How to use dict.get() with multidimensional dict?

Question

I have a multidimensional dict, and I’d like to be able to retrieve a value by a key:key pair, and return ‘NA’ if the first key doesn’t exist. All of the sub-dicts have the same keys.

d = {   'a': {'j':1,'k':2},
        'b': {'j':2,'k':3},
        'd': {'j':1,'k':3}
    }

I know I can use d.get('c','NA') to get the sub-dict if it exists and return ‘NA’ otherwise, but I really only need one value from the sub-dict. I’d like to do something like d.get('c['j']','NA') if that existed.

Right now I’m just checking to see if the top-level key exists and then assigning the sub-value to a variable if it exists or 'NA' if not. However, I’m doing this about 500k times and also retrieving/generating other information about each top-level key from elsewhere, and I’m trying to speed this up a little bit.

Asked By: user2034412

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Source

Answer 1

How about

d.get('a', {'j': 'NA'})['j']

?

If not all subdicts have a j key, then

d.get('a', {}).get('j', 'NA')

To cut down on identical objects created, you can devise something like

class DefaultNASubdict(dict):
    class NADict(object):
        def __getitem__(self, k):
            return 'NA'

    NA = NADict()

    def __missing__(self, k):
        return self.NA

nadict = DefaultNASubdict({
                'a': {'j':1,'k':2},
                'b': {'j':2,'k':3},
                'd': {'j':1,'k':3}
            })

print nadict['a']['j']  # 1
print nadict['b']['j']  # 2
print nadict['c']['j']  # NA

Same idea using defaultdict:

import collections

class NADict(object):
    def __getitem__(self, k):
        return 'NA'

    @staticmethod
    def instance():
        return NADict._instance

NADict._instance = NADict()


nadict = collections.defaultdict(NADict.instance, {
                'a': {'j':1,'k':2},
                'b': {'j':2,'k':3},
                'd': {'j':1,'k':3}
            })

Answered By: Pavel Anossov

Answer 2

Rather than a hierarchy of nested dict objects, you could use one dictionary whose keys are a tuple representing a path through the hierarchy.

In [34]: d2 = {(x,y):d[x][y] for x in d for y in d[x]}

In [35]: d2
Out[35]:
{('a', 'j'): 1,
 ('a', 'k'): 2,
 ('b', 'j'): 2,
 ('b', 'k'): 3,
 ('d', 'j'): 1,
 ('d', 'k'): 3}

In [36]: timeit [d[x][y] for x,y in d2.keys()]
100000 loops, best of 3: 2.37 us per loop

In [37]: timeit [d2[x] for x in d2.keys()]
100000 loops, best of 3: 2.03 us per loop

Accessing this way looks like it’s about 15% faster. You can still use the get method with a default value:

In [38]: d2.get(('c','j'),'NA')
Out[38]: 'NA'

Answered By: mtadd

Answer 3

Here’s a simple and efficient way to do it with ordinary dictionaries, nested an arbitrary number of levels. The example code works in both Python 2 and 3.

from __future__ import print_function
try:
    from functools import reduce
except ImportError:  # Assume it's built-in (Python 2.x)
    pass

def chained_get(dct, *keys):
    SENTRY = object()
    def getter(level, key):
        return 'NA' if level is SENTRY else level.get(key, SENTRY)
    return reduce(getter, keys, dct)


d = {'a': {'j': 1, 'k': 2},
     'b': {'j': 2, 'k': 3},
     'd': {'j': 1, 'k': 3},
    }

print(chained_get(d, 'a', 'j'))  # 1
print(chained_get(d, 'b', 'k'))  # 3
print(chained_get(d, 'k', 'j'))  # NA

It could also be done recursively:

# Recursive version.

def chained_get(dct, *keys):
    SENTRY = object()
    def getter(level, keys):
        return (level if keys[0] is SENTRY else
                    'NA' if level is SENTRY else
                        getter(level.get(keys[0], SENTRY), keys[1:]))
    return getter(dct, keys+(SENTRY,))

Although this way of doing it isn’t quite as efficient as the first.

Answered By: martineau

Answer 4

Another way to get multidimensional dict example ( use get method twice)

d.get('a', {}).get('j')

Answered By: Hafiz Muhammad Shafiq

Answer 5

For a functional approach very similar to martineau’s answer, I’ve gone with the following:

def chained_get(dictionary: dict, *args, default: Any = None) -> Any:
    """
    Get a value nested in a dictionary by its nested path.
    """
    value_path = list(args)
    dict_chain = dictionary
    while value_path:
        try:
            dict_chain = dict_chain.get(value_path.pop(0))
        except AttributeError:
            return default

    return dict_chain

It’s a slightly simpler implementation but is still recursive and optionally allows a default value.

The usage is identical to martineau’s answer:

from typing import Any


def chained_get(dictionary: dict, *args, default: Any = None) -> Any:
    """
    Get a value nested in a dictionary by its nested path.
    """
    value_path = list(args)
    dict_chain = dictionary
    while value_path:
        try:
            dict_chain = dict_chain.get(value_path.pop(0))
        except AttributeError:
            return default

    return dict_chain


def main() -> None:
    dct = {
        "a": {"j": 1, "k": 2},
        "b": {"j": 2, "k": 3},
        "d": {"j": 1, "k": 3},
    }

    print(chained_get(dct, "a", "j"))  # 1
    print(chained_get(dct, "b", "k"))  # 3
    print(chained_get(dct, "k", "j"))  # None
    print(chained_get(dct, "k", "j", default="NA"))  # NA


if __name__ == "__main__":
    main()

Answered By: Bill Wallis

How to use dict.get() with multidimensional dict?

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