Given a dictionary likeso:
map = { 'a': 1, 'b':2 }
How can one invert this map to get:
inv_map = { 1: 'a', 2: 'b' }
+5
A:
Try this :
inv_map = dict(zip(map.values(), map.keys()))
or alternatively
inv_map = dict((map[k], k) for k in map)
or using python 3.0's dict comprehensions
inv_map = {map[k] : k for k in map}
sykora
2009-01-27 14:49:46
Would this be slwoer than Heikogerlach's solution because it has to iterate through the map twice - once for the values, once for the keys? Thanks!
Brian M. Hunt
2009-01-27 14:56:09
Based on my own testing using the timeit class, surprisingly, my first and third methods take roughly half as long as my own second method and Heikogerlach's solution. I didn't do any major testing, just a few test cases.
sykora
2009-01-27 15:29:47
Do values() and keys() guarantee that they will return their items in the same order? If not, then you wouldn't end up with the mapping that you started with.
davidavr
2009-01-27 22:11:25
Both values() and keys() will return them in the same order. The order itself is according to python's internal storage of the keys, but the order will not change between values() and keys() (Unless of course the dictionary itself changes in between the calls).
sykora
2009-01-28 00:47:36
@sykora: how do you know that?
recursive
2009-01-28 01:16:46
@recursive: I read it somewhere, but when I searched again, the only link I could turn up was this one (http://www.python.org/dev/peps/pep-3106/), the original proposal to turn keys(), values() and items() into iterators. I'll keep looking.
sykora
2009-01-28 11:35:53
@sykora: unless it's documented in the dict class, I think you are depending on an implementation detail which is risky.
davidavr
2009-01-29 12:56:08
Argh, why do I always find the answer in the last place I look? http://docs.python.org/3.0/library/stdtypes.html#dictionary-view-objects . Everybody satisfied? :)
sykora
2009-01-29 13:28:00
+27
A:
Assuming that the values in the dict are unique:
dict((v,k) for k, v in map.iteritems())
unbeknown
2009-01-27 14:50:22
+7
A:
If the values in map aren't unique:
for k, v in map.iteritems():
inv_map[v] = inv_map.get(v, [])
inv_map[v].append(k)
Robert Rossney
2009-01-27 21:33:26
`from collections import defaultdict; inv_map = defaultdict(list)` will let you skip the second line and go straight to appending.
gotgenes
2009-11-20 21:16:48
+1
A:
Or if the values aren't unique and you're a little harcore:
inv_map = \
dict( \
(v, [k for (k, xx) in filter(lambda (key, value): value == v, map.items())]) \
for v in set(map.values()))
pcv
2010-04-17 20:14:06