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In Python 3, how to copy a key-value mapping from one dict to another, including remove if necessary?

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}

Comparing values in a dictionary and grouping them again in Python [duplicate]

Given a dictionary like so:
my_map = {'a': 1, 'b': 2}
How can one invert this map to get:
inv_map = {1: 'a', 2: 'b'}

Python 3+:
inv_map = {v: k for k, v in my_map.items()}
Python 2:
inv_map = {v: k for k, v in my_map.iteritems()}

Assuming that the values in the dict are unique:
Python 3:
dict((v, k) for k, v in my_map.items())
Python 2:
dict((v, k) for k, v in my_map.iteritems())

If the values in my_map aren't unique:
Python 3:
inv_map = {}
for k, v in my_map.items():
inv_map[v] = inv_map.get(v, []) + [k]
Python 2:
inv_map = {}
for k, v in my_map.iteritems():
inv_map[v] = inv_map.get(v, []) + [k]

To do this while preserving the type of your mapping (assuming that it is a dict or a dict subclass):
def inverse_mapping(f):
return f.__class__(map(reversed, f.items()))

Try this:
inv_map = dict(zip(my_map.values(), my_map.keys()))
(Note that the Python docs on dictionary views explicitly guarantee that .keys() and .values() have their elements in the same order, which allows the approach above to work.)
Alternatively:
inv_map = dict((my_map[k], k) for k in my_map)
or using python 3.0's dict comprehensions
inv_map = {my_map[k] : k for k in my_map}

Another, more functional, way:
my_map = { 'a': 1, 'b':2 }
dict(map(reversed, my_map.items()))

We can also reverse a dictionary with duplicate keys using defaultdict:
from collections import Counter, defaultdict
def invert_dict(d):
d_inv = defaultdict(list)
for k, v in d.items():
d_inv[v].append(k)
return d_inv
text = 'aaa bbb ccc ddd aaa bbb ccc aaa'
c = Counter(text.split()) # Counter({'aaa': 3, 'bbb': 2, 'ccc': 2, 'ddd': 1})
dict(invert_dict(c)) # {1: ['ddd'], 2: ['bbb', 'ccc'], 3: ['aaa']}
See here:
This technique is simpler and faster than an equivalent technique using dict.setdefault().

This expands upon the answer by Robert, applying to when the values in the dict aren't unique.
class ReversibleDict(dict):
# Ref: https://stackoverflow.com/a/13057382/
def reversed(self):
"""
Return a reversed dict, with common values in the original dict
grouped into a list in the returned dict.
Example:
>>> d = ReversibleDict({'a': 3, 'c': 2, 'b': 2, 'e': 3, 'd': 1, 'f': 2})
>>> d.reversed()
{1: ['d'], 2: ['c', 'b', 'f'], 3: ['a', 'e']}
"""
revdict = {}
for k, v in self.items():
revdict.setdefault(v, []).append(k)
return revdict
The implementation is limited in that you cannot use reversed twice and get the original back. It is not symmetric as such. It is tested with Python 2.6. Here is a use case of how I am using to print the resultant dict.
If you'd rather use a set than a list, and there could exist unordered applications for which this makes sense, instead of setdefault(v, []).append(k), use setdefault(v, set()).add(k).

Combination of list and dictionary comprehension. Can handle duplicate keys
{v:[i for i in d.keys() if d[i] == v ] for k,v in d.items()}

A case where the dictionary values is a set. Like:
some_dict = {"1":{"a","b","c"},
"2":{"d","e","f"},
"3":{"g","h","i"}}
The inverse would like:
some_dict = {vi: k for k, v in some_dict.items() for vi in v}
The output is like this:
{'c': '1',
'b': '1',
'a': '1',
'f': '2',
'd': '2',
'e': '2',
'g': '3',
'h': '3',
'i': '3'}

