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Represent a 1-1 relationship [duplicate]

This question already has answers here:
How to implement an efficient bidirectional hash table?
(8 answers)
Closed 2 years ago.
I'm doing this switchboard thing in python where I need to keep track of who's talking to whom, so if Alice --> Bob, then that implies that Bob --> Alice.
Yes, I could populate two hash maps, but I'm wondering if anyone has an idea to do it with one.
Or suggest another data structure.
There are no multiple conversations. Let's say this is for a customer service call center, so when Alice dials into the switchboard, she's only going to talk to Bob. His replies also go only to her.

You can create your own dictionary type by subclassing dict and adding the logic that you want. Here's a basic example:
class TwoWayDict(dict):
def __setitem__(self, key, value):
# Remove any previous connections with these values
if key in self:
del self[key]
if value in self:
del self[value]
dict.__setitem__(self, key, value)
dict.__setitem__(self, value, key)
def __delitem__(self, key):
dict.__delitem__(self, self[key])
dict.__delitem__(self, key)
def __len__(self):
"""Returns the number of connections"""
return dict.__len__(self) // 2
And it works like so:
>>> d = TwoWayDict()
>>> d['foo'] = 'bar'
>>> d['foo']
'bar'
>>> d['bar']
'foo'
>>> len(d)
1
>>> del d['foo']
>>> d['bar']
Traceback (most recent call last):
File "<stdin>", line 7, in <module>
KeyError: 'bar'
I'm sure I didn't cover all the cases, but that should get you started.

In your special case you can store both in one dictionary:
relation = {}
relation['Alice'] = 'Bob'
relation['Bob'] = 'Alice'
Since what you are describing is a symmetric relationship. A -> B => B -> A

I know it's an older question, but I wanted to mention another great solution to this problem, namely the python package bidict. It's extremely straight forward to use:
from bidict import bidict
map = bidict(Bob = "Alice")
print(map["Bob"])
print(map.inv["Alice"])

I would just populate a second hash, with
reverse_map = dict((reversed(item) for item in forward_map.items()))

Two hash maps is actually probably the fastest-performing solution assuming you can spare the memory. I would wrap those in a single class - the burden on the programmer is in ensuring that two the hash maps sync up correctly.

A less verbose way, still using reversed:
dict(map(reversed, my_dict.items()))

You have two separate issues.
You have a "Conversation" object. It refers to two Persons. Since a Person can have multiple conversations, you have a many-to-many relationship.
You have a Map from Person to a list of Conversations. A Conversion will have a pair of Persons.
Do something like this
from collections import defaultdict
switchboard= defaultdict( list )
x = Conversation( "Alice", "Bob" )
y = Conversation( "Alice", "Charlie" )
for c in ( x, y ):
switchboard[c.p1].append( c )
switchboard[c.p2].append( c )

No, there is really no way to do this without creating two dictionaries. How would it be possible to implement this with just one dictionary while continuing to offer comparable performance?
You are better off creating a custom type that encapsulates two dictionaries and exposes the functionality you want.

You may be able to use a DoubleDict as shown in recipe 578224 on the Python Cookbook.

Another possible solution is to implement a subclass of dict, that holds the original dictionary and keeps track of a reversed version of it. Keeping two seperate dicts can be useful if keys and values are overlapping.
class TwoWayDict(dict):
def __init__(self, my_dict):
dict.__init__(self, my_dict)
self.rev_dict = {v : k for k,v in my_dict.iteritems()}
def __setitem__(self, key, value):
dict.__setitem__(self, key, value)
self.rev_dict.__setitem__(value, key)
def pop(self, key):
self.rev_dict.pop(self[key])
dict.pop(self, key)
# The above is just an idea other methods
# should also be overridden.
Example:
>>> d = {'a' : 1, 'b' : 2} # suppose we need to use d and its reversed version
>>> twd = TwoWayDict(d) # create a two-way dict
>>> twd
{'a': 1, 'b': 2}
>>> twd.rev_dict
{1: 'a', 2: 'b'}
>>> twd['a']
1
>>> twd.rev_dict[2]
'b'
>>> twd['c'] = 3 # we add to twd and reversed version also changes
>>> twd
{'a': 1, 'c': 3, 'b': 2}
>>> twd.rev_dict
{1: 'a', 2: 'b', 3: 'c'}
>>> twd.pop('a') # we pop elements from twd and reversed version changes
>>> twd
{'c': 3, 'b': 2}
>>> twd.rev_dict
{2: 'b', 3: 'c'}

