I'm building some Python code to read and manipulate deeply nested dicts (ultimately for interacting with JSON services, however it would be great to have for other purposes) I'm looking for a way to easily read/set/update values deep within the dict, without needing a lot of code.
#see also Python: Recursively access dict via attributes as well as index access? -- Curt Hagenlocher's "DotDictify" solution is pretty eloquent. I also like what Ben Alman presents for JavaScript in http://benalman.com/projects/jquery-getobject-plugin/ It would be great to somehow combine the two.
Building off of Curt Hagenlocher and Ben Alman's examples, it would be great in Python to have a capability like:
>>> my_obj = DotDictify()
>>> my_obj.a.b.c = {'d':1, 'e':2}
>>> print my_obj
{'a': {'b': {'c': {'d': 1, 'e': 2}}}}
>>> print my_obj.a.b.c.d
1
>>> print my_obj.a.b.c.x
None
>>> print my_obj.a.b.c.d.x
None
>>> print my_obj.a.b.c.d.x.y.z
None
Any idea if this is possible, and if so, how to go about modifying the DotDictify solution?
Alternatively, the get method could be made to accept a dot notation (and a complementary set method added) however the object notation sure is cleaner.
>>> my_obj = DotDictify()
>>> my_obj.set('a.b.c', {'d':1, 'e':2})
>>> print my_obj
{'a': {'b': {'c': {'d': 1, 'e': 2}}}}
>>> print my_obj.get('a.b.c.d')
1
>>> print my_obj.get('a.b.c.x')
None
>>> print my_obj.get('a.b.c.d.x')
None
>>> print my_obj.get('a.b.c.d.x.y.z')
None
This type of interaction would be great to have for dealing with deeply nested dicts. Does anybody know another strategy (or sample code snippet/library) to try?
Attribute Tree
The problem with your first specification is that Python can't tell in __getitem__ if, at my_obj.a.b.c.d, you will next proceed farther down a nonexistent tree, in which case it needs to return an object with a __getitem__ method so you won't get an AttributeError thrown at you, or if you want a value, in which case it needs to return None.
I would argue that in every case you have above, you should expect it to throw a KeyError instead of returning None. The reason being that you can't tell if None means "no key" or "someone actually stored None at that location". For this behavior, all you have to do is take dotdictify, remove marker, and replace __getitem__ with:
def __getitem__(self, key):
return self[key]
Because what you really want is a dict with __getattr__ and __setattr__.
There may be a way to remove __getitem__ entirely and say something like __getattr__ = dict.__getitem__, but I think this may be over-optimization, and will be a problem if you later decide you want __getitem__ to create the tree as it goes like dotdictify originally does, in which case you would change it to:
def __getitem__(self, key):
if key not in self:
dict.__setitem__(self, key, dotdictify())
return dict.__getitem__(self, key)
I don't like the marker business in the original dotdictify.
Path Support
The second specification (override get() and set()) is that a normal dict has a get() that operates differently from what you describe and doesn't even have a set (though it has a setdefault() which is an inverse operation to get()). People expect get to take two parameters, the second being a default if the key isn't found.