For instance, you have the following dictionary:
my_dict = {'a': 'fire', 'b': 'ice', 'c': 'fire', 'd': 'water'}
And you wanna get it in such an inverted form:
inverted_dict = {'fire': ['a', 'c'], 'ice': ['b'], 'water': ['d']}
First Solution. For inverting key-value pairs in your dictionary use a for-loop approach:
# Use this code to invert dictionaries that have non-unique values
inverted_dict = dict()
for key, value in my_dict.items():
inverted_dict.setdefault(value, list()).append(key)
Second Solution. Use a dictionary comprehension approach for inversion:
# Use this code to invert dictionaries that have unique values
inverted_dict = {value: key for key, value in my_dict.items()}
Third Solution. Use reverting the inversion approach (relies on the second solution):
# Use this code to invert dictionaries that have lists of values
my_dict = {value: key for key in inverted_dict for value in my_map[key]}

Lot of answers but didn't find anything clean in case we are talking about a dictionary with non-unique values.
A solution would be:
from collections import defaultdict
inv_map = defaultdict(list)
for k, v in my_map.items():
inv_map[v].append(k)
Example:
If initial dict my_map = {'c': 1, 'd': 5, 'a': 5, 'b': 10}
then, running the code above will give:
{5: ['a', 'd'], 1: ['c'], 10: ['b']}

I found that this version is more than 10% faster than the accepted version of a dictionary with 10000 keys.
d = {i: str(i) for i in range(10000)}
new_d = dict(zip(d.values(), d.keys()))

In addition to the other functions suggested above, if you like lambdas:
invert = lambda mydict: {v:k for k, v in mydict.items()}
Or, you could do it this way too:
invert = lambda mydict: dict( zip(mydict.values(), mydict.keys()) )

I think the best way to do this is to define a class. Here is an implementation of a "symmetric dictionary":
class SymDict:
def __init__(self):
self.aToB = {}
self.bToA = {}
def assocAB(self, a, b):
# Stores and returns a tuple (a,b) of overwritten bindings
currB = None
if a in self.aToB: currB = self.bToA[a]
currA = None
if b in self.bToA: currA = self.aToB[b]
self.aToB[a] = b
self.bToA[b] = a
return (currA, currB)
def lookupA(self, a):
if a in self.aToB:
return self.aToB[a]
return None
def lookupB(self, b):
if b in self.bToA:
return self.bToA[b]
return None
Deletion and iteration methods are easy enough to implement if they're needed.
This implementation is way more efficient than inverting an entire dictionary (which seems to be the most popular solution on this page). Not to mention, you can add or remove values from your SymDict as much as you want, and your inverse-dictionary will always stay valid -- this isn't true if you simply reverse the entire dictionary once.

If the values aren't unique, and you're a little hardcore:
inv_map = dict(
(v, [k for (k, xx) in filter(lambda (key, value): value == v, my_map.items())])
for v in set(my_map.values())
)
Especially for a large dict, note that this solution is far less efficient than the answer Python reverse / invert a mapping because it loops over items() multiple times.

This handles non-unique values and retains much of the look of the unique case.
inv_map = {v:[k for k in my_map if my_map[k] == v] for v in my_map.itervalues()}
For Python 3.x, replace itervalues with values.

I am aware that this question already has many good answers, but I wanted to share this very neat solution that also takes care of duplicate values:
def dict_reverser(d):
seen = set()
return {v: k for k, v in d.items() if v not in seen or seen.add(v)}
This relies on the fact that set.add always returns None in Python.

