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How to create a reverse dictionary that takes in account repeated values?

I am trying to create a function that takes in a dictionary and returns a reverse of it while taking care of repeated values. That is, if the original dictionary would be
original_dict = {'first': ['a'], 'second': ['b', 'c'], 'third': ['d'], 'fourth': ['d']}
the function should return
{'a': ['first'], 'b': ['second'], 'c': ['second'], 'd': ['third', 'fourth']}
I've written
def reversed_dict(d):
new_dict = {}
for keys,values in d.items():
new_dict[values]=keys
but when I try it out with the original dictionary, I get an error "unhashable type: 'list'" when I try out the function. Are there any hints what might be causing it?

You have to iterate over the values in the list as well:
def reversed_dict(d):
new_dict = {}
for keys,values in d.items():
for val in values:
new_dict.setdefault(val, []).append(keys)
return new_dict

You have to iterate over the values and add them as keys. You also have to take into account the possibility that you may have already added a value as a key.
def reversed_dict(d):
new_dict = {}
for keys,values in d.items():
for v in values:
if v in new_dict:
new_dict[v].append(keys)
else:
new_dict[v] = [keys]
return new_dict

Use collections.defaultdict:
from collections import defaultdict
def reversed_dict(d):
new_dict = defaultdict(list)
for key, values in d.items():
for value in values:
new_dict[value].append(key)
return new_dict
The problem with your approach is you're using the entire list as the key of the dictionary. Instead you need to iterate over the list (i.e. for value in values: in the code above.)
defaultdict just makes it simpler to read.

You are getting this error because any of your original_dict values is a mutable type which is, as the error suggests, an unhashable type thus not
avalid candidate for a key in the reversed_dict.
You can workaround this problem by type-checking and casting mutable types into an immutable equivalent, e.g. a list into a tuple.
(also I find dict comp a way more elegant and concise approach):
def reversed_dict(d):
return {v if not isinstance(v, list) else tuple(v): k for k, v in d.items()}

changing a dictionary in python and make a new one

I have a dictionary in python like this example:
small example:
d = {'chr12:112843810-112843817': 'CGTAGAG', 'chr16:48394513-48394520': 'TTTAAAC'}
the value of this dictionary is a sequence of characters and the characters can only be A, C, G or T. but I want to make a new dictionary in which the keys are similar but values are different. in fact in new dictionary the sequence of characters would have 2 changes:
1- it must be reversed (direction). for example "AGT" would be "TGA".
2- I want to convert: A -> T , T -> A , C -> G , G -> C.
considering 2 above items, AGTCC would be converted to GGACT
here is the expected output for the small example:
expected output:
d2 = {'chr12:112843810-112843817': 'CTCTACG', 'chr16:48394513-48394520': 'GTTTAAA'}
I am trying to do that in python using the following code:
def reverse_complement(dict):
for v in dict:
complement = {'A': 'T', 'C': 'G', 'G': 'C', 'T': 'A'}
return [complement[base] for base in v[::-1]]
but it does not return what would like to get. do you know how to fix it?

for v in dict iterates the keys, not the values. You should use items to iterate over key-value pairs.
The return statement causes the function to return after the first iteration.
dict is a bad variable name (d is probably worse, but at least it does not shadow a built-in name)
You can use a dict comphrension:
def reverse_complement(d):
complement = {'A': 'T', 'C': 'G', 'G': 'C', 'T': 'A'}
return {k: ''.join(complement[base] for base in v[::-1]) for k, v in d.items()}
or with str.translate:
def reverse_complement(d):
table = str.maketrans('ACGT', 'TGCA')
return {k: v[::-1].translate(table) for k, v in d.items()}

Here another solution with string replacement methods. This might be faster.
the complement method does the inversion by fist setting the string to lower case and then doing the renaming based on lower to upper case.
The "[::-1]" reverses the order and the rest is a dictionary comprehension, that creates a new dictionary
complement = lambda s: s.lower().replace("g","C").replace("c","G").replace("t","A").replace("a","T")
d2 = { k:complement(v)[::-1] for k,v in d.items() }

Similar to first answer by #DeepSpace, you could map the complements in-place with reversed():
d = {'chr12:112843810-112843817': 'CGTAGAG', 'chr16:48394513-48394520': 'TTTAAAC'}
complements = {'A': 'T', 'T': 'A', 'C': 'G', 'G': 'C'}
for k, v in d.items():
d[k] = ''.join(map(complements.get, reversed(v)))
print(d)
# {'chr12:112843810-112843817': 'CTCTACG', 'chr16:48394513-48394520': 'GTTTAAA'}
You can also rebuild new dictionary:
d2 = {k: ''.join(map(complements.get, reversed(v))) for k, v in d.items()}
print(d2)
# {'chr12:112843810-112843817': 'CTCTACG', 'chr16:48394513-48394520': 'GTTTAAA'}
Note: ''.join(map(complements.get, reversed(v))) is just another way of writing ''.join(complemements.get(x) for x in reversed(v)).

