How to Sort the Odd in Python

The challenge

You will be given an array of numbers. You have to sort the odd numbers in ascending order while leaving the even numbers at their original positions.

Examples

[7, 1]  =>  [1, 7]
[5, 8, 6, 3, 4]  =>  [3, 8, 6, 5, 4]
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]  =>  [1, 8, 3, 6, 5, 4, 7, 2, 9, 0]

The solution in Python code

Option 1:

def sort_array(arr):
  odds = sorted((x for x in arr if x%2 != 0), reverse=True)
  return [x if x%2==0 else odds.pop() for x in arr]

Option 2:

def sort_array(source_array):
    result = sorted([l for l in source_array if l % 2 == 1])
    for index, item in enumerate(source_array):
        if item % 2 == 0:
            result.insert(index, item)
    return result

Option 3:

def sort_array(source_array):
    odd = sorted(list(filter(lambda x: x % 2, source_array)))
    l, c = [], 0
    for i in source_array:
        if i in odd:
            l.append(odd[c])
            c += 1
        else:
            l.append(i)    
    return l

Test cases to validate our solution

test.assert_equals(sort_array([5, 3, 2, 8, 1, 4]), [1, 3, 2, 8, 5, 4])
test.assert_equals(sort_array([5, 3, 1, 8, 0]), [1, 3, 5, 8, 0])
test.assert_equals(sort_array([]),[])