反转/反转字典映射

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'}

For Python 2.7.x

inv_map = {v: k for k, v in my_map.iteritems()}

For Python 3+:

inv_map = {v: k for k, v in my_map.items()}

Assuming that the values in the dict are unique:

dict((v, k) for k, v in my_map.iteritems())

If the values in my_map aren’t unique:

inv_map = {}
for k, v in my_map.iteritems():
    inv_map[v] = inv_map.get(v, [])
    inv_map[v].append(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()))

This expands upon the answer by Robert, applying to when the values in the dict aren’t unique.

class ReversibleDict(dict):

    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.iteritems():
            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).

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().

For instance, you have the following dictionary:

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 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 dict.items()}

Third Solution. Use reverting the inversion approach (relies on second solution):

# Use this code to invert dictionaries that have lists of values

dict = {value: key for key in inverted_dict for value in my_map[key]}

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()}

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.

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.

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.

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

I would do it that way in python 2.

inv_map = {my_map[x] : x for x in my_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’]}

  def reverse_dictionary(input_dict):
      out = {}
      for v in input_dict.values():  
          for value in v:
              if value not in out:
                  out[value.lower()] = []

      for i in input_dict:
          for j in out:
              if j in map (lambda x : x.lower(),input_dict[i]):
                  out[j].append(i.lower())
                  out[j].sort()
      return out

this code do like this:

r = reverse_dictionary({'Accurate': ['exact', 'precise'], 'exact': ['precise'], 'astute': ['Smart', 'clever'], 'smart': ['clever', 'bright', 'talented']})

print(r)

{'precise': ['accurate', 'exact'], 'clever': ['astute', 'smart'], 'talented': ['smart'], 'bright': ['smart'], 'exact': ['accurate'], 'smart': ['astute']}

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

I wrote this with the help of cycle ‘for’ and method ‘.get()’ and I changed the name ‘map’ of the dictionary to ‘map1’ because ‘map’ is a function.

def dict_invert(map1):
    inv_map = {} # new dictionary
    for key in map1.keys():
        inv_map[map1.get(key)] = key
    return inv_map

If values aren’t unique AND may be a hash (one dimension):

for k, v in myDict.items():
    if len(v) > 1:
        for item in v:
            invDict[item] = invDict.get(item, [])
            invDict[item].append(k)
    else:
        invDict[v] = invDict.get(v, [])
        invDict[v].append(k)

And with a recursion if you need to dig deeper then just one dimension:

def digList(lst):
    temp = []
    for item in lst:
        if type(item) is list:
            temp.append(digList(item))
        else:
            temp.append(item)
    return set(temp)

for k, v in myDict.items():
    if type(v) is list:
        items = digList(v)
        for item in items:
            invDict[item] = invDict.get(item, [])
            invDict[item].append(k)
    else:
        invDict[v] = invDict.get(v, [])
        invDict[v].append(k)