Unsuccessful append to an empty NumPy array


I am trying to fill an empty(not np.empty!) array with values using append but I am gettin error:

My code is as follows:

import numpy as np

And I am getting:

ValueError: could not broadcast input array from shape (2) into shape (0)
Asked By: Cupitor



numpy.append always copies the array before appending the new values. Your code is equivalent to the following:

import numpy as np
result = np.zeros((2,0))
new_result = np.append([result[0]],[1,2])
result[0] = new_result # ERROR: has shape (2,0), new_result has shape (2,)

Perhaps you mean to do this?

import numpy as np
result = np.zeros((2,0))
result = np.append([result[0]],[1,2])
Answered By: Stuart Berg

numpy.append is pretty different from list.append in python. I know that’s thrown off a few programers new to numpy. numpy.append is more like concatenate, it makes a new array and fills it with the values from the old array and the new value(s) to be appended. For example:

import numpy

old = numpy.array([1, 2, 3, 4])
new = numpy.append(old, 5)
print old
# [1, 2, 3, 4]
print new
# [1, 2, 3, 4, 5]
new = numpy.append(new, [6, 7])
print new
# [1, 2, 3, 4, 5, 6, 7]

I think you might be able to achieve your goal by doing something like:

result = numpy.zeros((10,))
result[0:2] = [1, 2]

# Or
result = numpy.zeros((10, 2))
result[0, :] = [1, 2]


If you need to create a numpy array using loop, and you don’t know ahead of time what the final size of the array will be, you can do something like:

import numpy as np

a = np.array([0., 1.])
b = np.array([2., 3.])

temp = []
while True:
    rnd = random.randint(0, 100)
    if rnd > 50:
    if rnd == 0:

 result = np.array(temp)

In my example result will be an (N, 2) array, where N is the number of times the loop ran, but obviously you can adjust it to your needs.

new update

The error you’re seeing has nothing to do with types, it has to do with the shape of the numpy arrays you’re trying to concatenate. If you do np.append(a, b) the shapes of a and b need to match. If you append an (2, n) and (n,) you’ll get a (3, n) array. Your code is trying to append a (1, 0) to a (2,). Those shapes don’t match so you get an error.

Answered By: Bi Rico

This error arise from the fact that you are trying to define an object of shape (0,) as an object of shape (2,). If you append what you want without forcing it to be equal to result[0] there is no any issue:

b = np.append([result[0]], [1,2])

But when you define result[0] = b you are equating objects of different shapes, and you can not do this. What are you trying to do?

Answered By: Alejandro

Here’s the result of running your code in Ipython. Note that result is a (2,0) array, 2 rows, 0 columns, 0 elements. The append produces a (2,) array. result[0] is (0,) array. Your error message has to do with trying to assign that 2 item array into a size 0 slot. Since result is dtype=float64, only scalars can be assigned to its elements.

In [65]: result=np.asarray([np.asarray([]),np.asarray([])])

In [66]: result
Out[66]: array([], shape=(2, 0), dtype=float64)

In [67]: result[0]
Out[67]: array([], dtype=float64)

In [68]: np.append(result[0],[1,2])
Out[68]: array([ 1.,  2.])

np.array is not a Python list. All elements of an array are the same type (as specified by the dtype). Notice also that result is not an array of arrays.

Result could also have been built as

ll = [[],[]]
result = np.array(ll)


ll[0] = [1,2]
# ll = [[1,2],[]]

the same is not true for result.

np.zeros((2,0)) also produces your result.

Actually there’s another quirk to result.

result[0] = 1

does not change the values of result. It accepts the assignment, but since it has 0 columns, there is no place to put the 1. This assignment would work in result was created as np.zeros((2,1)). But that still can’t accept a list.

But if result has 2 columns, then you can assign a 2 element list to one of its rows.

result = np.zeros((2,2))
result[0] # == [0,0]
result[0] = [1,2]

What exactly do you want result to look like after the append operation?

Answered By: hpaulj

SO thread ‘Multiply two arrays element wise, where one of the arrays has arrays as elements‘ has an example of constructing an array from arrays. If the subarrays are the same size, numpy makes a 2d array. But if they differ in length, it makes an array with dtype=object, and the subarrays retain their identity.

Following that, you could do something like this:

In [5]: result=np.array([np.zeros((1)),np.zeros((2))])

In [6]: result
Out[6]: array([array([ 0.]), array([ 0.,  0.])], dtype=object)

In [7]: np.append([result[0]],[1,2])
Out[7]: array([ 0.,  1.,  2.])

In [8]: result[0]
Out[8]: array([ 0.])

In [9]: result[0]=np.append([result[0]],[1,2])

In [10]: result
Out[10]: array([array([ 0.,  1.,  2.]), array([ 0.,  0.])], dtype=object)

However, I don’t offhand see what advantages this has over a pure Python list or lists. It does not work like a 2d array. For example I have to use result[0][1], not result[0,1]. If the subarrays are all the same length, I have to use np.array(result.tolist()) to produce a 2d array.

Answered By: hpaulj

I might understand the question incorrectly, but if you want to declare an array of a certain shape but with nothing inside, the following might be helpful:

Initialise empty array:

>>> a = np.zeros((0,3)) #or np.empty((0,3)) or np.array([]).reshape(0,3)
>>> a
array([], shape=(0, 3), dtype=float64)

Now you can use this array to append rows of similar shape to it. Remember that a numpy array is immutable, so a new array is created for each iteration:

>>> for i in range(3):
...     a = np.vstack([a, [i,i,i]])
>>> a
array([[ 0.,  0.,  0.],
       [ 1.,  1.,  1.],
       [ 2.,  2.,  2.]])

np.vstack and np.hstack is the most common method for combining numpy arrays, but coming from Matlab I prefer np.r_ and np.c_:

Concatenate 1d:

>>> a = np.zeros(0)
>>> for i in range(3):
...     a = np.r_[a, [i, i, i]]
>>> a
array([ 0.,  0.,  0.,  1.,  1.,  1.,  2.,  2.,  2.])

Concatenate rows:

>>> a = np.zeros((0,3))
>>> for i in range(3):
...     a = np.r_[a, [[i,i,i]]]
>>> a
array([[ 0.,  0.,  0.],
       [ 1.,  1.,  1.],
       [ 2.,  2.,  2.]])

Concatenate columns:

>>> a = np.zeros((3,0))
>>> for i in range(3):
...     a = np.c_[a, [[i],[i],[i]]]
>>> a
array([[ 0.,  1.,  2.],
       [ 0.,  1.,  2.],
       [ 0.,  1.,  2.]])
Answered By: Simon Streicher
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