Safe method to get value of nested dictionary

Question

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 ?

Answer 1

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')