I have the following code to create a dictionary whose elements are lists from another dictionary:
olddict={'Fiesta': {'key':'Ford'},'Golf': {'key':'Volkswagen'}, 'Bora': {'key':'Volkswagen'} }
newdict = {}
for key, value in olddict.items():
newkey = value['key']
if newkey in newdict:
newdict[newkey].append(key)
else:
newdict[newkey] = [key]
The code works fine, but seems utterly non-pythonic. Maybe I am a bit tired, but a one-line solution would be great...
Use defaultdict():
>>> olddict={'Fiesta': {'key':'Ford'},'Golf': {'key':'Volkswagen'}, 'Bora': {'key':'Volkswagen'} }
>>> from collections import defaultdict
>>> result = defaultdict(list)
>>> for k, v in olddict.items():
... result[v['key']].append(k)
...
>>> result
defaultdict(<type 'list'>, {'Ford': ['Fiesta'], 'Volkswagen': ['Golf', 'Bora']})
This method initializes an empty list when a new key is found. Essentially getting rid of the else part in your code. I am not sure you can combine a defaultdict with list comprehension to make it a one-liner.
Using itertools.groupby.
from itertools import groupby as g
keyfunc = lambda x: olddict[x]['key']
newdict = dict( (k, list(vs)) for k, vs in g(sorted(olddict, key=keyfunc), keyfunc))
If you are using Python 2.7 or later, you can also use a dict comprehension
newdict = { k: list(vs) for k, vs in g(sorted(olddict, key=keyfunc), keyfunc) }