There's the collections-extended library on pypi: https://pypi.python.org/pypi/collections-extended/0.6.0
Using the bijection class is as easy as:
RESPONSE_TYPES = bijection({
0x03 : 'module_info',
0x09 : 'network_status_response',
0x10 : 'trust_center_device_update'
})
>>> RESPONSE_TYPES[0x03]
'module_info'
>>> RESPONSE_TYPES.inverse['network_status_response']
0x09

I like the suggestion of bidict in one of the comments.
pip install bidict
Useage:
# This normalization method should save hugely as aDaD ~ yXyX have the same form of smallest grammar.
# To get back to your grammar's alphabet use trans
def normalize_string(s, nv=None):
if nv is None:
nv = ord('a')
trans = bidict()
r = ''
for c in s:
if c not in trans.inverse:
a = chr(nv)
nv += 1
trans[a] = c
else:
a = trans.inverse[c]
r += a
return r, trans
def translate_string(s, trans):
res = ''
for c in s:
res += trans[c]
return res
if __name__ == "__main__":
s = "bnhnbiodfjos"
n, tr = normalize_string(s)
print(n)
print(tr)
print(translate_string(n, tr))
Since there aren't much docs about it. But I've got all the features I need from it working correctly.
Prints:
abcbadefghei
bidict({'a': 'b', 'b': 'n', 'c': 'h', 'd': 'i', 'e': 'o', 'f': 'd', 'g': 'f', 'h': 'j', 'i': 's'})
bnhnbiodfjos

The kjbuckets C extension module provides a "graph" data structure which I believe gives you what you want.

Here's one more two-way dictionary implementation by extending pythons dict class in case you didn't like any of those other ones:
class DoubleD(dict):
""" Access and delete dictionary elements by key or value. """
def __getitem__(self, key):
if key not in self:
inv_dict = {v:k for k,v in self.items()}
return inv_dict[key]
return dict.__getitem__(self, key)
def __delitem__(self, key):
if key not in self:
inv_dict = {v:k for k,v in self.items()}
dict.__delitem__(self, inv_dict[key])
else:
dict.__delitem__(self, key)
Use it as a normal python dictionary except in construction:
dd = DoubleD()
dd['foo'] = 'bar'

A way I like to do this kind of thing is something like:
{my_dict[key]: key for key in my_dict.keys()}

How to look up keys using values in python [duplicate]

Python dict is a very useful data-structure:
d = {'a': 1, 'b': 2}
d['a'] # get 1
Sometimes you'd also like to index by values.
d[1] # get 'a'
Which is the most efficient way to implement this data-structure? Any official recommend way to do it?

