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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]