If you want to extend __getitem__ and __setitem__ to handle dotted-key notation, you'll need to modify doctictify to:
class dotdictify(dict):
def __init__(self, value=None):
if value is None:
pass
elif isinstance(value, dict):
for key in value:
self.__setitem__(key, value[key])
else:
raise TypeError, 'expected dict'
def __setitem__(self, key, value):
if '.' in key:
myKey, restOfKey = key.split('.', 1)
target = self.setdefault(myKey, dotdictify())
if not isinstance(target, dotdictify):
raise KeyError, 'cannot set "%s" in "%s" (%s)' % (restOfKey, myKey, repr(target))
target[restOfKey] = value
else:
if isinstance(value, dict) and not isinstance(value, dotdictify):
value = dotdictify(value)
dict.__setitem__(self, key, value)
def __getitem__(self, key):
if '.' not in key:
return dict.__getitem__(self, key)
myKey, restOfKey = key.split('.', 1)
target = dict.__getitem__(self, myKey)
if not isinstance(target, dotdictify):
raise KeyError, 'cannot get "%s" in "%s" (%s)' % (restOfKey, myKey, repr(target))
return target[restOfKey]
def __contains__(self, key):
if '.' not in key:
return dict.__contains__(self, key)
myKey, restOfKey = key.split('.', 1)
target = dict.__getitem__(self, myKey)
if not isinstance(target, dotdictify):
return False
return restOfKey in target
def setdefault(self, key, default):
if key not in self:
self[key] = default
return self[key]
__setattr__ = __setitem__
__getattr__ = __getitem__
Test code:
>>> life = dotdictify({'bigBang': {'stars': {'planets': {}}}})
>>> life.bigBang.stars.planets
{}
>>> life.bigBang.stars.planets.earth = { 'singleCellLife' : {} }
>>> life.bigBang.stars.planets
{'earth': {'singleCellLife': {}}}
>>> life['bigBang.stars.planets.mars.landers.vikings'] = 2
>>> life.bigBang.stars.planets.mars.landers.vikings
2
>>> 'landers.vikings' in life.bigBang.stars.planets.mars
True
>>> life.get('bigBang.stars.planets.mars.landers.spirit', True)
True
>>> life.setdefault('bigBang.stars.planets.mars.landers.opportunity', True)
True
>>> 'landers.opportunity' in life.bigBang.stars.planets.mars
True
>>> life.bigBang.stars.planets.mars
{'landers': {'opportunity': True, 'vikings': 2}}
The older answers have some pretty good tips in them, but they all require replacing standard Python data structures (dicts, etc.) with custom ones, and would not work with keys that are not valid attribute names.
These days we can do better, using a pure-Python, Python 2/3-compatible library, built for exactly this purpose, called glom. Using your example:
import glom
target = {} # a plain dictionary we will deeply set on
glom.assign(target, 'a.b.c', {'d': 1, 'e': 2}, missing=dict)
# {'a': {'b': {'c': {'e': 2, 'd': 1}}}}
Notice the missing=dict, used to autocreate dictionaries. We can easily get the value back using glom's deep-get:
glom.glom(target, 'a.b.c.d')
# 1
There's a lot more you can do with glom, especially around deep getting and setting. I should know, since (full disclosure) I created it. That means if you find a gap, you should let me know!
To fellow googlers: we now have addict:
pip install addict
and
mapping.a.b.c.d.e = 2
mapping
{'a': {'b': {'c': {'d': {'e': 2}}}}}
I used it extensively.
To work with dotted paths, I found dotted:
obj = DottedDict({'hello': {'world': {'wide': 'web'}}})
obj['hello.world.wide'] == 'web' # true
I had used something similar in order to build somithing similar Trie for an application. I hope it helps.
class Trie:
"""
A Trie is like a dictionary in that it maps keys to values.
However, because of the way keys are stored, it allows
look up based on the longest prefix that matches.
"""
def __init__(self):
# Every node consists of a list with two position. In
# the first one,there is the value while on the second
# one a dictionary which leads to the rest of the nodes.
self.root = [0, {}]
def insert(self, key):
"""
Add the given value for the given key.
>>> a = Trie()
>>> a.insert('kalo')
>>> print(a)
[0, {'k': [1, {'a': [1, {'l': [1, {'o': [1, {}]}]}]}]}]
>>> a.insert('kalo')
>>> print(a)
[0, {'k': [2, {'a': [2, {'l': [2, {'o': [2, {}]}]}]}]}]
>>> b = Trie()
>>> b.insert('heh')
>>> b.insert('ha')
>>> print(b)
[0, {'h': [2, {'a': [1, {}], 'e': [1, {'h': [1, {}]}]}]}]
"""
# find the node to append the new value.
curr_node = self.root
for k in key:
curr_node = curr_node[1].setdefault(k, [0, {}])
curr_node[0] += 1
def find(self, key):
"""
Return the value for the given key or None if key not
found.
>>> a = Trie()
>>> a.insert('ha')
>>> a.insert('ha')
>>> a.insert('he')
>>> a.insert('ho')
>>> print(a.find('h'))
4
>>> print(a.find('ha'))
2
>>> print(a.find('he'))
1
"""
curr_node = self.root
for k in key:
try:
curr_node = curr_node[1][k]
except KeyError:
return 0
return curr_node[0]
def __str__(self):
return str(self.root)
def __getitem__(self, key):
curr_node = self.root
for k in key:
try:
curr_node = curr_node[1][k]
except KeyError:
yield None
for k in curr_node[1]:
yield k, curr_node[1][k][0]
if __name__ == '__main__':
a = Trie()
a.insert('kalo')
a.insert('kala')
a.insert('kal')
a.insert('kata')
print(a.find('kala'))
for b in a['ka']:
print(b)
print(a)
Not a full-fledged solution, but a simple approach with no dependencies, and which doesn't require replacing/modifying the built-in dictionary type. Might fit the bill for some:
def get(nested_dict: dict, key: str):
return reduce(lambda d, k: d[k], key.split('.'), nested_dict)
my_dict = {'a': {'b': {'c': 123}}}
get(my_dict, "a.b.c") # 123
The setter is not quite as nice, but works:
def set(nested_dict: dict, key: str, value):
*keys, last_key = key.split('.')
