Related
I have a dict of nested dicts and key-values.
dicta = {'key1': {'keya': 'a', 'keyb': 'b'},
'key2': {'key3':{'keyc': 'c'}}, 'key4': 4}}
Is there a way I can use a for loop to get to 'c'?
I tried,
print(dicta['key2'])
for x in dicta['key2']:
for y in x['key3']:
print(y)
For the first print, I get
{'key3': {'keyc': 'c'}}
But I get TypeError: string indices must be integers for the second print.
Thanks in advance!
*I edited replacing n with dicta; I copied and pasted it wrong the first time.
In general, no. Dicts can be arbitrarily recursively deep, and there's no good way to traverse them using nothing but a for loop. (You could implement your own stack using a list and simulate recursion, but that's not "good".)
There's some recursive code for traversing dictionaries (counting the depth) in this question.
In specific, sure. Knowing the structure in advance, you can use the right number of for loops:
n = {'key1': {'keya': 'a', 'keyb': 'b'},
'key2': {'key3':{'keyc': 'c'}}, 'key4': 4}
for k1,v1 in n.items():
try:
for k2,v2 in v1.items():
try:
for k3,v3 in v2.items():
print(v3)
except AttributeError:
pass
except AttributeError:
pass
for i in somedict is to loop all the keys from a dictionary.
See more details from dict views
When looping through dictionaries, the key and corresponding value can
be retrieved at the same time using the items() method.
items() returns a new view of the dictionary’s items ((key, value) pairs).
So you can just try this:
for k,v in dicta['key2'].items():
if isinstance(v,dict):
for k1,v1 in v.items():
print(v1)
isinstance(v,dict) return true if the object argument is an instance of the dict argument,so you don't need to catch exception.
Have a look at looping-techniques
Hope this helps.
for k in n:
v = n.get(k)
if isinstance(v, dict):
for j in v:
print(v.get(j))
You can use recursion to dive into given dictionary and print the key and its values.
def drill(dicta):
for k in dicta.keys():
if isinstance(dicta[k],dict):
drill(dicta[k])
else:
print ('dicta[',k,']=',dicta[k])
return
Using nested loop you can get the required results.
for k,v in dicta.items():
if k=="key2":
new_dict=v
for k1, v1 in new_dict.items():
print v1["keyc"]
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
I have question about Dictionaries in Python.
here it is:
I have a dict like dict = { 'abc':'a', 'cdf':'b', 'gh':'a', 'fh':'g', 'hfz':'g' }
Now i want to get all Key-Elements by the same value and save it in a new dict.
The new Dict should be look like:
new_dict = { 'b':('cdf'), 'a':('abc','gh'), 'g':('fh','hfz')}
If you are fine with lists instead of tuples in the new dictionary, you can use
from collections import defaultdict
some_dict = { 'abc':'a', 'cdf':'b', 'gh':'a', 'fh':'g', 'hfz':'g' }
new_dict = defaultdict(list)
for k, v in some_dict.iteritems():
new_dict[v].append(k)
If you want to avoid the use of defaultdict, you could also do
new_dict = {}
for k, v in some_dict.iteritems():
new_dict.setdefault(v, []).append(k)
Here's a naive implementation. Someone with better Python skills can probably make it more concise and awesome.
dict = { 'abc':'a', 'cdf':'b', 'gh':'a', 'fh':'g', 'hfz':'g' }
new_dict = {}
for pair in dict.items():
if pair[1] not in new_dict.keys():
new_dict[pair[1]] = []
new_dict[pair[1]].append(pair[0])
print new_dict
This produces
{'a': ['abc', 'gh'], 'b': ['cdf'], 'g': ['fh', 'hfz']}
If you do specifically want tuples as the values in your new dictionary, you can still use defaultdict, and use tuple concatenation. This solution works in Python 3.4+:
from collections import defaultdict
source = {'abc': 'a', 'cdf': 'b', 'gh': 'a', 'fh': 'g', 'hfz': 'g'}
target = defaultdict(tuple)
for key in source:
target[source[key]] += (key, )
print(target)
Which will produce
defaultdict(<class 'tuple'>, {'a': ('abc', 'gh'), 'g': ('fh', 'hfz'), 'b': ('cdf',)})
This will probably be slower than generating a dictionary by list insertion, and will create more objects to be collected. So, you can build your dictionary out of lists, and then map it into tuples:
target2 = defaultdict(list)
for key in source:
target2[source[key]].append(key)
for key in target2:
target2[key] = tuple(target2[key])
print(target2)
Which will give the same result as above.
It can be done this way too, without using any extra functions .
some_dict = { 'abc':'a', 'cdf':'b', 'gh':'a', 'fh':'g', 'hfz':'g' }
new_dict = { }
for keys in some_dict:
new_dict[some_dict[keys]] = [ ]
for keys in some_dict:
new_dict[some_dict[keys]].append(keys)
print(new_dict)
This question already has answers here:
Reverse / invert a dictionary mapping
(32 answers)
Closed 10 months ago.
