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

Asked By: maxv15

||

Source

Answer 1

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]])

Answered By: behzad.nouri

Answer 2

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 By: Sturla Molden

Answer 3

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 arraylist - 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

Answered By: YoniChechik

Answer 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 contactenate

Answered By: Ming

Answer 5

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]])

Answered By: WurmD

Answer 6

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)

Answered By: mon

Answer 7

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 By: EmT

Concatenating empty array in Numpy

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