Here is a class for a bidirectional dict, inspired by Finding key from value in Python dictionary and modified to allow the following 2) and 3).
Note that :
The inverse directory bd.inverse auto-updates itself when the standard dict bd is modified.
The inverse directory bd.inverse[value] is always a list of key such that bd[key] == value.
Unlike the bidict module from https://pypi.python.org/pypi/bidict, here we can have 2 keys having same value, this is very important.
Code:
class bidict(dict):
def __init__(self, *args, **kwargs):
super(bidict, self).__init__(*args, **kwargs)
self.inverse = {}
for key, value in self.items():
self.inverse.setdefault(value, []).append(key)
def __setitem__(self, key, value):
if key in self:
self.inverse[self[key]].remove(key)
super(bidict, self).__setitem__(key, value)
self.inverse.setdefault(value, []).append(key)
def __delitem__(self, key):
self.inverse.setdefault(self[key], []).remove(key)
if self[key] in self.inverse and not self.inverse[self[key]]:
del self.inverse[self[key]]
super(bidict, self).__delitem__(key)
Usage example:
bd = bidict({'a': 1, 'b': 2})
print(bd) # {'a': 1, 'b': 2}
print(bd.inverse) # {1: ['a'], 2: ['b']}
bd['c'] = 1 # Now two keys have the same value (= 1)
print(bd) # {'a': 1, 'c': 1, 'b': 2}
print(bd.inverse) # {1: ['a', 'c'], 2: ['b']}
del bd['c']
print(bd) # {'a': 1, 'b': 2}
print(bd.inverse) # {1: ['a'], 2: ['b']}
del bd['a']
print(bd) # {'b': 2}
print(bd.inverse) # {2: ['b']}
bd['b'] = 3
print(bd) # {'b': 3}
print(bd.inverse) # {2: [], 3: ['b']}

You can use the same dict itself by adding key,value pair in reverse order.
d={'a':1,'b':2}
revd=dict([reversed(i) for i in d.items()])
d.update(revd)

A poor man's bidirectional hash table would be to use just two dictionaries (these are highly tuned datastructures already).
There is also a bidict package on the index:
https://pypi.python.org/pypi/bidict
The source for bidict can be found on github:
https://github.com/jab/bidict

The below snippet of code implements an invertible (bijective) map:
class BijectionError(Exception):
"""Must set a unique value in a BijectiveMap."""
def __init__(self, value):
self.value = value
msg = 'The value "{}" is already in the mapping.'
super().__init__(msg.format(value))
class BijectiveMap(dict):
"""Invertible map."""
def __init__(self, inverse=None):
if inverse is None:
inverse = self.__class__(inverse=self)
self.inverse = inverse
def __setitem__(self, key, value):
if value in self.inverse:
raise BijectionError(value)
self.inverse._set_item(value, key)
self._set_item(key, value)
def __delitem__(self, key):
self.inverse._del_item(self[key])
self._del_item(key)
def _del_item(self, key):
super().__delitem__(key)
def _set_item(self, key, value):
super().__setitem__(key, value)
The advantage of this implementation is that the inverse attribute of a BijectiveMap is again a BijectiveMap. Therefore you can do things like:
>>> foo = BijectiveMap()
>>> foo['steve'] = 42
>>> foo.inverse
{42: 'steve'}
>>> foo.inverse.inverse
{'steve': 42}
>>> foo.inverse.inverse is foo
True

Something like this, maybe:
import itertools
class BidirDict(dict):
def __init__(self, iterable=(), **kwargs):
self.update(iterable, **kwargs)
def update(self, iterable=(), **kwargs):
if hasattr(iterable, 'iteritems'):
iterable = iterable.iteritems()
for (key, value) in itertools.chain(iterable, kwargs.iteritems()):
self[key] = value
def __setitem__(self, key, value):
if key in self:
del self[key]
if value in self:
del self[value]
dict.__setitem__(self, key, value)
dict.__setitem__(self, value, key)
def __delitem__(self, key):
value = self[key]
dict.__delitem__(self, key)
dict.__delitem__(self, value)
def __repr__(self):
return '%s(%s)' % (type(self).__name__, dict.__repr__(self))
You have to decide what you want to happen if more than one key has a given value; the bidirectionality of a given pair could easily be clobbered by some later pair you inserted. I implemented one possible choice.
Example :
bd = BidirDict({'a': 'myvalue1', 'b': 'myvalue2', 'c': 'myvalue2'})
print bd['myvalue1'] # a
print bd['myvalue2'] # b

