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?
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.
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) 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']}