Defining empty numpy array when we do not know the size

Question

From my understanding, when we want to define a numpy array, we have to define its size.

However, in my case, I want to define a numpy array and then extend it based on my values in the for loop. The shape of values might differ in each run. So I cannot define the numpy array shape in advance.

Is there any way to overcome this?

I would like to avoid using lists.

Thanks

NumPy arrays cannot be created without size. You can start with a small (1-element?) array and then add items to it. — DYZ
– DYZ, Commented Oct 19, 2021 at 5:01
Sorry that's not how we do things in the numpy. Lists are for growing. Arrays have a fixed size. — hpaulj
– hpaulj, Commented Oct 19, 2021 at 5:25
If you know the range for the for loop, you know the target array size. — hpaulj
– hpaulj, Commented Oct 19, 2021 at 5:38
Why do you want to avoid lists? Growing a list with append is a lot faster than growing an array by making a new one, and copying data from the old. You apparently have worked with MATLAB. There you can grow a matrix by indexing M[i,j]=0, but I suspect that's doing a whole lot of work in the background. It's not something I routinely did when I worked with MATLAB years ago. — hpaulj
– hpaulj, Commented Oct 19, 2021 at 7:03

Dadá · Accepted Answer · 2021-10-19 05:19:35Z

1

import numpy as np  
myArrayShape = 2
myArray = np.empty(shape=2)

Note that this generates random values for each element in the array.

answered Oct 19, 2021 at 5:19

Dadá

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hpaulj Over a year ago

But empty here does not mean the same thing as the emlty list, []

TaekYoung · Accepted Answer · 2021-10-19 05:06:08Z

1

I think numpy array is just like array in clang or c++, I mean when you make numpy array you allocate memory depend on your request(size and dtype). So it is better to make array after the size of array is determinated.

Or you can try numpy.append https://numpy.org/doc/stable/reference/generated/numpy.append.html But I don't think it is preferable way because it keeps generate new arrays.

answered Oct 19, 2021 at 5:06

TaekYoung

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hpaulj Over a year ago

I hate np.append with a passion. It is not a list append clone.

hpaulj · Accepted Answer · 2021-10-19 07:19:10Z

From the Octave (free-MATLAB) docs, https://octave.org/doc/v6.3.0/Advanced-Indexing.html

In cases where a loop cannot be avoided, or a number of values must be combined to form a larger matrix, it is generally faster to set the size of the matrix first (pre-allocate storage), and then insert elements using indexing commands. For example, given a matrix a,

[nr, nc] = size (a);
x = zeros (nr, n * nc);
for i = 1:n
  x(:,(i-1)*nc+1:i*nc) = a;
endfor
is considerably faster than

x = a;
for i = 1:n-1
  x = [x, a];
endfor
because Octave does not have to repeatedly resize the intermediate result.

The same idea applies in numpy. While you can start with a (0,n) shaped array, and grow by concatenating (1,n) arrays, that is a lot slower than starting with a (m,n) array, and assigning values.

There's a deleted answer that illustrates how to create an array by list append. That is highly recommended.

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