将列表列表转换为具有多个键值的字典

Question

I need to write a function that accepts a list of lists representing friends for each person and need to convert it into a dictionary. so an input of [['A','B'],['A','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']] should return {A:[B,C,D],B:[A],C:[B,D],D:[B],E:None}

Input:

[['A','B'],['A','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']]

Expected Output:

{A:[B,C,D],B:[A],C:[B,D],D:[B],E:None}

Currently I am trying the following:

s=[['A','B'],['A','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']]

output=dict.fromkeys((set([x[0] for x in s])),[ ])

for x in s:
    if len(x)>1:
        output[x[0]].append(x[1])
    else:
        output[x[0]].append(None)

But the output is giving me all values for every key rather than returning only the corresponding values

The output i am getting is:

{
'A': ['B', 'C', 'D', 'A', 'B', 'D', 'B', None],

 'B': ['B', 'C', 'D', 'A', 'B', 'D', 'B', None],

 'C': ['B', 'C', 'D', 'A', 'B', 'D', 'B', None],

 'D': ['B', 'C', 'D', 'A', 'B', 'D', 'B', None],

 'E': ['B', 'C', 'D', 'A', 'B', 'D', 'B', None]
}

Answer 1

You can iterate through the key-value pairs in the list of lists, but unpack the value as a list to accommodate the possible lack of a value:

s = [['A','B'],['A','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']]
output = {}
for k, *v in s:
    if v:
        output.setdefault(k, []).extend(v)
    else:
        output[k] = None

output becomes:

{'A': ['B', 'C', 'D'], 'B': ['A'], 'C': ['B', 'D'], 'D': ['B'], 'E': None}

Or if you don't mind that keys without a value get an empty list instead of None , you can simply do:

output = {}
for k, *v in s:
    output.setdefault(k, []).extend(v)

output would then become:

{'A': ['B', 'C', 'D'], 'B': ['A'], 'C': ['B', 'D'], 'D': ['B'], 'E': []}

Answer 2

You should use a common dict comprehension to initiate the dict :

output = {x[0]: [] for x in s}

dict.fromkeys gives all keys the identical referential value. With a mutable value that is a problem. The comprehension will give each key an independent list object, in addition to being more readable.

Answer 3

The issue is the list you feed to dict.keys is only one reference across keys.

Your desired result is inconsistent. I recommend you choose an empty list for 'E' , however much it seems None is more appropriate. With this adjusted requirement, you can use collections.defaultdict .

from collections import defaultdict

L = [['A','B'],['E','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']]

dd = defaultdict(list)

for lst in L:
    if len(lst) > 1:
        dd[lst[0]].append(lst[1])
    else:
        dd[lst[0]]

print(dd)

defaultdict(list,
            {'A': ['B', 'C', 'D'],
             'B': ['A'],
             'C': ['B', 'D'],
             'D': ['B'],
             'E': []})

Answer 4

One of the way to solve this is as given below:

friend_combi = [['A','B'],['A','C'],['A','D'],['B','A'],['C','B'],['C','D'],['D','B'],['E']]  # Input to be processed

final_dict = {} #Empty dict to store result
for i in friend_combi: # loop through each element in list
    if final_dict.get(i[0]):  #if data present in dict then append else add
        final_dict[i[0]].append(i[1])
    else:
        final_dict[i[0]] = [i[1]] if i[1:] else None #check if value exist in list else save None
print (final_dict)
#Output --> {'A': ['B', 'C', 'D'], 'B': ['A'], 'C': ['B', 'D'], 'D': ['B'], 'E': None}

I hope this helps :)

Answer 5

You can define a function named get_dictionary() as the below code shows.

>>> def get_dictionary(l):
...     d = {}
...     for arr in l:
...         if len(arr) == 2:
...             key = arr[0]
...             if key in d:
...                 d[key].append(arr[1])
...             else:
...                 d[key] = [arr[1]]
...         else:
...             d[key] = None
...     return d
...
>>> l = [['A','B'], ['A','C'], ['A','D'], ['B','A'], ['C','B'], ['C','D'], ['D','B'], ['E']]
>>>
>>> get_dictionary(l)
{'A': ['B', 'C', 'D'], 'B': ['A'], 'C': ['B', 'D'], 'D': None}
>>>

Pretty printing the dictionary as JSON.

>>> import json
>>>
>>> d = get_dictionary(l)
>>>
>>> print(json.dumps(d, indent=4))
{
    "A": [
        "B",
        "C",
        "D"
    ],
    "B": [
        "A"
    ],
    "C": [
        "B",
        "D"
    ],
    "D": null
}
>>>