Here is another way to do it.
my_map = {'a': 1, 'b': 2}
inv_map= {}
for key in my_map.keys() :
val = my_map[key]
inv_map[val] = key

dict([(value, key) for key, value in d.items()])

Function is symmetric for values of type list; Tuples are coverted to lists when performing reverse_dict(reverse_dict(dictionary))
def reverse_dict(dictionary):
reverse_dict = {}
for key, value in dictionary.iteritems():
if not isinstance(value, (list, tuple)):
value = [value]
for val in value:
reverse_dict[val] = reverse_dict.get(val, [])
reverse_dict[val].append(key)
for key, value in reverse_dict.iteritems():
if len(value) == 1:
reverse_dict[key] = value[0]
return reverse_dict

Since dictionaries require one unique key within the dictionary unlike values, we have to append the reversed values into a list of sort to be included within the new specific keys.
def r_maping(dictionary):
List_z=[]
Map= {}
for z, x in dictionary.iteritems(): #iterate through the keys and values
Map.setdefault(x,List_z).append(z) #Setdefault is the same as dict[key]=default."The method returns the key value available in the dictionary and if given key is not available then it will return provided default value. Afterward, we will append into the default list our new values for the specific key.
return Map

Fast functional solution for non-bijective maps (values not unique):
from itertools import imap, groupby
def fst(s):
return s[0]
def snd(s):
return s[1]
def inverseDict(d):
"""
input d: a -> b
output : b -> set(a)
"""
return {
v : set(imap(fst, kv_iter))
for (v, kv_iter) in groupby(
sorted(d.iteritems(),
key=snd),
key=snd
)
}
In theory this should be faster than adding to the set (or appending to the list) one by one like in the imperative solution.
Unfortunately the values have to be sortable, the sorting is required by groupby.

Try this for python 2.7/3.x
inv_map={};
for i in my_map:
inv_map[my_map[i]]=i
print inv_map

def invertDictionary(d):
myDict = {}
for i in d:
value = d.get(i)
myDict.setdefault(value,[]).append(i)
return myDict
print invertDictionary({'a':1, 'b':2, 'c':3 , 'd' : 1})
This will provide output as : {1: ['a', 'd'], 2: ['b'], 3: ['c']}

A lambda solution for current python 3.x versions:
d1 = dict(alice='apples', bob='bananas')
d2 = dict(map(lambda key: (d1[key], key), d1.keys()))
print(d2)
Result:
{'apples': 'alice', 'bananas': 'bob'}
This solution does not check for duplicates.
Some remarks:
The lambda construct can access d1 from the outer scope, so we only
pass in the current key. It returns a tuple.
The dict() constructor accepts a list of tuples. It
also accepts the result of a map, so we can skip the conversion to a
list.
This solution has no explicit for loop. It also avoids using a list comprehension for those who are bad at math ;-)

Taking up the highly voted answer starting If the values in my_map aren't unique:, I had a problem where not only the values were not unique, but in addition, they were a list, with each item in the list consisting again of a list of three elements: a string value, a number, and another number.
Example:
mymap['key1'] gives you:
[('xyz', 1, 2),
('abc', 5, 4)]
I wanted to switch only the string value with the key, keeping the two number elements at the same place. You simply need another nested for loop then:
inv_map = {}
for k, v in my_map.items():
for x in v:
# with x[1:3] same as x[1], x[2]:
inv_map[x[0]] = inv_map.get(x[0], []) + [k, x[1:3]]
Example:
inv_map['abc'] now gives you:
[('key1', 1, 2),
('key1', 5, 4)]

This works even if you have non-unique values in the original dictionary.
def dict_invert(d):
'''
d: dict
Returns an inverted dictionary
'''
# Your code here
inv_d = {}
for k, v in d.items():
if v not in inv_d.keys():
inv_d[v] = [k]
else:
inv_d[v].append(k)
inv_d[v].sort()
print(f"{inv_d[v]} are the values")
return inv_d

I would do it that way in python 2.
inv_map = {my_map[x] : x for x in my_map}

Not something completely different, just a bit rewritten recipe from Cookbook. It's futhermore optimized by retaining setdefault method, instead of each time getting it through the instance:
def inverse(mapping):
'''
A function to inverse mapping, collecting keys with simillar values
in list. Careful to retain original type and to be fast.
>> d = dict(a=1, b=2, c=1, d=3, e=2, f=1, g=5, h=2)
>> inverse(d)
{1: ['f', 'c', 'a'], 2: ['h', 'b', 'e'], 3: ['d'], 5: ['g']}
'''
res = {}
setdef = res.setdefault
for key, value in mapping.items():
setdef(value, []).append(key)
return res if mapping.__class__==dict else mapping.__class__(res)
Designed to be run under CPython 3.x, for 2.x replace mapping.items() with mapping.iteritems()
On my machine runs a bit faster, than other examples here

Python: Easily access deeply nested dict (get and set)

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.

