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I have two dictionaries, RFQDict and AwardsDict. I want to take the keys of RFQdict and search through AwardsDict values for matches.
So I tried this
RFQDict = {}
AwardsDict = {}
# Fetch RFQ
RFQref = db.reference('TestRFQ')
snapshot = RFQref.get()
for key, val in snapshot.items():
RFQDict[key] = val
print('{0} => {1}'.format(key, val))
Awardsref = db.reference('DibbsAwards')
dsnapshot = Awardsref.get()
for key, val in dsnapshot.items():
AwardsDict[key] = val
print('{0} => {1}'.format(key, val))
for key in RFQDict:
if key in AwardsDict.values():
print(key+ " Match found")
is this the way to do it or there is a better way and how could return the key and values where the match was found?
In python3 you can do AwardsDict.values() & RFQDict.keys() and you will get a set with the common key/values.
The '&' operator is used for set intersection and works with the dictionary views returned by values() and keys(). More information of the view returned by those methods: https://docs.python.org/3/library/stdtypes.html?highlight=dictview#dictionary-view-objects
If you want to store the keys and values that match, it would probably be best to store the key and value from the second dictionary since if you just store the matching key and value you will have elements like (a, a) which won't really tell you much about where they matched in the second dictionary. Maybe something like this
d1 = {'a': 1, 'b': 2, 'c': 3}
d2 = {'x': 'a', 'y': 1, 'z': 'c'}
res = [(i, {j: d2[j]}) for i in d1 for j in d2 if i == d2[j]]
print(res)
# [('a': {'x': 'a'}), ('c': {'z': 'c'})]
I would do a list comprehension:
result=[x for x in AwardsDict.values() if x in RFQDict.keys() ]
This way you get a list keeping the duplicates. That is, if a RFQ key is presented in more than one value in AwardsDict. With the & operator you loss that information (as sets only have unique elements).
For example:
RFQDict = {}
AwardsDict = {}
for i in range(5):
RFQDict[i]=0
for i in range(5):
AwardsDict[i]=i
for i in range(5,11):
AwardsDict[i]=i//2 #integer division, i=8 and i=9 get a value of 4
result=[x for x in AwardsDict.values() if x in RFQDict.keys() ]
print('{}'.format(result))
#output [0, 1, 2, 3, 4, 2, 3, 3, 4, 4]
I have a dictionary that consists of numbers and their value
dict = {1:5, 2:5, 3:5}
I have an array with some numbers
arr = [1,2]
What I want to do is:
iterate through the dict and the array
where the dictionary value is equal to the number in the array, set the dictionary value to zero
any value in the dictionary for which there isn't a value in the array matching it, add 1
so in the above example, I should end up with
arr = [1,2]
dict = {1:0, 2:0, 3:6}
The bit I am getting stuck on is creating a variable from the array value and accessing that particular number in the dictionary - using dict[i] for example
arr = [1,2]
data = {1:0, 2:0, 3:6} # don't call it dict because it shadow build-in class
unique = set(arr) # speed up search in case if arr is big
# readable
for k, v in data.items():
if k in unique:
data[k] = 0
else:
data[k] += 1
# oneliner
data = {k: (0 if k in unique else v + 1) for v, k in data.items()}
Additional example:
for a, b, c in [(1,2,3), (4,5,6)]:
print('-',a,b,c)
# will print:
# - 1 2 3
# - 4 5 6
You just need a dict-comprehension that will re-built your dictionary with an if condition for the value part.
my_dict = {1:5, 2:5, 3:5}
arr = [1,2]
my_dict = {k: (0 if k in arr else v+1) for k, v in my_dict.items()}
print(my_dict) # {1: 0, 2: 0, 3: 6}
Note that I have re-named the dictionary from dict to my_dict. That is because by using dict you are overwriting the Python built-in called dict. And you do not want to do that.
Theirs always the dict(map()) approach, which rebuilds a new dictionary with new values to each of the keys:
>>> d = {1:5, 2:5, 3:5}
>>> arr = {1, 2}
>>> dict(map(lambda x: (x[0], 0) if x[0] in arr else (x[0], x[1]+1), d.items()))
{1: 0, 2: 0, 3: 6}
This works because wrapping dict() will automatically convert mapped 2-tuples to a dictionary.
Also you should not use dict as a variable name, since it shadows the builtin dict.
Just use .update method :
dict_1 = {1:5, 2:5, 3:5}
arr = [1,2]
for i in dict_1:
if i in arr:
dict_1.update({i:0})
else:
dict_1.update({i:dict_1.get(i)+1})
print(dict_1)
output:
{1: 0, 2: 0, 3: 6}
P.S : don't use dict as variable
After trying to count the occurrences of an element in a list using the below code
from collections import Counter
A = ['a','a','a','b','c','b','c','b','a']
A = Counter(A)
min_threshold = 3
After calling Counter on A above, a counter object like this is formed:
>>> A
Counter({'a': 4, 'b': 3, 'c': 2})
From here, how do I filter only 'a' and 'b' using minimum threshold value of 3?
Build your Counter, then use a dict comprehension as a second, filtering step.
{x: count for x, count in A.items() if count >= min_threshold}
# {'a': 4, 'b': 3}
As covered by Satish BV, you can iterate over your Counter with a dictionary comprehension. You could use items (or iteritems for more efficiency and if you're on Python 2) to get a sequence of (key, value) tuple pairs.
And then turn that into a Counter.
my_dict = {k: v for k, v in A.iteritems() if v >= min_threshold}
filteredA = Counter(my_dict)
Alternatively, you could iterate over the original Counter and remove the unnecessary values.
for k, v in A.items():
if v < min_threshold:
A.pop(k)
This looks nicer:
{ x: count for x, count in A.items() if count >= min_threshold }
You could remove the keys from the dictionary that are below 3:
for key, cnts in list(A.items()): # list is important here
if cnts < min_threshold:
del A[key]
Which gives you:
>>> A
Counter({'a': 4, 'b': 3})
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
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