Reverse / invert a dictionary mapping

Question:

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'}
Asked By: Brian M. Hunt

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Answers:

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}
Answered By: sykora

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())
Answered By: unbeknown

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()}
Answered By: SilentGhost

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]
Answered By: Robert Rossney

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()))
Answered By: fs.

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.

Answered By: pcv

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).

Answered By: Asclepius

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()) )
Answered By: RussellStewart

Another, more functional, way:

my_map = { 'a': 1, 'b':2 }
dict(map(reversed, my_map.items()))
Answered By: Brendan Maguire

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.

Answered By: cjay

Try this for python 2.7/3.x

inv_map={};
for i in my_map:
    inv_map[my_map[i]]=i    
print inv_map
Answered By: dhvlnyk

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
Answered By: Alf

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.

Answered By: NcAdams

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
Answered By: eyoeldefare

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

Answered By: thodnev

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
Answered By: Taras Voitovych

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().

Answered By: irudyak

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)
Answered By: mveith

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.

Answered By: Ersatz Kwisatz

I would do it that way in python 2.

inv_map = {my_map[x] : x for x in my_map}
Answered By: genghiscrade
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’]}

Answered By: RVR
  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']}
Answered By: Shb8086

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()}
Answered By: SVJ

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]}
Answered By: Andy Jazz

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 😉
Answered By: mit

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()))
Answered By: nauer

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

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'}
Answered By: Mondaa

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.

Answered By: mss

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']}
Answered By: Ani Menon
dict([(value, key) for key, value in d.items()])
Answered By: algorytmus

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
Answered By: user42
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