014 at 18:46

Benjamin Loison · Accepted Answer · 2025-07-04 10:25:07Z

135

if you know the number of columns before hand:

>>> xs = np.array([[1,2,3,4,5],[10,20,30,40,50]])
>>> ys = np.array([], dtype=np.int64).reshape(0,5)
>>> ys
array([], shape=(0, 5), dtype=int64)
>>> np.vstack([ys, xs])
array([[  1.,   2.,   3.,   4.,   5.],
       [ 10.,  20.,  30.,  40.,  50.]])

if not:

>>> ys = np.array([])
>>> ys = np.vstack([ys, xs]) if ys.size else xs
array([[ 1,  2,  3,  4,  5],
       [10, 20, 30, 40, 50]])

edited Jul 4 at 10:25

Benjamin Loison

5,7514 gold badges20 silver badges37 bronze badges

answered Mar 29, 2014 at 15:14

behzad.nouri

78.5k18 gold badges130 silver badges127 bronze badges

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

6 Comments

Nirvedh Meshram Over a year ago

I dont understand how the second implementation works. It looks like ys.size should evaluate to False and then the statement is just ys=xs.

m13op22 Over a year ago

@NirvedhMeshram The second statement says that if ys.size is a value, any value that is not zero or False, then execute the statement.

flow2k Over a year ago

@NirvedhMeshram I think you're right. Logically, the second implementation doesn't need the first statement. It can just check if ys already exists: ys = np.vstack([ys, xs]) if 'ys' in vars() else xs.

flow2k Over a year ago

Also, I think vstack can be replaced with concatenate (and the use of () correspondingly); does anyone think otherwise? If I'm right - the essence of this method is to use reshape on an empty array.

hpaulj Over a year ago

This whole idea of creating a 0 element array just so you can vstack is wrong. As a one time operation it is useless, and as part of a repeated vstack it is inefficient. Stick with list append. Don't slavishly imitate MATLAB.

|

Benjamin Loison · Accepted Answer · 2025-07-04 10:24:35Z

4

In Python, if possible to work with the individual vectors, to append you should use list.append()

>>> E = []
>>> B = np.array([1,2,3,4,5])
>>> C = np.array([10,20,30,40,50])
>>> E = E.append(B)
>>> E = E.append(C)
[array([1, 2, 3, 4, 5]), array([10, 20, 30, 40, 50])]

and then after all append operations are done, return to np.array thusly

>>> E = np.array(E)
array([[ 1,  2,  3,  4,  5],
   [10, 20, 30, 40, 50]])

edited Jul 4 at 10:24

Benjamin Loison

5,7514 gold badges20 silver badges37 bronze badges

answered Sep 1, 2020 at 11:47

WurmD

1,4936 gold badges26 silver badges50 bronze badges

1 Comment

lpnorm Over a year ago

Do you know if this is efficient?

Benjamin Loison · Accepted Answer · 2025-07-04 10:33:12Z

4

If you wanna do this just because you cannot concatenate an array with an initialized empty array in a loop, then just use a conditional statement, e.g.

if (i == 0): 
   do the first assignment
else:  
   start your concatenate

edited Jul 4 at 10:33

Benjamin Loison

5,7514 gold badges20 silver badges37 bronze badges

answered Mar 18, 2019 at 17:22

Ming

8467 silver badges8 bronze badges

Comments

YoniChechik · Accepted Answer · 2018-12-25 20:03:46Z

Something that I've build to deal with this sort of problem. It's also deals with list input instead of np.array:

import numpy as np


def cat(tupleOfArrays, axis=0):
    # deals with problems of concating empty arrays
    # also gives better error massages

    # first check that the input is correct
    assert isinstance(tupleOfArrays, tuple), 'first var should be tuple of arrays'

    firstFlag = True
    res = np.array([])

    # run over each element in tuple
    for i in range(len(tupleOfArrays)):
        x = tupleOfArrays[i]
        if len(x) > 0:  # if an empty array\list - skip
            if isinstance(x, list):  # all should be ndarray
                x = np.array(x)
            if x.ndim == 1:  # easier to concat 2d arrays
                x = x.reshape((1, -1))
            if firstFlag:  # for the first non empty array, just swich the empty res array with it
                res = x
                firstFlag = False
            else:  # actual concatination

                # first check that concat dims are good
                if axis == 0:
                    assert res.shape[1] == x.shape[1], "Error concating vertically element index " + str(i) + \
                                                       " with prior elements: given mat shapes are " + \
                                                       str(res.shape) + " & " + str(x.shape)
                else:  # axis == 1:
                    assert res.shape[0] == x.shape[0], "Error concating horizontally element index " + str(i) + \
                                                       " with prior elements: given mat shapes are " + \
                                                       str(res.shape) + " & " + str(x.shape)

                res = np.concatenate((res, x), axis=axis)
    return res


if __name__ == "__main__":
    print(cat((np.array([]), [])))
    print(cat((np.array([1, 2, 3]), np.array([]), [1, 3, 54+1j]), axis=0))
    print(cat((np.array([[1, 2, 3]]).T, np.array([]), np.array([[1, 3, 54+1j]]).T), axis=1))
    print(cat((np.array([[1, 2, 3]]).T, np.array([]), np.array([[3, 54]]).T), axis=1))  # a bad one

EmT · Accepted Answer · 2021-10-14 10:15:17Z

1

A solution is to use the None object and np.concatenate, np.hstack or np.vstack.

>>> arr=None
>>> p=np.array([0,1,2,3])

>>> for i in range(0,2):
>>>     arr = (np.vstack((arr, p)) if (arr is not None) else p)

array([[ 0, 1, 2, 3],
      [[ 0, 1, 2, 3]])

answered Oct 14, 2021 at 10:15

EmT

111 bronze badge

Comments

Benjamin Loison · Accepted Answer · 2025-07-04 10:19:06Z

0

E = np.array([
    
]).reshape(0, 5)
print("E: \n{}\nShape {}\n".format(E, E.shape))

A = np.vstack([
    [1, 2, 3, 4, 5], 
    [10, 20, 30, 40, 50]]
)
print("A:\n{}\nShape {}\n".format(A, A.shape))

C = np.r_[
    E, 
    A
].astype(np.int32)

print("C:\n{}\nShape {}\n".format(C, C.shape))

E: 
[]
Shape (0, 5)

A:
[[ 1  2  3  4  5]
 [10 20 30 40 50]]
Shape (2, 5)

C:
[[ 1  2  3  4  5]
 [10 20 30 40 50]]
Shape (2, 5)

edited Jul 4 at 10:19

Benjamin Loison

5,7514 gold badges20 silver badges37 bronze badges

answered Jan 1, 2021 at 1:25

mon

23k33 gold badges152 silver badges261 bronze badges

Comments

Sturla Molden · Accepted Answer · 2014-03-29 20:26:20Z

-8

np.concatenate, np.hstack and np.vstack will do what you want. Note however that NumPy arrays are not suitable for use as dynamic arrays. Use Python lists for that purpose instead.

answered Mar 29, 2014 at 20:26

Sturla Molden

1,1447 silver badges7 bronze badges

Collectives™ on Stack Overflow

Concatenating empty array in Numpy

7 Answers 7

6 Comments

1 Comment

Comments

Comments

Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

7 Answers 7

6 Comments

1 Comment

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related