First, you have to make sure the key to value mapping is one to one, otherwise, it is not possible to build a bidirectional map.
Second, how large is the dataset? If there is not much data, just use 2 separate maps, and update both of them when updating. Or better, use an existing solution like Bidict, which is just a wrapper of 2 dicts, with updating/deletion built in.
But if the dataset is large, and maintaining 2 dicts is not desirable:
If both key and value are numeric, consider the possibility of using
Interpolation to approximate the mapping. If the vast majority of the
key-value pairs can be covered by the mapping function (and its
reverse function), then you only need to record the outliers in maps.
If most of access is uni-directional (key->value), then it is totally
ok to build the reverse map incrementally, to trade time for
space.
Code:
d = {1: "one", 2: "two" }
reverse = {}
def get_key_by_value(v):
if v not in reverse:
for _k, _v in d.items():
if _v == v:
reverse[_v] = _k
break
return reverse[v]

a better way is convert the dictionary to a list of tuples then sort on a specific tuple field
def convert_to_list(dictionary):
list_of_tuples = []
for key, value in dictionary.items():
list_of_tuples.append((key, value))
return list_of_tuples
def sort_list(list_of_tuples, field):
return sorted(list_of_tuples, key=lambda x: x[field])
dictionary = {'a': 9, 'b': 2, 'c': 3, 'd': 4, 'e': 5}
list_of_tuples = convert_to_list(dictionary)
print(sort_list(list_of_tuples, 1))
output
[('b', 2), ('c', 3), ('d', 4), ('e', 5), ('a', 9)]

Unfortunately, the highest rated answer, bidict does not work.
There are three options:
Subclass dict: You can create a subclass of dict, but beware. You need to write custom implementations ofupdate, pop, initializer, setdefault. The dict implementations do not call __setitem__. This is why the highest rated answer has issues.
Inherit from UserDict: This is just like a dict, except all the routines are made to call correctly. It uses a dict under the hood, in an item called data. You can read the Python Documentation, or use a simple implementation of a by directional list that works in Python 3. Sorry for not including it verbatim: I'm unsure of its copyright.
Inherit from Abstract Base Classes: Inheriting from collections.abc will help you get all the correct protocols and implementations for a new class. This is overkill for a bidirectional dictionary, unless it can also encrypt and cache to a database.
TL;DR -- Use this for your code. Read Trey Hunner's article for details.

Pythonic way to convert a dictionary into namedtuple or another hashable dict-like?

I have a dictionary like:
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
which I would like to convert to a namedtuple.
My current approach is with the following code
namedTupleConstructor = namedtuple('myNamedTuple', ' '.join(sorted(d.keys())))
nt= namedTupleConstructor(**d)
which produces
myNamedTuple(a=1, b=2, c=3, d=4)
This works fine for me (I think), but am I missing a built-in such as...
nt = namedtuple.from_dict() ?
UPDATE: as discussed in the comments, my reason for wanting to convert my dictionary to a namedtuple is so that it becomes hashable, but still generally useable like a dict.
UPDATE2: 4 years after I've posted this question, TLK posts a new answer recommending using the dataclass decorator that I think is really great. I think that's now what I would use going forward.

To create the subclass, you may just pass the keys of a dict directly:
MyTuple = namedtuple('MyTuple', d)
Now to create tuple instances from this dict, or any other dict with matching keys:
my_tuple = MyTuple(**d)
Beware: namedtuples compare on values only (ordered). They are designed to be a drop-in replacement for regular tuples, with named attribute access as an added feature. The field names will not be considered when making equality comparisons. It may not be what you wanted nor expected from the namedtuple type! This differs from dict equality comparisons, which do take into account the keys and also compare order agnostic.
For readers who don't really need a type which is a subclass of tuple, there probably isn't much point to use a namedtuple in the first place. If you just want to use attribute access syntax on fields, it would be simpler and easier to create namespace objects instead:
>>> from types import SimpleNamespace
>>> SimpleNamespace(**d)
namespace(a=1, b=2, c=3, d=4)
my reason for wanting to convert my dictionary to a namedtuple is so that it becomes hashable, but still generally useable like a dict
For a hashable "attrdict" like recipe, check out a frozen box:
>>> from box import Box
>>> b = Box(d, frozen_box=True)
>>> hash(b)
7686694140185755210
>>> b.a
1
>>> b["a"]
1
>>> b["a"] = 2
BoxError: Box is frozen
There may also be a frozen mapping type coming in a later version of Python, watch this draft PEP for acceptance or rejection:
PEP 603 -- Adding a frozenmap type to collections

