Vfrdtyky

I am asked to define a function that takes in a list and returns another list, all that while using recursion. However when I run the else command and I print the lst_ , the output shows that on every run the list contains a single element, instead of having the doubles added one by one.Also I try not to use append() Thoughts?

def double(lst, lst_ = ):

    """

    parameters : lst of type list;

    returns : another list with lst's elements doubled

    """

    if len(lst) == 0:

        return lst_

    else:

        lst[0] = int(lst[0]) + int(lst[0])

        lst_ = lst_ + lst[0:1]            

        print(lst_)

        return double(lst[1:])



print(double([1,2,3,4,5,6,7,8]))

This is the output

[2]

[4]

[6]

[8]

[10]

[12]

[14]

[16]

If the idea is to return a copy without modifying the original, I would not recommend using the mutable default argument.

Instead,

def double(lst):

    if not lst:

        return 

    return [2*lst[0], *double(lst[1:])] # [2*lst[0]] + double(lst[1:]) 

The recursive case must return a fresh list, and the base case will check for, and return an empty list.

lst1 = double([1,2,3,4,5,6,7,8]) 

print(lst1)

[2, 4, 6, 8, 10, 12, 14, 16]

If you want to have a little fun, you can attempt a generator-based recursive solution using yield from (generator delegation):

def double(lst):

    if lst:

        yield 2*lst[0]

        yield from double(lst[1:])



lst = list(double([1,2,3,4,5,6,7,8]) )

print(lst)

[2, 4, 6, 8, 10, 12, 14, 16]

Use .append() on list to add elements at the end:

def double(lst, lst_ = ):

    """

    parameters : lst of type list;

    returns : another list with lst's elements doubled

    """

    if len(lst) == 0:

        return lst_

    else:

        lst[0] += lst[0]

        lst_.append(lst[0])          

        return double(lst[1:])



print(double([1,2,3,4,5,6,7,8]))

# [2, 4, 6, 8, 10, 12, 14, 16]

Also, note that this line lst[0] = int(lst[0]) + int(lst[0]) in you code can be shortened to lst[0] += lst[0], because you are dealing with integers only and that explicit casting is redundant.

If you don't want to use append(). Then you can use this solution:

def double(lst, lst_ = ):

    if not lst:

        return lst_

    else:

        return [lst[0] * 2 , *double(lst[1:])]



print(double([1,2,3,4,5,6,7,8]))

The output will be : [2, 4, 6, 8, 10, 12, 14, 16]

Just in case you are wondering with the *double(lst[1:]) call: * is used for unpacking argument list. Read more here.
If you call without the * you will get an output like:

[2, [4, [6, [8, [10, [12, [14, [16, ]]]]]]]]

Another easy solution will be:

def double(lst, lst_ = ):

    if not lst:

        return lst_

    else:

        lst[0] = lst[0] * 2

        lst_ = lst_.append(lst[0])

        return double(lst[1:])



print(double([1,2,3,4,5,6,7,8]))

Try this way:

def double(lst, lst_ = ):

  if len(lst) == 0:

    return lst_

  else:

    lst[0] = int(lst[0]) + int(lst[0])

    lst_.extend(lst[:1])

    return double(lst[1:])



print(double([1,2,3,4,5,6,7,8]))



#=> [2, 4, 6, 8, 10, 12, 14, 16]