Building a dictionary with lists using recursion

Question

I was trying to build a dictionary with recursion for a school project. Right now I think I have the general structure figured out, but I can't figure out how to get the return statement to concatenate pieces of the dictionary together.

I realize this would probably be easier by constructing an empty dictionary then adding to it, but I wanted to see if there were any tricks I could use.

The output I was looking for is:

print(recur_join([1,2,3], ['a', 'b', 'c']))
>>> {1: 'a', 2 : 'b', 3 : 'c'}

I have tried .update() and something of the form dict(basket_one, **basket_two) from another answer. I may have used them wrong. I am using python 3.

Here is my code as of now:

def recur_join(list1, list2):

    if len(list1) == len(list2):

        if len(list1) == 1:
            return {list1[0]: list2[0]}

        elif len(list1) > 1:
            # dict(basket_one, **basket_two)
            #return dict({list1[0]: list2[0]}, **recur_join(list1[1:], 
    list2[1:]))
            return {list1[0]: list2[0]}.update(recur_join(list1[1:], list2[1:]))

    else:
        print('lists do not match in size')
        return 0

Any help would be appreciated, sorry if this was answered before. Thanks!

Primusa · Accepted Answer · 2018-04-12 01:04:17Z

2

I suggest you don't use recursion and use dict comprehensions instead:

def recur_join(list1, list2):
    if len(list1) != len(list2):
        print('lists do not match in size')
        return
    else: return {list1[i]:list2[i] for i in range(len(list1))}

For the recursive route (warning: very ugly):

def recur_join(list1, list2):
    if len(list1) != len(list2):
        print('lists do not match in size')
        return
    elif list1 == [] and list2 == []:
        return {}
    else:
        return dict(list({list1[0]: list2[0]}.items()) + list(recur_join(list1[1:], list2[1:]).items()))

answered Apr 12, 2018 at 1:04

Primusa

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1 Comment

Connorr.0 Over a year ago

Thanks, I was using recursion as more of a puzzle and because I had to write a program with recursion for a class, Thanks though. I will keep that dict comprehension in mind!

138 · Accepted Answer · 2018-04-12 01:11:46Z

1

"Cleanish" recursive solution. I would personally use Primusa's dictionary comprehension solution.

def recur_join(list1, list2):
    cur = {}
    if len(list1) == len(list2) and len(list1) > 0:
        cur = {list1[0]: list2[0]}
        if len(list1) > 1:
            cur.update(recur_join(list1[1:], list2[1:]))
    else:
        print('lists do not match in size')
    return cur

answered Apr 12, 2018 at 1:11

138

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2 Comments

Connorr.0 Over a year ago

I picked yours because I was looking specifically for the recursive option, the dict comprehension looks far cleaner, but I wanted to see how to do it with recursion for fun (and my class.)

138 Over a year ago

If you are interested in developing a knack for recursive solutions I highly recommend picking up Scheme (or Lisp), it is an interestingly fun language!

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