Accessing values nested within dictionaries

Question

I have a dictionary which contains dictionaries, which may also contain dictionaries, e.g.

dictionary = {'ID': 0001, 'Name': 'made up name', 'Transactions':
               {'Transaction Ref': 'a1', 'Transaction Details':
                  {'Bill To': 'abc', 'Ship To': 'def', 'Product': 'Widget A'
                      ...} ...} ... }

Currently I'm unpacking to get the 'Bill To' for ID 001, 'Transaction Ref' a1 as follows:

if dictionary['ID'] == 001:
    transactions = dictionary['Transactions']
        if transactions['Transaction Ref'] == 'a1':
            transaction_details = transactions['Transaction Details']
            bill_to = transaction_details['Bill To']

I can't help but think this is is a little clunky, especially the last two lines - I feel like something along the lines of the following should work:

bill_to = transactions['Transaction Details']['Bill To']

Is there a simpler approach for drilling down into nested dictionaries without having to unpack into interim variables?

Note that 0001 is not valid in Python3, neither is any number with leading zeroes. — wjandrea
– wjandrea, Commented Aug 6, 2019 at 2:00

Rob.Kachmar · Accepted Answer · 2015-03-09 13:53:32Z

20

You can use something like this:

>>> def lookup(dic, key, *keys):
...     if keys:
...         return lookup(dic.get(key, {}), *keys)
...     return dic.get(key)
...
>>> d = {'a':{'b':{'c':5}}}
>>> print lookup(d, 'a', 'b', 'c')
5
>>> print lookup(d, 'a', 'c')
None

Additionally, if you don't want to define your search keys as individual parameters, you can just pass them in as a list like this:

>>> print lookup(d, *['a', 'b', 'c'])
5
>>> print lookup(d, *['a', 'c'])
None

edited Mar 9, 2015 at 13:53

Rob.Kachmar

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answered Jul 28, 2012 at 13:43

Aleksei astynax Pirogov

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1 Comment

Rob.Kachmar Over a year ago

I hope you don't mind, I added an edit to clarify the true dynamic search capability of this function. The ability to pass in a list of search keys on-the-fly makes this a cut above the other techniques.

Fred Foo · Accepted Answer · 2012-07-28 12:10:43Z

bill_to = transactions['Transaction Details']['Bill To']

actually works. transactions['Transaction Details'] is an expression denoting a dict, so you can do lookup in it. For practical programs, I would prefer an OO approach to nested dicts, though. collections.namedtuple is particularly useful for quickly setting up a bunch of classes that only contain data (and no behavior of their own).

There's one caveat: in some settings, you might want to catch KeyError when doing lookups, and in this setting, that works too, it's hard to tell which dictionary lookup failed:

try:
    bill_to = transactions['Transaction Details']['Bill To']
except KeyError:
    # which of the two lookups failed?
    # we don't know unless we inspect the exception;
    # but it's easier to do the lookup and error handling in two steps

blackfyre · Accepted Answer · 2014-12-18 14:18:00Z

2

Following is another way of accessing nested dictionaries

>>> dbo={'m':{'d':{'v':{'version':1}}}}
>>> name='m__d__v__version' # it'll refer to 'dbo['m']['d']['v']['version']', '__' is the separator
>>> version = reduce(dict.get, name.split('__'), dbo)
>>> print version
1
>>>

Here, variable 'name' refers to 'dbo['m']['d']['v']['version']', which seems much shorter and neat.

This method will not throw KeyError. If a key is not found then you'll get 'None'.

Ref.: http://code.activestate.com/recipes/475156-using-reduce-to-access-deeply-nested-dictionaries/

edited Dec 18, 2014 at 14:18

answered Dec 16, 2014 at 11:51

blackfyre

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1 Comment

krasnaya Over a year ago

It will throw a TypeError if you try name='m__foo__v__foo': TypeError: descriptor 'get' requires a 'dict' object but received a 'NoneType'

edd313 · Accepted Answer · 2022-03-08 21:42:53Z

0

You can access nested dictionaries with a tuple using NestedDict.

>>> from ndicts.ndicts import NestedDict
>>> nested_dict = {"a": {"a": 0, "b": 1},
...                "b": {"a": 2, "b": 3}}
>>> nd = NestedDict(nested_dict)
>>> nd["a", "a"]
0
>>> nd["b", "a"]
2
>>> nd["a"]
{"a": 0, "b": 1}

To install ndicts:

pip install ndicts

answered Mar 8, 2022 at 21:42

edd313

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