for k in keys:
if k not in nested_dict:
nested_dict[k] = dict()
nested_dict = nested_dict[k]
nested_dict[last_key] = value
set(my_dict, "very.very.many.levels", True)
A more full-fledged solution should probably check the keys accessed along the way. Probably other stuff I haven't though about at the moment.
Related
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()}
pop is a great little function that, when used on dictionaries (given a known key) removes the item with that key from the dictionary and also returns the corresponding value. But what if I want the key as well?
Obviously, in simple cases I could probably just do something like this:
pair = (key, some_dict.pop(key))
But if, say, I wanted to pop the key-value pair with the lowest value, following the above idea I would have to do this...
pair = (min(some_dict, key=some.get), some_dict.pop(min(some_dict, key=some_dict.get)))
... which is hideous as I have to do the operation twice (obviously I could store the output from min in a variable, but I'm still not completely happy with that). So my question is: Is there an elegant way to do this? Am I missing an obvious trick here?
You can define yourself dictionary object using python ABCs which provides the infrastructure for defining abstract base classes. And then overload the pop attribute of python dictionary objects based on your need:
from collections import Mapping
class MyDict(Mapping):
def __init__(self, *args, **kwargs):
self.update(dict(*args, **kwargs))
def __setitem__(self, key, item):
self.__dict__[key] = item
def __getitem__(self, key):
return self.__dict__[key]
def __delitem__(self, key):
del self.__dict__[key]
def pop(self, k, d=None):
return k,self.__dict__.pop(k, d)
def update(self, *args, **kwargs):
return self.__dict__.update(*args, **kwargs)
def __iter__(self):
return iter(self.__dict__)
def __len__(self):
return len(self.__dict__)
def __repr__(self):
return repr(self.__dict__)
Demo:
d=MyDict()
d['a']=1
d['b']=5
d['c']=8
print d
{'a': 1, 'c': 8, 'b': 5}
print d.pop(min(d, key=d.get))
('a', 1)
print d
{'c': 8, 'b': 5}
Note : As #chepner suggested in comment as a better choice you can override popitem, which already returns a key/value pair.
A heap supports the pop-min operation you describe. You'll need to create a heap from your dictionary first, though.
import heapq
# Must be two steps; heapify modifies its argument in-place.
# Reversing the key and the value because the value will actually be
# the "key" in the heap. (Or rather, tuples are compared
# lexicographically, so put the value in the first position.)
heap = [(v, k) for k, v in some_dict.items()]
heapq.heapify(heap)
# Get the smallest item from the heap
value, key = heapq.heappop(heap)
here is a simpler implementation
class CustomDict(dict):
def pop_item(self, key):
popped = {key:self[key]} #save "snapshot" of the value of key before popping
self.pop(key)
return popped
a = CustomDict()
b = {"hello":"wassup", "lol":"meh"}
a.update(b)
print(a.pop_item("lol"))
print(a)
So here we create a custom dict that pops the item you want and gives out the key-value pair
In Python 3, how to copy a key-value mapping from one dict to another, including remove if necessary? Here's some ugly code to do this:
if key in dict2:
dict1[key] = dict2[key]
elif key in dict1:
del dict1[key]
I'm hoping someone can reply with a cleaner and hopefully one-liner way to do this. (And I don't mean "just put that ugly code inside a function" because I don't want to add a hundred little functions to my code.) TIA.
