Avoid emptying dictionary recursively-CodePudding

I am trying to create a dictionary from a list recursively and my code only works when there is only one item in the list. It fails for multiple items and I suspect that this is because the dictionary is being recreated through each instance of the recursion instead of adding to it after the first instance. How can I avoid doing this so that the whole list is converted to a dictionary?

Note: the list is a list of tuples containing two items.

def poncePlanner(restaurantChoices):
    if len(restaurantChoices) == 0:
        return {}
    else:
        name, resto = restaurantChoices[0][0], restaurantChoices[0][1]
        try:
            dic[name] = resto
            poncePlanner(restaurantChoices[1:])
            return dic
        except:
            dic = {name: resto}
            poncePlanner(restaurantChoices[1:])
            return dic

Intended input and output:

>>> restaurantChoice = [("Paige", "Dancing Goats"), ("Fareeda", "Botiwala"),
                  ("Ramya", "Minero"), ("Jane", "Pancake Social")]
>>> poncePlanner(restaurantChoice)
{'Jane': 'Pancake Social',
 'Ramya': 'Minero',
 'Fareeda': 'Botiwala',
 'Paige': 'Dancing Goats'}

CodePudding user response：

You have the edge condition, so you need to define what to do when you have more than one. Here you just take the first tuple, make a dict, and then add the results of recursion into that dict:

restaurantChoice = [("Paige", "Dancing Goats"), ("Fareeda", "Botiwala"),
                    ("Ramya", "Minero"), ("Jane", "Pancake Social")]

def poncePlanner(restaurantChoice):
    if not restaurantChoice:
        return {}
    head, *rest = restaurantChoice
    return {head[0]: head[1], **poncePlanner(rest)}

poncePlanner(restaurantChoice)

Returning:

{'Jane': 'Pancake Social',
 'Ramya': 'Minero',
 'Fareeda': 'Botiwala',
 'Paige': 'Dancing Goats'}

CodePudding user response：

Since restaurantChoices are already (key,value) pairs, you can simply use the built-in function dict to create the dictionary:

def poncePlanner(restaurantChoices):
    return dict(restaurantChoices)

Without built-in functions, you can also use a simple for-loop to return the desired transformation:

def poncePlanner(restaurantChoices):
    result = {}
    for key,value in restaurantChoices:
            result[key] = value
    return result

If you really want recursion, I would do something like this, but it doesn't make sense because the lists are not nested:

def poncePlanner(restaurantChoices):
    def recursion(i, result):
        if i<len(restaurantChoices):
            key, value  = restaurantChoices[i]
            result[key] = value
            recursion(i 1, result)
        return result
    return recursion(0,{})

This recursive function has O(1) time/space complexity per call, so it runs with optimal efficiency.

The main problem with the original code is that the dictionary dic is not passed to the deeper recursion calls, so the new contents are never added to the final dictionary. (They contents added to new dictionaries and forgotten).