python - sorting numbers using recursion -

the purpous of using recursion instead of using for in range(11): because advantagous start top trying solve specific mathematical problem. function changed returns [n]that matches criteria.

print(numbers)=[[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]] why there brackets? print(numbers[7])=indexerror: list index out of range has brackets?

# function supposed sorting numbers in list def sorting_numbers(n):     if n > 1:         return [n] + sorting_numbers(n-1)     else:         return [1]  numbers = [] n = 10 numbers = (sorting_numbers(n)) print(numbers) 

print(numbers)=[[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]] why there brackets? print(numbers[7])=indexerror: list index out of range has brackets?

well if really want use recursion such simple problem - warned it's terribly inefficient , cause maximum recursion depth exceeded error values of n close 1000 (python not designed handle recursion efficiently):

def storing_numbers(n):     if n > 1:         return [n] + storing_numbers(n-1)     else:         return [1] 

or bit shorter:

def storing_numbers(n):     return [] if n <= 0 else [n] + storing_numbers(n-1) 

notice how build list along way, , base case returns list. use this:

numbers = storing_numbers(10) numbers => [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] 

of course, it'd unrealistic use recursive function building list beyond academic exercise. real-life, practical implementation instead:

list(range(10, 0, -1)) => [10, 9, 8, 7, 6, 5, 4, 3, 2, 1] 

and here's how you'd use list comprehension filtering numbers matching criteria, let's say, numbers:

[x x in range(10, 0, -1) if x % 2 == 0] => [10, 8, 6, 4, 2]