The problem

You're appending the same array a to your list0 4 times. Arrays like a are mutable objects, which means, among other things, that when you assign values to them the underlying object changes. Since the array is present in your list 4 times, those changes (seem to) show up in 4 different places.

Solution

You can fix the code you have with one small change. Append a copy of the array to your list, instead of the array itself:

import numpy as npa = np.empty((3), int)list0 = []for idx in range(4):        for i in range(3):        a[i] = idx*10 + i    print("idx =",idx,"; a =",a)    list0.append(a.copy())print("list0 =",list0)

Output:

idx = 0 ; a = [0 1 2]idx = 1 ; a = [10 11 12]idx = 2 ; a = [20 21 22]idx = 3 ; a = [30 31 32]list0 = [array([0, 1, 2]), array([10, 11, 12]), array([20, 21, 22]), array([30, 31, 32])]

Optimized solution

Python/Numpy offer many better ways (both in terms of using fewer lines of code and running faster) to initialize arrays. For a bunch of ranges like this, here is a reasonable approach:

list0 = [np.arange(n*10, n*10+3) for n in range(4)]print(list0)

Output:

[array([0, 1, 2]), array([10, 11, 12]), array([20, 21, 22]), array([30, 31, 32])]

You might also consider just using a single 2D array in place of a list of arrays. One single array is often easier to work with than a heterogenous mix of arrays in a list. Here's how you do that:

arr0 = np.array([np.arange(n*10, n*10+3) for n in range(4)])print(arr0)

Output:

[[ 0  1  2] [10 11 12] [20 21 22] [30 31 32]]

Just do this:

list_to_append.append(np_array.copy())

In a nutshell, numpy arrays or lists are mutable objects, which means that you when you assign a numpy array or list to a variable, what you are really assigning are references to memory locations aka pointers.

In your case, "a" is a pointer, so what you are really doing is appending to list0 an address to the memory location pointed by "a", and not the actual values pointed by the pointer. Thus it means that each new position of "list0", after appending, turns out to be the same address of memory: "a".

So, instead of:

list0.append(a)

You call the copy() method of "a" that creates a new memory location for the new values of "a" and returns it:

list0.append(a.copy())