Deleting dictionary keys from a provided list in Python [duplicate]

I know how to remove an entry, 'key' from my dictionary d, safely. You do:
if d.has_key('key'):
del d['key']
However, I need to remove multiple entries from a dictionary safely. I was thinking of defining the entries in a tuple as I will need to do this more than once.
entities_to_remove = ('a', 'b', 'c')
for x in entities_to_remove:
if x in d:
del d[x]
However, I was wondering if there is a smarter way to do this?

Using dict.pop:
d = {'some': 'data'}
entries_to_remove = ('any', 'iterable')
for k in entries_to_remove:
d.pop(k, None)

Using Dict Comprehensions
final_dict = {key: value for key, value in d if key not in [key1, key2]}
where key1 and key2 are to be removed.
In the example below, keys "b" and "c" are to be removed & it's kept in a keys list.
>>> a
{'a': 1, 'c': 3, 'b': 2, 'd': 4}
>>> keys = ["b", "c"]
>>> print {key: a[key] for key in a if key not in keys}
{'a': 1, 'd': 4}
>>>

Why not like this:
entries = ('a', 'b', 'c')
the_dict = {'b': 'foo'}
def entries_to_remove(entries, the_dict):
for key in entries:
if key in the_dict:
del the_dict[key]
A more compact version was provided by mattbornski using dict.pop()

a solution is using map and filter functions
python 2
d={"a":1,"b":2,"c":3}
l=("a","b","d")
map(d.__delitem__, filter(d.__contains__,l))
print(d)
python 3
d={"a":1,"b":2,"c":3}
l=("a","b","d")
list(map(d.__delitem__, filter(d.__contains__,l)))
print(d)
you get:
{'c': 3}

If you also need to retrieve the values for the keys you are removing, this would be a pretty good way to do it:
values_removed = [d.pop(k, None) for k in entities_to_remove]
You could of course still do this just for the removal of the keys from d, but you would be unnecessarily creating the list of values with the list comprehension. It is also a little unclear to use a list comprehension just for the function's side effect.

Found a solution with pop and map
d = {'a': 'valueA', 'b': 'valueB', 'c': 'valueC', 'd': 'valueD'}
keys = ['a', 'b', 'c']
list(map(d.pop, keys))
print(d)
The output of this:
{'d': 'valueD'}
I have answered this question so late just because I think it will help in the future if anyone searches the same. And this might help.
Update
The above code will throw an error if a key does not exist in the dict.
DICTIONARY = {'a': 'valueA', 'b': 'valueB', 'c': 'valueC', 'd': 'valueD'}
keys = ['a', 'l', 'c']
def remove_key(key):
DICTIONARY.pop(key, None)
list(map(remove_key, keys))
print(DICTIONARY)
output:
DICTIONARY = {'b': 'valueB', 'd': 'valueD'}

Some timing tests for cpython 3 shows that a simple for loop is the fastest way, and it's quite readable. Adding in a function doesn't cause much overhead either:
timeit results (10k iterations):
all(x.pop(v) for v in r) # 0.85
all(map(x.pop, r)) # 0.60
list(map(x.pop, r)) # 0.70
all(map(x.__delitem__, r)) # 0.44
del_all(x, r) # 0.40
<inline for loop>(x, r) # 0.35
def del_all(mapping, to_remove):
"""Remove list of elements from mapping."""
for key in to_remove:
del mapping[key]
For small iterations, doing that 'inline' was a bit faster, because of the overhead of the function call. But del_all is lint-safe, reusable, and faster than all the python comprehension and mapping constructs.

I have no problem with any of the existing answers, but I was surprised to not find this solution:
keys_to_remove = ['a', 'b', 'c']
my_dict = {k: v for k, v in zip("a b c d e f g".split(' '), [0, 1, 2, 3, 4, 5, 6])}
for k in keys_to_remove:
try:
del my_dict[k]
except KeyError:
pass
assert my_dict == {'d': 3, 'e': 4, 'f': 5, 'g': 6}
Note: I stumbled across this question coming from here. And my answer is related to this answer.

I have tested the performance of three methods:
# Method 1: `del`
for key in remove_keys:
if key in d:
del d[key]
# Method 2: `pop()`
for key in remove_keys:
d.pop(key, None)
# Method 3: comprehension
{key: v for key, v in d.items() if key not in remove_keys}
Here are the results of 1M iterations:
del: 2.03s 2.0 ns/iter (100%)
pop(): 2.38s 2.4 ns/iter (117%)
comprehension: 4.11s 4.1 ns/iter (202%)
So both del and pop() are the fastest. Comprehensions are 2x slower.
But anyway, we speak nanoseconds here :) Dicts in Python are ridiculously fast.