from collections import namedtuple
nt = namedtuple('x', d.keys())(*d.values())

If you want an easier approach, and you have the flexibility to use another approach other than namedtuple I would like to suggest using SimpleNamespace (docs).
from types import SimpleNamespace as sn
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
dd= sn(**d)
# dd.a>>1
# add new property
dd.s = 5
#dd.s>>5
PS: SimpleNamespace is a type, not a class

I'd like to recommend the dataclass for this type of situation. Similar to a namedtuple, but with more flexibility.
https://docs.python.org/3/library/dataclasses.html
from dataclasses import dataclass
#dataclass
class InventoryItem:
"""Class for keeping track of an item in inventory."""
name: str
unit_price: float
quantity_on_hand: int = 0
def total_cost(self) -> float:
return self.unit_price * self.quantity_on_hand

You can use this function to handle nested dictionaries:
def create_namedtuple_from_dict(obj):
if isinstance(obj, dict):
fields = sorted(obj.keys())
namedtuple_type = namedtuple(
typename='GenericObject',
field_names=fields,
rename=True,
)
field_value_pairs = OrderedDict(
(str(field), create_namedtuple_from_dict(obj[field]))
for field in fields
)
try:
return namedtuple_type(**field_value_pairs)
except TypeError:
# Cannot create namedtuple instance so fallback to dict (invalid attribute names)
return dict(**field_value_pairs)
elif isinstance(obj, (list, set, tuple, frozenset)):
return [create_namedtuple_from_dict(item) for item in obj]
else:
return obj

use the dictionary keys as the fieldnames to the namedtuple
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
def dict_to_namedtuple(d):
return namedtuple('GenericDict', d.keys())(**d)
result=dict_to_namedtuple(d)
print(result)
output
GenericDict(a=1, b=2, c=3, d=4)

def toNametuple(dict_data):
return namedtuple(
"X", dict_data.keys()
)(*tuple(map(lambda x: x if not isinstance(x, dict) else toNametuple(x), dict_data.values())))
d = {
'id': 1,
'name': {'firstName': 'Ritesh', 'lastName':'Dubey'},
'list_data': [1, 2],
}
obj = toNametuple(d)
Access as obj.name.firstName, obj.id
This will work for nested dictionary with any data types.

I find the following 4-liner the most beautiful. It supports nested dictionaries as well.
def dict_to_namedtuple(typename, data):
return namedtuple(typename, data.keys())(
*(dict_to_namedtuple(typename + '_' + k, v) if isinstance(v, dict) else v for k, v in data.items())
)
The output will look good also:
>>> nt = dict_to_namedtuple('config', {
... 'path': '/app',
... 'debug': {'level': 'error', 'stream': 'stdout'}
... })
>>> print(nt)
config(path='/app', debug=config_debug(level='error', stream='stdout'))

