How To Get All The Keys In A 2d Dict Python

I have a dictionary of form: d = {123:{2:1,3:1}, 124:{3:1}, 125:{2:1},126:{1:1}} So, lets look into 2nd degree keys.. 123--> 2,3 124--> 3 125--> 2 126--> 1 So total n

Solution 1:

keyset = set()
forkin d:
    keyset.update(d[k])

forkin d:
    forkkin keyset:
        d[k].setdefault(kk, 0)

Solution 2:

In [25]: d = {123:{2:1,3:1}, 124:{3:1}, 125:{2:1},126:{1:1}}

In [26]: se=set(y for x in d for y in d[x])

In [27]: for x in d:
    foo=se.difference(d[x])
    d[x].update(dict(zip(foo,[0]*len(foo))))
   ....:     
   ....:     

In [30]: d
Out[30]: 
{123: {1: 0, 2: 1, 3: 1},
 124: {1: 0, 2: 0, 3: 1},
 125: {1: 0, 2: 1, 3: 0},
 126: {1: 1, 2: 0, 3: 0}}

here use set difference to get the missing keys and then update() the dict:

In [39]: for x in d:
    foo=se.difference(d[x])
    print foo                # missing keys per dictset([1])
set([1, 2])
set([1, 3])
set([2, 3])

Solution 3:

I like the solution of Ashwini Chaudhary.

I edited it to incorporate all the suggestions in the comments with other minor changes for it to look how I would prefer it:

Edited (incorporates the suggestion of Steven Rumbalski to this answer).

all_second_keys = set(keyfor value in d.itervalues() forkeyin value)

for value in d.itervalues():
    value.update((key,0) forkeyin all_second_keys ifkeynotin value)

Solution 4:

importoperator

second_order_keys = reduce(operator.__or__,
                           (set(v.iterkeys()) for v in d.itervalues()))
for v in d.itervalues():
    for k in second_order_keys:
        v.setdefault(k, 0)

Or, in Python 3:

from functools import reduce
importoperator

second_order_keys = reduce(operator.__or__,
                           (v.keys() for v in d.values()))
for v in d.values():
    for k in second_order_keys:
        v.setdefault(k, 0)