How to get multiple dictionary values?

There already exists a function for this:

from operator import itemgetter

my_dict = {x: x**2 for x in range(10)}

itemgetter(1, 3, 2, 5)(my_dict)
#>>> (1, 9, 4, 25)

itemgetter will return a tuple if more than one argument is passed. To pass a list to itemgetter, use

itemgetter(*wanted_keys)(my_dict)

Keep in mind that itemgetter does not wrap its output in a tuple when only one key is requested, and does not support zero keys being requested.

You can use At from pydash:

from pydash import at
my_dict = {'a': 1, 'b': 2, 'c': 3}
my_list = at(my_dict, 'a', 'b')
my_list == [1, 2]

If the fallback keys are not too many you can do something like this

value = my_dict.get('first_key') or my_dict.get('second_key')

I’d suggest the very useful map function, which allows a function to operate element-wise on a list:

mydictionary = {'a': 'apple', 'b': 'bear', 'c': 'castle'}
keys = ['b', 'c']

values = list( map(mydictionary.get, keys) )

# values = ['bear', 'castle']

If you have pandas installed you can turn it into a series with the keys as the index. So something like

import pandas as pd

s = pd.Series(my_dict)

s[['key1', 'key3', 'key2']]

As I see no similar answer here – it is worth pointing out that with the usage of a (list / generator) comprehension, you can unpack those multiple values and assign them to multiple variables in a single line of code:

first_val, second_val = (myDict.get(key) for key in [first_key, second_key])

I think list comprehension is one of the cleanest ways that doesn’t need any additional imports:

>>> d={"foo": 1, "bar": 2, "baz": 3}
>>> a = [d.get(k) for k in ["foo", "bar", "baz"]]
>>> a
[1, 2, 3]

Or if you want the values as individual variables then use multiple-assignment:

>>> a,b,c = [d.get(k) for k in ["foo", "bar", "baz"]]
>>> a,b,c
(1, 2, 3)

def get_all_values(nested_dictionary):
    for key, value in nested_dictionary.items():
        if type(value) is dict:
            get_all_values(value)
        else:
            print(key, ":", value)

nested_dictionary = {'ResponseCode': 200, 'Data': {'256': {'StartDate': '2022-02-07', 'EndDate': '2022-02-27', 'IsStoreClose': False, 'StoreTypeMsg': 'Manual Processing Stopped', 'is_sync': False}}}

get_all_values(nested_dictionary)

def get_all_values(nested_dictionary):
    for key, val in nested_dictionary.items():
        data_list = []
        if type(val) is dict:
            for key1, val1 in val.items():
                data_list.append(val1)

    return data_list

If you want to retain the mapping of the values to the keys, you should use a dict comprehension instead:

{key: myDictionary[key] for key in [
  'firstKey',
  'secondKey',
  'thirdKey',
  'fourthKey',
  'fifthKey'
]}

Slighlty different variation of list comprehension approach.

#doc
[dict[key] for key in (tuple_of_searched_keys)]

#example
my_dict = {x: x**2 for x in range(10)}
print([my_dict[key] for key in (8,9)])

%timeit response of all the answers listed above. My apologies if missed some of the solutions, and I used my judgment to club similar answers. itemgetter seems to be the winner to me. pydash reports much lesser time but I don’t know why it ran lesser loops and don’t know if I can call it a fastest. Your thoughts?

from operator import itemgetter

my_dict = {x: x**2 for x in range(10)}
req_keys = [1, 3, 2, 5]
%timeit itemgetter(1, 3, 2, 5)(my_dict)
257 ns ± 4.61 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%timeit [my_dict.get(key) for key in req_keys]
604 ns ± 6.94 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)


%timeit list( map(my_dict.get, req_keys) )
529 ns ± 34.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)


!pip install pydash
from pydash import at

%timeit at(my_dict, 1, 3, 2, 5)
22.2 µs ± 572 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)


%timeit (my_dict.get(key) for key in req_keys)
308 ns ± 6.53 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

s = pd.Series(my_dict)

%timeit s[req_keys]
334 µs ± 58.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)