Check this out:
def fill_tuple(NamedTupleType, container):
if container is None:
args = [None] * len(NamedTupleType._fields)
return NamedTupleType(*args)
if isinstance(container, (list, tuple)):
return NamedTupleType(*container)
elif isinstance(container, dict):
return NamedTupleType(**container)
else:
raise TypeError("Cannot create '{}' tuple out of {} ({}).".format(NamedTupleType.__name__, type(container).__name__, container))
Exceptions for incorrect names or invalid argument count is handled by __init__ of namedtuple.
Test with py.test:
def test_fill_tuple():
A = namedtuple("A", "aa, bb, cc")
assert fill_tuple(A, None) == A(aa=None, bb=None, cc=None)
assert fill_tuple(A, [None, None, None]) == A(aa=None, bb=None, cc=None)
assert fill_tuple(A, [1, 2, 3]) == A(aa=1, bb=2, cc=3)
assert fill_tuple(A, dict(aa=1, bb=2, cc=3)) == A(aa=1, bb=2, cc=3)
with pytest.raises(TypeError) as e:
fill_tuple(A, 2)
assert e.value.message == "Cannot create 'A' tuple out of int (2)."

Although I like #fuggy_yama answer, before read it I got my own function, so I leave it here just to show a different approach. It also handles nested namedtuples
def dict2namedtuple(thedict, name):
thenametuple = namedtuple(name, [])
for key, val in thedict.items():
if not isinstance(key, str):
msg = 'dict keys must be strings not {}'
raise ValueError(msg.format(key.__class__))
if not isinstance(val, dict):
setattr(thenametuple, key, val)
else:
newname = dict2namedtuple(val, key)
setattr(thenametuple, key, newname)
return thenametuple

Reaching into a nested dictionary several levels deep (that might not exist)

I have an API that I call that returns a dictionary. Part of that dictionary is itself another dictionary. In that inside dictionary, there are some keys that might not exist, or they might. Those keys could reference another dictionary.
To give an example, say I have the following dictionaries:
dict1 = {'a': {'b': {'c':{'d':3}}}}
dict2 = {'a': {'b': {''f': 2}}}
I would like to write a function that I can pass in the dictionary, and a list of keys that would lead me to the 3 in dict1, and the 2 in dict2. However, it is possible that b and c might not exist in dict1, and b and f might not exist in dict2.
I would like to have a function that I could call like this:
get_value(dict1, ['a', 'b', 'c'])
and that would return a 3, or if the keys are not found, then return a default value of 0.
I know that I can use something like this:
val = dict1.get('a', {}).get('b', {}).get('c', 0)
but that seems to be quite wordy to me.
I can also flatten the dict (see https://stackoverflow.com/a/6043835/1758023), but that can be a bit intensive since my dictionary is actually fairly large, and has about 5 levels of nesting in some keys. And, I only need to get two things from the dict.
Right now I am using the flattenDict function in the SO question, but that seems a bit of overkill for my situation.

You can use a recursive function:
def get_value(mydict, keys):
if not keys:
return mydict
if keys[0] not in mydict:
return 0
return get_value(mydict[keys[0]], keys[1:])
If keys can not only be missing, but be other, non-dict types, you can handle this like so:
def get_value(mydict, keys):
if not keys:
return mydict
key = keys[0]
try:
newdict = mydict[key]
except (TypeError, KeyError):
return 0
return get_value(newdict, keys[1:])

Without recursion, just iterate through the keys and go down one level at a time. Putting that inside a try/except allows you to handle the missing key case. KeyError will be raised when the key is not there, and TypeError will be raised if you hit the "bottom" of the dict too soon and try to apply the [] operator to an int or something.
def get_value(d, ks):
for k in ks:
try:
d = d[k] # descend one level
except (KeyError, TypeError):
return 0 # when any lookup fails, return 0
return d # return the final element

Here is a recursive function that should work for general cases
def recursive_get(d, k):
if len(k) == 0:
return 0
elif len(k) == 1:
return d.get(k[0], 0)
else:
value = d.get(k[0], 0)
if isinstance(value, dict):
return recursive_get(value, k[1:])
else:
return value
It takes arguments of the dict to search, and a list of keys, which it will check one per level
>>> dict1 = {'a': {'b': {'c':{'d':3}}}}
>>> recursive_get(dict1, ['a', 'b', 'c'])
{'d': 3}
>>> dict2 = {'a': {'b': {'f': 2}}}
>>> recursive_get(dict2, ['a', 'b', 'c'])
0