UPDATE:
Since one comment asked for context, and others have tried to give answers that don't actually do what the question stated, I'm going to give an example context -- this isn't exactly what I'm trying to do, but it'll give you a good idea. I just wrote up this code quickly without testing, so hopefully it's not so wrong that I don't get the idea across. Note the ugly code I asked about originally is toward the bottom of this expanded example. TIA.
class TimestampedDict(dict):
def __init__(self):
self._ts = {} # from key to timestamp
def __setitem__(self, key, val):
super().__setitem__(key, val)
self._ts[key] = datetime.now()
def __delitem__(self, key):
super().__delitem__(key)
# When deleting, timestamp is updated by 1 microsecond so that it wins
# against the original but loses to any later changes.
self._ts[key] = self._ts[key] + timedelta(microseconds=1)
def update(self, other):
for key in other._ts:
if key not in self._ts or self._ts[key] < other._ts[key]:
# Other wins, so replace the mapping (including del if necessary).
if key in other:
self[key] = other[key]
elif key in self:
del self[key]
self._ts[key] = other._ts[key]
I might be misunderstanding, but you are looking to create update dict1 with the values from dict2 and remove those that are missing? In that case couldn't you simply set dict1 = dict2?
Otherwise if you are looking to combine two dictionaries and update shared keys, maybe the following will work:
dict1 = {1:2,3:4,5:6}
dict2 = {1:7}
dict(dict1.items() + dict2.items())
The addition of the items of the 2 dictionaries, will produce a new dictionary with all keys, and the values of the second dictionary if overlapping. However after rereading your question I believe you are looking for the former solution (dict1 = dict2).
Have you tried using update:
If this isn't what you want, can you please provide sample data and desired result?
d1 = {'a': 1, 'b': 2, 'c': 3}
d2 = {'a': 10, 'c': 30, 'd': 40}
d1.update(d2)
>>> d1
{'a': 10, 'b': 2, 'c': 30, 'd': 40}
As I mentioned I don't think there is much to improve with your logic but I find it interesting to see how twisted one can get, so here's a really ugly way of doing it using update():
def update(other):
u = {k: v for k, v in other._ts.items() if k not in self._ts or self._ts[k] < v}
self._ts.update(u)
dict.update({k: other[k] for k in u
if k in other or (self.pop(k, 0) and False})
You are guaranteed (self.pop(k, 0) and False) evaluates to False but also has the side effect of removing k from self. This will only be evaluated if k in other evaluates to False.
>>> a = {1:1, 2:2, 3:3, 4:4}
>>> b = {1:2, 2:3}
>>> a.update({k: b[k] for k in [1,2,3] if k in b or (a.pop(k, 0) and False)})
>>> a
{1: 2, 2: 3, 4: 4}
I need a function to change one item in composite dictionary.
I've tried something like..
def SetItem(keys, value):
item = self.dict
for key in keys:
item = item[key]
item = value
and
SetItem(['key1', 'key2'], 86)
It should be equivalent to self.dict['key1']['key2'] = 86, but this function has no effect.
Almost. You actually want to do something like:
def set_keys(d, keys, value):
item = d
for key in keys[:-1]:
item = item[key]
item[keys[-1]] = value
Or recursively like this:
def set_key(d, keys, value):
if len(keys) == 1:
d[keys[0]] = value
else:
set_key(d[keys[0]], keys[1:], value)
Marcin's right though. You would really want to incorporate something more rigorous, with some error handling for missing keys/missing dicts.
setItem = lambda self,names,value: map((lambda name: setattr(self,name,value)),names)
You don't have a self parameter
Just use the line of working code you have.
If you insist, here's a way:
def setitem(self, keys, value):
reduce(dict.get, # = lambda dictionary, key: dictionary[key]
keys[:-1], self.dictionary)[keys[-1]] = value
Obviously, this will break if the list of keys hits a non-dict value. You'll want to handle that. In fact, an explicit loop would probably be better for that reason, but you get the idea.
An idea involving recursion and EAFP, both of which I always like:
def set_item(d, keys, value):
key = keys.pop(0)
try:
set_item(d[key], keys, value)
# IndexError happens when the pop fails (empty list), KeyError happens when it's not a dict.
# Assume both mean we should finish recursing
except (IndexError, KeyError):
d[key] = value
Example:
>>> d = {'a': {'aa':1, 'ab':2}, 'b':{'ba':1, 'bb':2}}
>>> set_item(d, ['a', 'ab'], 50)
>>> print d
{'a': {'aa': 1, 'ab': 50}, 'b': {'ba': 1, 'bb': 2}}
Edit: As Marcin points out below, this will not work for arbitrarily nested dicts since Python has a recursion limit. It's also not for highly performance-sensitive situations (recursion in Python generally isn't). Nonetheless, outside of these two situations I find this to be somewhat more explicit than something involving reduce or lambda.
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()}