Destructuring-bind dictionary contents

I am trying to 'destructure' a dictionary and associate values with variables names after its keys. Something like
params = {'a':1,'b':2}
a,b = params.values()
But since dictionaries are not ordered, there is no guarantee that params.values() will return values in the order of (a, b). Is there a nice way to do this?

from operator import itemgetter
params = {'a': 1, 'b': 2}
a, b = itemgetter('a', 'b')(params)
Instead of elaborate lambda functions or dictionary comprehension, may as well use a built in library.

One way to do this with less repetition than Jochen's suggestion is with a helper function. This gives the flexibility to list your variable names in any order and only destructure a subset of what is in the dict:
pluck = lambda dict, *args: (dict[arg] for arg in args)
things = {'blah': 'bleh', 'foo': 'bar'}
foo, blah = pluck(things, 'foo', 'blah')
Also, instead of joaquin's OrderedDict you could sort the keys and get the values. The only catches are you need to specify your variable names in alphabetical order and destructure everything in the dict:
sorted_vals = lambda dict: (t[1] for t in sorted(dict.items()))
things = {'foo': 'bar', 'blah': 'bleh'}
blah, foo = sorted_vals(things)

How come nobody posted the simplest approach?
params = {'a':1,'b':2}
a, b = params['a'], params['b']

Python is only able to "destructure" sequences, not dictionaries. So, to write what you want, you will have to map the needed entries to a proper sequence. As of myself, the closest match I could find is the (not very sexy):
a,b = [d[k] for k in ('a','b')]
This works with generators too:
a,b = (d[k] for k in ('a','b'))
Here is a full example:
>>> d = dict(a=1,b=2,c=3)
>>> d
{'a': 1, 'c': 3, 'b': 2}
>>> a, b = [d[k] for k in ('a','b')]
>>> a
1
>>> b
2
>>> a, b = (d[k] for k in ('a','b'))
>>> a
1
>>> b
2

Here's another way to do it similarly to how a destructuring assignment works in JS:
params = {'b': 2, 'a': 1}
a, b, rest = (lambda a, b, **rest: (a, b, rest))(**params)
What we did was to unpack the params dictionary into key values (using **) (like in Jochen's answer), then we've taken those values in the lambda signature and assigned them according to the key name - and here's a bonus - we also get a dictionary of whatever is not in the lambda's signature so if you had:
params = {'b': 2, 'a': 1, 'c': 3}
a, b, rest = (lambda a, b, **rest: (a, b, rest))(**params)
After the lambda has been applied, the rest variable will now contain:
{'c': 3}
Useful for omitting unneeded keys from a dictionary.
Hope this helps.

Maybe you really want to do something like this?
def some_func(a, b):
print a,b
params = {'a':1,'b':2}
some_func(**params) # equiv to some_func(a=1, b=2)

If you are afraid of the issues involved in the use of the locals dictionary and you prefer to follow your original strategy, Ordered Dictionaries from python 2.7 and 3.1 collections.OrderedDicts allows you to recover you dictionary items in the order in which they were first inserted

(Ab)using the import system
The from ... import statement lets us desctructure and bind attribute names of an object. Of course, it only works for objects in the sys.modules dictionary, so one could use a hack like this:
import sys, types
mydict = {'a':1,'b':2}
sys.modules["mydict"] = types.SimpleNamespace(**mydict)
from mydict import a, b
A somewhat more serious hack would be to write a context manager to load and unload the module:
with obj_as_module(mydict, "mydict_module"):
from mydict_module import a, b
By pointing the __getattr__ method of the module directly to the __getitem__ method of the dict, the context manager can also avoid using SimpleNamespace(**mydict).
See this answer for an implementation and some extensions of the idea.
One can also temporarily replace the entire sys.modules dict with the dict of interest, and do import a, b without from.

