Invert keys and values of the original dictionary

Question

For example, I call this function by passing a dictionary as parameter:

>>> inv_map({'a':1, 'b':2, 'c':3, 'd':2})
{1: ['a'], 2: ['b', 'd'], 3: ['c']}
>>> inv_map({'a':3, 'b':3, 'c':3})
{3: ['a', 'c', 'b']}
>>> inv_map({'a':2, 'b':1, 'c':2, 'd':1})
{1: ['b', 'd'], 2: ['a', 'c']}

If

map = { 'a': 1, 'b':2 }

I can only invert this map to get:

inv_map = { 1: 'a', 2: 'b' }

by using this

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

Anyone knows how to do that for my case?

Answer 1

You can use a defaultdict with list:

>>> from collections import defaultdict
>>> m = {'a': 2, 'b': 1, 'c': 2, 'd': 1}
>>> dd = defaultdict(list)
>>> for k, v in m.iteritems():
...     dd[v].append(k)
... 
>>> dict(dd)
{1: ['b', 'd'], 2: ['a', 'c']}

If you don't care if you have an dict or defaultdict, you can omit the last step und use the defaultdict directly.

Answer 2

You can probably use defaultdict or setdefault here.

def invertDictionary(orig_dict):
    result = {} # or change to defaultdict(list)
    for k, v in orig_dict.iteritems():
        result.setdefault(v, []).append(k)

Answer 3

EDIT In python 2.7:

from itertools import groupby
def inv_map(d):
    return {k : [i[0] for i in list(v)] for k, v in groupby(d.items(),lambda x:x[1])}

print inv_map({'a':1, 'b':2, 'c':3, 'd':2})
print inv_map({'a':3, 'b':3, 'c':3})
print inv_map({'a':2, 'b':1, 'c':2, 'd':1})

Output:

{1: ['a'], 2: ['b', 'd'], 3: ['c']}
{3: ['a', 'c', 'b']}
{1: ['b', 'd'], 2: ['a', 'c']}