I receive a dictionary as input, and would like to to return a dictionary whose keys will be the input's values and whose value will be the corresponding input keys. Values are unique.
For example, say my input is:
a = dict()
a['one']=1
a['two']=2
I would like my output to be:
{1: 'one', 2: 'two'}
To clarify I would like my result to be the equivalent of the following:
res = dict()
res[1] = 'one'
res[2] = 'two'
Any neat Pythonic way to achieve this?
Python 2:
res = dict((v,k) for k,v in a.iteritems())
Python 3 (thanks to #erik):
res = dict((v,k) for k,v in a.items())
new_dict = dict(zip(my_dict.values(), my_dict.keys()))
From Python 2.7 on, including 3.0+, there's an arguably shorter, more readable version:
>>> my_dict = {'x':1, 'y':2, 'z':3}
>>> {v: k for k, v in my_dict.items()}
{1: 'x', 2: 'y', 3: 'z'}
You can make use of dict comprehensions:
Python 3
res = {v: k for k, v in a.items()}
Python 2
res = {v: k for k, v in a.iteritems()}
Edited: For Python 3, use a.items() instead of a.iteritems(). Discussions about the differences between them can be found in iteritems in Python on SO.
In [1]: my_dict = {'x':1, 'y':2, 'z':3}
Python 3
In [2]: dict((value, key) for key, value in my_dict.items())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
Python 2
In [2]: dict((value, key) for key, value in my_dict.iteritems())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
The current leading answer assumes values are unique which is not always the case. What if values are not unique? You will loose information!
For example:
d = {'a':3, 'b': 2, 'c': 2}
{v:k for k,v in d.iteritems()}
returns {2: 'b', 3: 'a'}.
The information about 'c' was completely ignored.
Ideally it should had be something like {2: ['b','c'], 3: ['a']}. This is what the bottom implementation does.
Python 2.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.iteritems():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
Python 3.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.items():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
You could try:
Python 3
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.items())
d2
{'two': 2, 'one': 1}
Python 2
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.iteritems())
d2
{'two': 2, 'one': 1}
Beware that you cannot 'reverse' a dictionary if
More than one key shares the same value. For example {'one':1,'two':1}. The new dictionary can only have one item with key 1.
One or more of the values is unhashable. For example {'one':[1]}. [1] is a valid value but not a valid key.
See this thread on the python mailing list for a discussion on the subject.
res = dict(zip(a.values(), a.keys()))
new_dict = dict( (my_dict[k], k) for k in my_dict)
or even better, but only works in Python 3:
new_dict = { my_dict[k]: k for k in my_dict}
Another way to expand on Ilya Prokin's response is to actually use the reversed function.
dict(map(reversed, my_dict.items()))
In essence, your dictionary is iterated through (using .items()) where each item is a key/value pair, and those items are swapped with the reversed function. When this is passed to the dict constructor, it turns them into value/key pairs which is what you want.
Suggestion for an improvement for Javier answer :
dict(zip(d.values(),d))
Instead of d.keys() you can write just d, because if you go through dictionary with an iterator, it will return the keys of the relevant dictionary.
Ex. for this behavior :
d = {'a':1,'b':2}
for k in d:
k
'a'
'b'
Can be done easily with dictionary comprehension:
{d[i]:i for i in d}
dict(map(lambda x: x[::-1], YourDict.items()))
.items() returns a list of tuples of (key, value). map() goes through elements of the list and applies lambda x:[::-1] to each its element (tuple) to reverse it, so each tuple becomes (value, key) in the new list spitted out of map. Finally, dict() makes a dict from the new list.
Hanan's answer is the correct one as it covers more general case (the other answers are kind of misleading for someone unaware of the duplicate situation). An improvement to Hanan's answer is using setdefault:
mydict = {1:a, 2:a, 3:b}
result = {}
for i in mydict:
result.setdefault(mydict[i],[]).append(i)
print(result)
>>> result = {a:[1,2], b:[3]}
Using loop:-
newdict = {} #Will contain reversed key:value pairs.
for key, value in zip(my_dict.keys(), my_dict.values()):
# Operations on key/value can also be performed.
newdict[value] = key
If you're using Python3, it's slightly different:
res = dict((v,k) for k,v in a.items())
Adding an in-place solution:
>>> d = {1: 'one', 2: 'two', 3: 'three', 4: 'four'}
>>> for k in list(d.keys()):
... d[d.pop(k)] = k
...
>>> d
{'two': 2, 'one': 1, 'four': 4, 'three': 3}
In Python3, it is critical that you use list(d.keys()) because dict.keys returns a view of the keys. If you are using Python2, d.keys() is enough.
I find this version the most comprehensive one:
a = {1: 'one', 2: 'two'}
swapped_a = {value : key for key, value in a.items()}
print(swapped_a)
output :
{'one': 1, 'two': 2}
An alternative that is not quite as readable (in my opinion) as some of the other answers:
new_dict = dict(zip(*list(zip(*old_dict.items()))[::-1]))
where list(zip(*old_dict.items()))[::-1] gives a list of 2 tuples, old_dict's values and keys, respectively.
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