Warning 1: as stated in the docs, this is not guaranteed to work on all Python implementations:
CPython implementation detail: This function relies on Python stack frame support
in the interpreter, which isn’t guaranteed to exist in all implementations
of Python. If running in an implementation without Python stack frame support
this function returns None.
Warning 2: this function does make the code shorter, but it probably contradicts the Python philosophy of being as explicit as you can. Moreover, it doesn't address the issues pointed out by John Christopher Jones in the comments, although you could make a similar function that works with attributes instead of keys. This is just a demonstration that you can do that if you really want to!
def destructure(dict_):
if not isinstance(dict_, dict):
raise TypeError(f"{dict_} is not a dict")
# the parent frame will contain the information about
# the current line
parent_frame = inspect.currentframe().f_back
# so we extract that line (by default the code context
# only contains the current line)
(line,) = inspect.getframeinfo(parent_frame).code_context
# "hello, key = destructure(my_dict)"
# -> ("hello, key ", "=", " destructure(my_dict)")
lvalues, _equals, _rvalue = line.strip().partition("=")
# -> ["hello", "key"]
keys = [s.strip() for s in lvalues.split(",") if s.strip()]
if missing := [key for key in keys if key not in dict_]:
raise KeyError(*missing)
for key in keys:
yield dict_[key]
In [5]: my_dict = {"hello": "world", "123": "456", "key": "value"}
In [6]: hello, key = destructure(my_dict)
In [7]: hello
Out[7]: 'world'
In [8]: key
Out[8]: 'value'
This solution allows you to pick some of the keys, not all, like in JavaScript. It's also safe for user-provided dictionaries

With Python 3.10, you can do:
d = {"a": 1, "b": 2}
match d:
case {"a": a, "b": b}:
print(f"A is {a} and b is {b}")
but it adds two extra levels of indentation, and you still have to repeat the key names.

Look for other answers as this won't cater to the unexpected order in the dictionary. will update this with a correct version sometime soon.
try this
data = {'a':'Apple', 'b':'Banana','c':'Carrot'}
keys = data.keys()
a,b,c = [data[k] for k in keys]
result:
a == 'Apple'
b == 'Banana'
c == 'Carrot'

Well, if you want these in a class you can always do this:
class AttributeDict(dict):
def __init__(self, *args, **kwargs):
super(AttributeDict, self).__init__(*args, **kwargs)
self.__dict__.update(self)
d = AttributeDict(a=1, b=2)

Based on #ShawnFumo answer I came up with this:
def destruct(dict): return (t[1] for t in sorted(dict.items()))
d = {'b': 'Banana', 'c': 'Carrot', 'a': 'Apple' }
a, b, c = destruct(d)
(Notice the order of items in dict)

An old topic, but I found this to be a useful method:
data = {'a':'Apple', 'b':'Banana','c':'Carrot'}
for key in data.keys():
locals()[key] = data[key]
This method loops over every key in your dictionary and sets a variable to that name and then assigns the value from the associated key to this new variable.
Testing:
print(a)
print(b)
print(c)
Output
Apple
Banana
Carrot

An easy and simple way to destruct dict in python:
params = {"a": 1, "b": 2}
a, b = [params[key] for key in ("a", "b")]
print(a, b)
# Output:
# 1 2

I don't know whether it's good style, but
locals().update(params)
will do the trick. You then have a, b and whatever was in your params dict available as corresponding local variables.

Since dictionaries are guaranteed to keep their insertion order in Python >= 3.7, that means that it's complete safe and idiomatic to just do this nowadays:
params = {'a': 1, 'b': 2}
a, b = params.values()
print(a)
print(b)
Output:
1
2

Two way/reverse map [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()}