Menu
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']
609
A defaultdict works exactly like a normal dict, but it is initialized with a function (“default factory”) that takes no arguments and provides the default value for a nonexistent key. A defaultdict will never raise a KeyError. Any key that does not exist gets the value returned by the default factory.
When the int class is passed as the default_factory argument, then a defaultdict is created with default value as zero.
The Python defaultdict type behaves almost exactly like a regular Python dictionary, but if you try to access or modify a missing key, then defaultdict will automatically create the key and generate a default value for it. This makes defaultdict a valuable option for handling missing keys in dictionaries.
get method and the experiment shows that defaultdict more that two times faster than dict. get method.
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
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
