Python | Convert list of tuple into dictionary

Given a list containing all the element and second list of tuple depicting the relation between indices, the task is to output a dictionary showing the relation of every element from the first list to every other element in the list.
These type of problems are often encountered in Coding competition.
Below are some ways to achieve the above task.

Input:
indices = ['x','y','z','w','t','r']
relation =[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')]
Output:
{'x': ['y', 'z', 'w'], 'y': ['x', 'z', 'w'], 'z': ['x', 'y', 'w'], 
'w': ['x', 'y', 'z'], 't': [], 'r': []}

Method #1: Using iteration is the easiest way to solve any task

#Python code to convert list of tuple into dictionary showing  
#relation of every element from first list to every other element in the list. 
  
#List initialization 
indices = ['x','y','z','w','t','r'] 
relation =[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')] 
  
#dictionary initialization 
Output = {} 
  
#Iteration  
for elem in indices: 
     temp= [] 
     for rel in relation: 
          if elem in rel: 
               if elem == rel[0]: 
                    temp.append(rel[1]) 
               else: 
                    temp.append(rel[0]) 
       
     Output[elem] = temp 
     temp = [] 
  
print("Initial list of tuple is :") 
print(relation) 
print("Converted dictionary of list :") 
print(Output) 

Output :
Initial list of tuple is :
[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')]
Converted dictionary of list :
{'w': ['x', 'y', 'z'], 'r': [], 'x': ['y', 'z', 'w'], 't': [], 
'y': ['x', 'z', 'w'], 'z': ['x', 'y', 'w']}

Method #2: Using networkx is the most simplest and shortest
way to convert list of tuple into dictionary

#Python code to convert list of tuple into dictionary showing  
#relation of every element from first list to every other element in the list. 
  
#Importing 
import networkx as nx  
  
#List initialization 
indices = ['x','y','z','w','t','r'] 
relation =[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')] 
  
#dictionary initialization 
Output = {} 
  
#Using networkx to solve  
temp=nx.Graph(relation) 
temp.add_nodes_from(indices) 
Output = nx.to_dict_of_lists(temp) 
  
#Printing 
print("Initial list of tuple is :") 
print(relation) 
print("Converted dictionary of list :") 
print(Output) 

Output :
Initial list of tuple is :
[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')]
Converted dictionary of list :
{'w': ['x', 'y', 'z'], 'r': [], 'x': ['y', 'z', 'w'], 't': [], 
'y': ['x', 'z', 'w'], 'z': ['x', 'y', 'w']}

Method #3: Using itertools and groupby is another way to convert list of tuple into dictionary.

#Python code to convert list of tuple into dictionary showing  
#relation of every element from first list to every other element in the list. 
  
#Importing 
from itertools import groupby 
from operator import itemgetter 
  
#List initialization 
indices = ['x','y','z','w','t','r'] 
relation =[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')] 
  
#Using itertools.groupby and maps 
edge = relation + [tuple(reversed(pair)) for pair in relation] 
st = itemgetter(0) 
end = itemgetter(1) 
groups = groupby(sorted(edge), key=st) 
mapping = {vertex: list(map(end, edges)) for vertex, edges in groups} 
from collections import defaultdict 
  
#Output list 
Output = defaultdict(list, mapping) 
Output = dict(mapping) 
Output.update({vertex: [] for vertex in indices if vertex not in mapping}) 
  
#Printing 
print("Initial list of tuple is :") 
print(relation) 
print("Converted dictionary of list :") 
print(Output) 

Output :
Initial list of tuple is :
[('x', 'y'), ('x', 'z'), ('x', 'w'), ('y', 'z'), ('y', 'w'), ('z', 'w')]
Converted dictionary of list :
{'w': ['x', 'y', 'z'], 'r': [], 'x': ['y', 'z', 'w'], 't': [], 
'y': ['x', 'z', 'w'], 'z': ['x', 'y', 'w']}

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