Assist with dict comprehension

Question

I am trying to write a dictionary comprehension.

I have a dictionary like this one:

main_dict = {
    'A' : {'key1' : 'valueA1', 'key2' : 'valueA2'},
    'B' : {'key2' : 'valueB2', 'key3' : 'valueB3'},
    'C' : {'key3' : 'valueC3', 'key1' : 'valueC1'}}

I want to perform the following logic:

d = {}
for k_outer, v_outer in main_dict.items():
    for k_inner, v_inner in v_outer.items():
        if k_inner in d.keys():
            d[k_inner].append([k_outer, v_inner])
        else:
            d[k_inner] = [[k_outer, v_inner]]

Which yields the following result:

{'key3': [['C', 'valueC3'], ['B', 'valueB3']], 
 'key2': [['A', 'valueA2'], ['B', 'valueB2']], 
 'key1': [['A', 'valueA1'], ['C', 'valueC1']]}

(I know I could use defaultdict(list) but this is just an example)

I want to perform the logic using a dict-comprehension, so far I have the following:

d = {k : [m, v] for m, x in main_dict.items() for k, v in x.items()}

This does not work, it only gives me the following output:

{'key3' : ['B', 'valueB3'],
'key2' : ['B', 'valueB2'],
'key1' : ['C', 'valueC1']}

Which is the last instance found for each inner_key...

I am at a loss of how to perform this nested list-comprehension correctly. I have tried multiple variations, all worse than the last.

Answer 1

in the end, this is what i used:

from collections import defaultdict

d = defaultdict(list)
for m, x in main_dict.items():
    for k, v in x.items():
        d[k].append((m, v))

Answer 2

You can try something like this:

In [61]: main_dict
Out[61]: 
{'A': {'key1': 'valueA1', 'key2': 'valueA2'},
 'B': {'key2': 'valueB2', 'key3': 'valueB3'},
 'C': {'key1': 'valueC1', 'key3': 'valueC3'}}

In [62]: keys=set(chain(*[x for x in main_dict.values()]))

In [64]: keys
Out[64]: set(['key3', 'key2', 'key1'])

In [63]: {x:[[y,main_dict[y][x]] for y in main_dict if x in main_dict[y]] for x in keys}
Out[63]: 
{'key1': [['A', 'valueA1'], ['C', 'valueC1']],
 'key2': [['A', 'valueA2'], ['B', 'valueB2']],
 'key3': [['C', 'valueC3'], ['B', 'valueB3']]}

A more readable solution using dict.setdefault:

In [81]: d={}

In [82]: for x in keys:
    for y in main_dict:
       if x in main_dict[y]:
           d.setdefault(x,[]).append([y,main_dict[y][x]])
   ....:            

In [83]: d
Out[83]: 
{'key1': [['A', 'valueA1'], ['C', 'valueC1']],
 'key2': [['A', 'valueA2'], ['B', 'valueB2']],
 'key3': [['C', 'valueC3'], ['B', 'valueB3']]}

Answer 3

Using three dictionary comprehensions to achieve such task, the third dict-comprehension is to combine the first two dicts:

e = {k : [m, v] for m, x in main_dict.items() for k, v in x.items()}
f = {k : [m, v] for m, x in main_dict.items() for k, v in x.items() if [m,v] not in e.values()}
g = {k1 : [m, v] for k1,m in e.items() for k2,v in f.items() if k1==k2}

Answer 4

One option is to use itertools.groupby

from itertools import groupby
from operator import itemgetter

main_dict = {
    'A' : {'key1' : 'valueA1', 'key2' : 'valueA2'},
    'B' : {'key2' : 'valueB2', 'key3' : 'valueB3'},
    'C' : {'key3' : 'valueC3', 'key1' : 'valueC1'}}

## Pull inner key, outer key and value and sort by key (prior to grouping)
x = sorted([(k2, [k1, v2]) for k1, v1 in main_dict.items() for k2, v2, in v1.items()])

## Group by key. This creates an itertools.groupby object that can be iterated
## to get key and value iterables
xx = groupby(x, key=itemgetter(0))

for k, g in xx:
    print('{0} : {1}'.format(k, [r[1] for r in list(g)]))

Depending on your data and performance requirements, sorting may not be ideal so it's worth profiling.

Further, it does not result in a dict as specified, but a groupby object. That might be 'dict-like' enough for your needs; Iterating it yields key and iterables.