Covert nested list in dictionaries(every element in list had to be a key in dictionary)

Question

I have trouble converting the nested list into dictionaries.

For example nested list with [[5, 0], [6, 0], [7, 1, 0], [8, 1, 0], [9, 1]] . I hope to get a result of

{5:[0], 6:[0], 0:[5,6,7,8,1], 1:[7,8,9], 7:[1,0], 8:[1,0], 9:[1]}

I have tried method such as using nested for loop and list comprehension but can't achieve the above solution.

Is there any way that I can achieve such results?

Answer 1

Imo, you're looking for permutations which are implemented in itertools :

from collections import defaultdict
from itertools import permutations

lst = [[5, 0], [6, 0], [7, 1, 0], [8, 1, 0], [9, 1]]

dd = defaultdict(set)

for sublst in lst:
    for key, value in permutations(sublst, 2):
        dd[key].add(value)

print(dd)

This yields

defaultdict(<class 'set'>, {5: {0}, 0: {1, 5, 6, 7, 8}, 6: {0}, 7: {0, 1},
                            1: {0, 8, 9, 7}, 8: {0, 1}, 9: {1}})

Answer 2

I managed to make this but the complexity of the algorithm is O(n^3). May be there is a way to do it with a lower complexity but I didn't find how to.

mylist = [[5, 0], [6, 0], [7, 1, 0], [8, 1, 0], [9, 1]]
dictionnary = {}
for l in mylist:
    # For each list, select one element and iterate through all  
    # the list's elements in order to populate the dict.
    for currentElem in l:
        for otherElem in l:
            if currentElem != otherElem:
                # create the key and empty list if not exists
                if currentElem not in dictionnary.keys():
                    dictionnary[currentElem] = []
                # add value to key only if not already there
                if otherElem not in dictionnary[currentElem]:
                    dictionnary[currentElem].append(otherElem)

print(dictionnary)

Generated output:

{5: [0], 0: [5, 6, 7, 1, 8], 6: [0], 7: [1, 0], 1: [7, 0, 8, 9], 8: [1, 0], 9: [1]}

This is pretty much what you want except for the key 1 . Either it's a typo in your post or I didn't managed to do what you want.

Answer 3

My algorithm produces sorted output, both dict keys and list-values are sorted.

Algorithm complexity is O(N * M) where N is total count of numbers and M is the average sub-list length.

Try it online!

def to_dict(l):
    d = {}
    for e0 in l:
        s = set(e0)
        for e1 in s:
            if e1 not in d:
                d[e1] = set()
            d[e1].update(s - {e1})
    return dict(sorted([(k, sorted(v)) for k, v in d.items()]))
    
print(to_dict([[5, 0], [6, 0], [7, 1, 0], [8, 1, 0], [9, 1]]))

Output:

{0: [1, 5, 6, 7, 8], 1: [0, 7, 8, 9], 5: [0], 6: [0], 7: [0, 1], 8: [0, 1], 9: [1]}

Answer 4

arr = [[5, 0], [6, 0], [7, 1, 0], [8, 1, 0], [9, 1]]
my_dict = {}
for i in arr:
  for j in range(len(i)):
    if i[j] in my_dict.keys():
      temp = set(i[:j] + i[j+1:])
      temp_2 = set(my_dict[i[j]])
      new_temp = temp - temp_2
      my_dict[i[j]] += new_temp
    else:
      my_dict[i[j]] = i[:j] + i[j+1:]
    
print(my_dict)