Safe method to get value of nested dictionary
I have a nested dictionary. Is there only one way to get values out safely?
try:
example_dict['key1']['key2']
except KeyError:
pass
Or maybe python has a method like get()
for nested dictionary ?
You could use get
twice:
example_dict.get('key1', {}).get('key2')
This will return None
if either key1
or key2
does not exist.
Note that this could still raise an AttributeError
if example_dict['key1']
exists but is not a dict (or a dict-like object with a get
method). The try..except
code you posted would raise a TypeError
instead if example_dict['key1']
is unsubscriptable.
Another difference is that the try...except
short-circuits immediately after the first missing key. The chain of get
calls does not.
If you wish to preserve the syntax, example_dict['key1']['key2']
but do not want it to ever raise KeyErrors, then you could use the Hasher recipe:
class Hasher(dict):
# https://stackoverflow.com/a/3405143/190597
def __missing__(self, key):
value = self[key] = type(self)()
return value
example_dict = Hasher()
print(example_dict['key1'])
# {}
print(example_dict['key1']['key2'])
# {}
print(type(example_dict['key1']['key2']))
# <class '__main__.Hasher'>
Note that this returns an empty Hasher when a key is missing.
Since Hasher
is a subclass of dict
you can use a Hasher in much the same way you could use a dict
. All the same methods and syntax is available, Hashers just treat missing keys differently.
You can convert a regular dict
into a Hasher
like this:
hasher = Hasher(example_dict)
and convert a Hasher
to a regular dict
just as easily:
regular_dict = dict(hasher)
Another alternative is to hide the ugliness in a helper function:
def safeget(dct, *keys):
for key in keys:
try:
dct = dct[key]
except KeyError:
return None
return dct
So the rest of your code can stay relatively readable:
safeget(example_dict, 'key1', 'key2')
You could also use python reduce:
def deep_get(dictionary, *keys):
return reduce(lambda d, key: d.get(key) if d else None, keys, dictionary)