Recursive dictionary modification in python

What would be the easiest way to go about turning this dictionary:
{'item':{'w':{'c':1, 'd':2}, 'x':120, 'y':240, 'z':{'a':100, 'b':200}}}
into this one:
{'item':{'y':240, 'z':{'b':200}}}
given only that you need the vars y and b while maintaining the structure of the dictionary? The size or number of items or the depth of the dictionary should not matter, as the one I'm working with can be anywhere from 2 to 5 levels deep.
EDIT: I apologize for the type earlier, and to clarify, I am given an array of strings (eg ['y', 'b']) which I need to find in the dictionary and then keep ONLY 'y' and 'b' as well as any other keys in order to maintain the structure of the original dictionary, in this case, it would be 'z'
A better example can be found here where I need Chipset Model, VRAM, and Resolution.
In regards to the comment, the input would be the above link as the starting dictionary along with an array of ['chipset model', 'vram', 'resolution'] as the keep list. It should return this:
{'Graphics/Displays':{'NVIDIA GeForce 7300 GT':{'Chipset Model':'NVIDIA GeForce 7300 GT', 'Displays':{'Resolution':'1440 x 900 # 75 Hz'}, 'VRAM (Total)':'256 Mb'}}

Assuming that the dictionary you want to assign to an element of a super-dictionary is foo, you could just do this:
my_dictionary['keys']['to']['subdict']=foo
Regarding your edit—where you need to eliminate all keys except those on a certain list—this function should do the trick:
def drop_keys(recursive_dict,keep_list):
key_list=recursive_dict.keys()
for key in key_list:
if(type(recursive_dict[key]) is dict):
drop_keys(recursive_dict[key], keep_list)
elif(key not in keep_list):
del recursive_dict[key]

Something like this?
d = {'item': {'w': {'c': 1, 'd': 2}, 'x': 120, 'y': 240, 'z': {'a': 100, 'b': 200}}}
l = ['y', 'z']
def do_dict(d, l):
return {k: v for k, v in d['item'].items() if k in l}

Here's what I arrived at for a recursive solution, which ended up being similar to what #Dan posted:
def recursive_del(d,keep):
for k in d.copy():
if type(d[k]) == dict:
recursive_del(d[k],keep)
if len(d[k]) == 0: #all keys were deleted, clean up empty dict
del d[k]
elif k not in keep:
del d[k]
demo:
>>> keepset = {'y','b'}
>>> a = {'item':{'w':{'c':1, 'd':2}, 'x':120, 'y':240, 'z':{'a':100, 'b':200}}}
>>> recursive_del(a,keepset)
>>> a
{'item': {'z': {'b': 200}, 'y': 240}}
The only thing I think he missed is that you will need to sometimes need to clean up dicts which had all their keys deleted; i.e. without that adjustment you would end up with a vestigial 'w':{} in your example output.

Using your second example I made something like this, it's not exactly pretty but it should be easy to extend. If your tree starts to get big, you can define some sets of rules to parse the dict.
Each rule here are actually pretty much "what should I do when i'm in which state".
def rule2(key, value):
if key == 'VRAM (Total)':
return (key, value)
elif key == 'Chipset Model':
return (key, value)
def rule1(key, value):
if key == "Graphics/Displays":
if isinstance(value, dict):
return (key, recursive_checker(value, rule1))
else:
return (key, value)
else:
return (key, recursive_checker(value, rule2))
def recursive_checker(dat, rule):
def inner(item):
key = item[0]
value = item[1]
return rule(key, value)
return dict(filter(lambda x: x!= None, map(inner, dat.items())))
# Important bits
print recursive_checker(data, rule1)
In your case, as there is not many states, it isn't worth doing it but in case you have multiple cards and you don't necessarly know which key should be traversed but only know that you want certain keys from the tree. This method could be used to search the tree easily. It can be applied to many things.