Is there a way to get the original/consistent list of keys from defaultdict even when non existing keys were requested?
from collections import defaultdict
>>> d = defaultdict(lambda: 'default', {'key1': 'value1', 'key2' :'value2'})
>>>
>>> d.keys()
['key2', 'key1']
>>> d['bla']
'default'
>>> d.keys() # how to get the same: ['key2', 'key1']
['key2', 'key1', 'bla']
You have to exclude. the keys that has the default value!
>>> [i for i in d if d[i]!=d.default_factory()]
['key2', 'key1']
Time comparison with method suggested by Jean,
>>> def funct(a=None,b=None,c=None):
... s=time.time()
... eval(a)
... print time.time()-s
...
>>> funct("[i for i in d if d[i]!=d.default_factory()]")
9.29832458496e-05
>>> funct("[k for k,v in d.items() if v!=d.default_factory()]")
0.000100135803223
>>> ###storing the default value to a variable and using the same in the list comprehension reduces the time to a certain extent!
>>> defa=d.default_factory()
>>> funct("[i for i in d if d[i]!=defa]")
8.82148742676e-05
>>> funct("[k for k,v in d.items() if v!=defa]")
9.79900360107e-05