Sort 3D array by specific dimension and element

A way to do this is by using the list.sort method or the sorted function together with an appropriate value of the key parameter (see the documentation:howto/sorting).

The Python documentation does a great job explaining the purpose of key parameter:

"Both list.sort() and sorted() have a key parameter to specify a function to be called on each list element prior to making comparisons."

For example, let us sort the first item of your list:

first=[[10, 2], [5, 3], [4, 4]]

def by_first(element):

    """
    Sort a two-dimensional list by the first element 

    Param: element of the list i.e [10, 2]
    Return: first item of element
    """

    return element[0]

So, to sort the above list we do this

sorted(first,key=by_first)

Finally, to solve the initial problem(three-dimensional list) we just have to do the above for each item of your list

list_numbers = [[[10, 2], [5, 3], [4, 4]], [[7, 6], [4, 2], [5, 8]]]

[sorted(entry, key=by_first) for entry in list_numbers]

The code below is a simple workaround to achieve the desired output.

list= [[[10,2],[5,3],[4,4]],[[7,6],[4,2],[5,8]]]

for i in range(len(list)):
    list[:][:][i].sort(key=lambda x: x[:][:][0])

print(list)

Through each iteration of the for loop, the elements are sorted one list at a time.

If you are looking for something more related to arrays (not lists) you can try the following:

import numpy as np

a = np.array([[[10,2],
               [5,3],
               [4,4]],

              [[7,6],
               [4,2],
               [5,8]]])
b = np.zeros(a.shape)

sorted_idx = np.argsort(a, axis=1)

for i in range(a.shape[0]):
    for j in range(a.shape[1]):
        b[i, j] = a[i, sorted_idx[i, j, 0]]

print(b.astype(int))

# [[[ 4  4]
#   [ 5  3]
#   [10  2]]
# 
#  [[ 4  2]
#   [ 5  8]
#   [ 7  6]]]

Note that actually, according to your expected output, you need to sort by the first element of axis=1 not the last axis.