Create 2D numpy array with append function

Question

I want to create a 2D numpy array of dimensions Nx2, in python. I want to create this by using 2 for loops. I can easily build this array in Matlab with the following code

matrix = [];

for i = 1:3
    for j = 1:4
        temp = [i, j];
        matrix = [matrix; temp];
    end
end

I have tried already numerous times, but failed. Usually, the error that I get is related to the sizes of arrays not matching when I run the for loops.

The output of the code is

np.zeros((100,100)) create a 2-D array filled with zeros. It would be clear if you add expected output to the question. — Ch3steR
– Ch3steR, Commented Mar 23, 2020 at 17:12
You are right, I am going to add the expected output. What you mentioned might be the first step. Thank you — Arman Mojaver
– Arman Mojaver, Commented Mar 23, 2020 at 17:19
Remember in MATLAB everything is 2d. The [] is 0x0; [[]; [1,2]] joins that with a 1x2,expanding shapes as needed. In numpy arrays may be 0d,1d etc. np.concatenate is picky about matching dimensions. np.append is a poorly conceived front end to concatenate. — hpaulj
– hpaulj, Commented Mar 23, 2020 at 19:07

B-L · Accepted Answer · 2020-03-23 21:14:41Z

1

Here is a function that does what you want and uses two for loops. The reason that N is +1 and M is +1 is because in python the range() function returns a sequence of numbers, starting from 0 by default, and increments by 1 (by default), and ends at a specified number. So since we are starting at 1 instead of 0 we need to add 1 so that the actual range is correct. ie. 0,4 is a range of 4, but 1,4 is only really a range of 3. Hope that helps! -Brack

import numpy as np


def Nx2(N, M):
    matrix=[]
    n = N+ 1
    m= M + 1
    for i in range(1,n):
        for j in range(1,m):
            temp=[i,j]
            matrix.append(temp)
    final_matrix = np.array(matrix)
    return final_matrix
print(Nx2(3, 4))

edited Mar 23, 2020 at 21:14

answered Mar 23, 2020 at 18:38

B-L

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7 Comments

Mad Physicist Over a year ago

You're setting a global matrix outside the function (bad), but then returning it (good).

Mad Physicist Over a year ago

Also, this is not materially different from the answer posted by SurajSubramanian a quarter hour earlier.

B-L Over a year ago

In this instance, is it better to put the matrix=[] inside of the function? (always learning)

Mad Physicist Over a year ago

Absolutely. It's simply wrong not to. Try running your function twice.

Mad Physicist Over a year ago

It's nice to have those moments. By the way, don't forget to fix your indentation.

|

Suraj · Accepted Answer · 2020-03-23 18:22:35Z

1

matrix= []

for i in range(1,4):
    for j in range(1,5):
        temp= [i,j]
        matrix.append(temp)

Finally you can convert the list of lists into a numpy array using matrix = np.array(matrix)

answered Mar 23, 2020 at 18:22

Suraj

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Comments

Mad Physicist · Accepted Answer · 2020-03-23 18:33:06Z

0

You really don't need a loop in either MATLAB or numpy for this. In both languages, explicit loops tend to be slow compared to the under-the-hood loops that run when you vectorize code. The additional problem with appending inside a loop is that each operation re-allocates the entire array to add the new element. You want to avoid appending and pre-allocate as much as possible.

Both packages have a meshgrid function. In numpy you could do:

ii = np.arange(3) + 1
jj = np.arange(4) + 1
j, i = np.meshgrid(jj, ii)

matrix = np.stack((i.ravel(), j.ravel()), axis=1)

In MATLAB, it would be very similar:

ii = 1:3;
jj = 1:4;
[i, j] = meshgrid(ii, jj);

matrix = [i(:) j(:)];

If you absolutely need to do this in a loop with constant reallocation, the MATLAB loop translates to numpy very easily. Instead of doing implicit vertcat, use np.append, or np.concatenate:

matrix = np.array((0, 2));

for i in range(1, 4)
    for j in range(1, 5)
        temp = [[i, j]]
        matrix = np.append(matrix, temp, axis=0)

edited Mar 23, 2020 at 18:33

answered Mar 23, 2020 at 18:22

Mad Physicist

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1 Comment

Arman Mojaver Over a year ago

I understand that with 'for' loops it is inefficient. I understand that I might be creating a bad habit, but the data dimensions that I am managing is not large, so it does not matter. The last piece of code you have put is what I have been trying to do. I get the following error, as I have so far. ValueError: all the input arrays must have same number of dimensions, but the array at index 0 has 1 dimension(s) and the array at index 1 has 2 dimension(s)

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