Why not:
entriestoremove = (2,5,1)
for e in entriestoremove:
if d.has_key(e):
del d[e]
I don't know what you mean by "smarter way". Surely there are other ways, maybe with dictionary comprehensions:
entriestoremove = (2,5,1)
newdict = {x for x in d if x not in entriestoremove}

inline
import functools
#: not key(c) in d
d = {"a": "avalue", "b": "bvalue", "d": "dvalue"}
entitiesToREmove = ('a', 'b', 'c')
#: python2
map(lambda x: functools.partial(d.pop, x, None)(), entitiesToREmove)
#: python3
list(map(lambda x: functools.partial(d.pop, x, None)(), entitiesToREmove))
print(d)
# output: {'d': 'dvalue'}

I think using the fact that the keys can be treated as a set is the nicest way if you're on python 3:
def remove_keys(d, keys):
to_remove = set(keys)
filtered_keys = d.keys() - to_remove
filtered_values = map(d.get, filtered_keys)
return dict(zip(filtered_keys, filtered_values))
Example:
>>> remove_keys({'k1': 1, 'k3': 3}, ['k1', 'k2'])
{'k3': 3}

It would be nice to have full support for set methods for dictionaries (and not the unholy mess we're getting with Python 3.9) so that you could simply "remove" a set of keys. However, as long as that's not the case, and you have a large dictionary with potentially a large number of keys to remove, you might want to know about the performance. So, I've created some code that creates something large enough for meaningful comparisons: a 100,000 x 1000 matrix, so 10,000,00 items in total.
from itertools import product
from time import perf_counter
# make a complete worksheet 100000 * 1000
start = perf_counter()
prod = product(range(1, 100000), range(1, 1000))
cells = {(x,y):x for x,y in prod}
print(len(cells))
print(f"Create time {perf_counter()-start:.2f}s")
clock = perf_counter()
# remove everything above row 50,000
keys = product(range(50000, 100000), range(1, 100))
# for x,y in keys:
# del cells[x, y]
for n in map(cells.pop, keys):
pass
print(len(cells))
stop = perf_counter()
print(f"Removal time {stop-clock:.2f}s")
10 million items or more is not unusual in some settings. Comparing the two methods on my local machine I see a slight improvement when using map and pop, presumably because of fewer function calls, but both take around 2.5s on my machine. But this pales in comparison to the time required to create the dictionary in the first place (55s), or including checks within the loop. If this is likely then its best to create a set that is a intersection of the dictionary keys and your filter:
keys = cells.keys() & keys
In summary: del is already heavily optimised, so don't worry about using it.

Another map() way to remove list of keys from dictionary
and avoid raising KeyError exception
dic = {
'key1': 1,
'key2': 2,
'key3': 3,
'key4': 4,
'key5': 5,
}
keys_to_remove = ['key_not_exist', 'key1', 'key2', 'key3']
k = list(map(dic.pop, keys_to_remove, keys_to_remove))
print('k=', k)
print('dic after = \n', dic)
**this will produce output**
k= ['key_not_exist', 1, 2, 3]
dic after = {'key4': 4, 'key5': 5}
Duplicate keys_to_remove is artificial, it needs to supply defaults values for dict.pop() function.
You can add here any array with len_ = len(key_to_remove)
For example
dic = {
'key1': 1,
'key2': 2,
'key3': 3,
'key4': 4,
'key5': 5,
}
keys_to_remove = ['key_not_exist', 'key1', 'key2', 'key3']
k = list(map(dic.pop, keys_to_remove, np.zeros(len(keys_to_remove))))
print('k=', k)
print('dic after = ', dic)
** will produce output **
k= [0.0, 1, 2, 3]
dic after = {'key4': 4, 'key5': 5}

def delete_keys_from_dict(dictionary, keys):
"""
Deletes the unwanted keys in the dictionary
:param dictionary: dict
:param keys: list of keys
:return: dict (modified)
"""
from collections.abc import MutableMapping
keys_set = set(keys)
modified_dict = {}
for key, value in dictionary.items():
if key not in keys_set:
if isinstance(value, list):
modified_dict[key] = list()
for x in value:
if isinstance(x, MutableMapping):
modified_dict[key].append(delete_keys_from_dict(x, keys_set))
else:
modified_dict[key].append(x)
elif isinstance(value, MutableMapping):
modified_dict[key] = delete_keys_from_dict(value, keys_set)
else:
modified_dict[key] = value
return modified_dict
_d = {'a': 1245, 'b': 1234325, 'c': {'a': 1245, 'b': 1234325}, 'd': 98765,
'e': [{'a': 1245, 'b': 1234325},
{'a': 1245, 'b': 1234325},
{'t': 767}]}
_output = delete_keys_from_dict(_d, ['a', 'b'])
_expected = {'c': {}, 'd': 98765, 'e': [{}, {}, {'t': 767}]}
print(_expected)
print(_output)

I'm late to this discussion but for anyone else. A solution may be to create a list of keys as such.
k = ['a','b','c','d']
Then use pop() in a list comprehension, or for loop, to iterate over the keys and pop one at a time as such.
new_dictionary = [dictionary.pop(x, 'n/a') for x in k]
The 'n/a' is in case the key does not exist, a default value needs to be returned.

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

How to swap keys for values in a dictionary [duplicate]

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.