Sorting a Python Dictionary: Values, Keys, and More

Using the sorted() Function

The critical function that you’ll use to sort dictionaries is the built-in sorted() function. This function takes an iterable as the main argument, with two optional keyword-only arguments—a key function and a reverse Boolean value.

To illustrate the sorted() function’s behavior in isolation, examine its use on a list of numbers:

>>> numbers = [5, 3, 4, 3, 6, 7, 3, 2, 3, 4, 1]
>>> sorted(numbers)
[1, 2, 3, 3, 3, 3, 4, 4, 5, 6, 7]

As you can see, the sorted() function takes an iterable, sorts comparable elements like numbers in ascending order, and returns a new list. With strings, it sorts them in alphabetical order:

>>> words = ["aa", "ab", "ac", "ba", "cb", "ca"]
>>> sorted(words)
['aa', 'ab', 'ac', 'ba', 'ca', 'cb']

Sorting by numerical or alphabetical precedence is the most common way to sort elements, but maybe you need more control.

Say you want to sort on the second character of each word in the last example. To customize what the sorted() function uses to sort the elements, you can pass in a callback function to the key parameter.

A callback function is a function that’s passed as an argument to another function. For sorted(), you pass it a function that acts as a sort key. The sorted() function will then call back the sort key for every element.

In the following example, the function passed as the key accepts a string and will return the second character of that string:

>>> def select_second_character(word):
...     return word[1]
...
>>> sorted(words, key=select_second_character)
['aa', 'ba', 'ca', 'ab', 'cb', 'ac']

The sorted() function passes every element of the words iterable to the key function and uses the return value for comparison. Using the key means that the sorted() function will compare the second letter instead of comparing the whole string directly.

More examples and explanations of the key parameter will come later in the tutorial when you use it to sort dictionaries by values or nested elements.

If you take another look at the results of this last sorting, you may notice the stability of the sorted() function. The three elements, aa, ba and ca, are equivalent when sorted by their second character. Because they’re equal, the sorted() function conserves their original order. Python guarantees this stability.

>>> list(reversed([3, 2, 1]))
[1, 2, 3]

So, how about dictionaries? You can actually take the dictionary and feed it straight into the sorted() function:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}
>>> sorted(people)
[1, 2, 3, 4]

But the default behavior of passing in a dictionary directly to the sorted() function is to take the keys of the dictionary, sort them, and return a list of the keys only. That’s probably not the behavior you had in mind! To preserve all the information in a dictionary, you’ll need to be acquainted with dictionary views.

Getting Keys, Values, or Both From a Dictionary

If you want to conserve all the information from a dictionary when sorting it, the typical first step is to call the .items() method on the dictionary. Calling .items() on the dictionary will provide an iterable of tuples representing the key-value pairs:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}
>>> people.items()
dict_items([(3, 'Jim'), (2, 'Jack'), (4, 'Jane'), (1, 'Jill')])

The .items() method returns a read-only dictionary view object, which serves as a window into the dictionary. This view is not a copy or a list—it’s a read-only iterable that’s actually linked to the dictionary it was generated from:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}
>>> view = people.items()
>>> people[2] = "Elvis"
>>> view
dict_items([(3, 'Jim'), (2, 'Elvis'), (4, 'Jane'), (1, 'Jill')])

You’ll notice that any updates to the dictionary also get reflected in the view because they’re linked. A view represents a lightweight way to iterate over a dictionary without generating a list first.

Crucially, you can use the sorted() function with dictionary views. You call the .items() method and use the result as an argument to the sorted() function. Using .items() keeps all the information from the dictionary:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}
>>> sorted(people.items())
[(1, 'Jill'), (2, 'Jack'), (3, 'Jim'), (4, 'Jane')]

This example results in a sorted list of tuples, with each tuple representing a key-value pair of the dictionary.

If you want to end up with a dictionary sorted by values, then you’ve still got two issues. The default behavior still seems to sort by key and not value. The other issue is that you end up with a list of tuples, not a dictionary. First, you’ll figure out how to sort by value.

Understanding How Python Sorts Tuples

When using the .items() method on a dictionary and feeding it into the sorted() function, you’re passing in an iterable of tuples, and the sorted() function compares the entire tuple directly.

When comparing tuples, Python behaves a lot like it’s sorting strings alphabetically. That is, it sorts them lexicographically.

Lexicographical sorting means that if you have two tuples, (1, 2, 4) and (1, 2, 3), then you start by comparing the first item of each tuple. The first item is 1 in both cases, which is equal. The second element, 2, is also identical in both cases. The third elements are 4 and 3, respectively. Since 3 is less than 4, you’ve found which item is less than the other.

So, to order the tuples (1, 2, 4) and (1, 2, 3) lexicographically, you would switch their order to (1, 2, 3) and (1, 2, 4).

Because of Python’s lexicographic sorting behavior for tuples, using the .items() method with the sorted() function will always sort by keys unless you use something extra.

Using the key Parameter and Lambda Functions

For example, if you want to sort by value, then you have to specify a sort key. A sort key is a way to extract a comparable value. For instance, if you have a pile of books, then you might use the author surname as the sort key. With the sorted() function, you can specify a sort key by passing a callback function as a key argument.

To see a sort key in action, take a look at this example, which is similar to the one you saw in the section introducing the sorted() function:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}

>>> # Sort key
>>> def value_getter(item):
...     return item[1]
...

>>> sorted(people.items(), key=value_getter)
[(2, 'Jack'), (4, 'Jane'), (1, 'Jill'), (3, 'Jim')]

>>> # Or with a lambda function
>>> sorted(people.items(), key=lambda item: item[1])
[(2, 'Jack'), (4, 'Jane'), (1, 'Jill'), (3, 'Jim')]

In this example, you try out two ways of passing a key parameter. The key parameter accepts a callback function. The function can be a normal function identifier or a lambda function. The lambda function in the example is the exact equivalent of the value_getter() function.

# With a normal function
def value_getter(item):
    return item[1]

sorted(people.items(), key=value_getter)

# With a lambda function
sorted(people.items(), key=lambda item: item[1])

The key callback function will receive each element of the iterable that it’s sorting. The callback function’s job is to return something that can be compared, such as a number or a string. In this example, you named the function value_getter() because all it does is get the value from a key-value tuple.

Since the default behavior of sorted() with tuples is to sort lexicographically, the key parameter allows you to select a value from the element that it’s comparing.

In the next section, you’ll take sort keys a bit further and use them to sort by a nested value.

Selecting a Nested Value With a Sort Key

You can also go further and use a sort key to select nested values that may or may not be present and return a default value if they’re not present:

data = {
    193: {"name": "John", "age": 30, "skills": {"python": 8, "js": 7}},
    209: {"name": "Bill", "age": 15, "skills": {"python": 6}},
    746: {"name": "Jane", "age": 58, "skills": {"js": 2, "python": 5}},
    109: {"name": "Jill", "age": 83, "skills": {"java": 10}},
    984: {"name": "Jack", "age": 28, "skills": {"c": 8, "assembly": 7}},
    765: {"name": "Penelope", "age": 76, "skills": {"python": 8, "go": 5}},
    598: {"name": "Sylvia", "age": 62, "skills": {"bash": 8, "java": 7}},
    483: {"name": "Anna", "age": 24, "skills": {"js": 10}},
    277: {"name": "Beatriz", "age": 26, "skills": {"python": 2, "js": 4}},
}

def get_relevant_skills(item):
    """Get the sum of Python and JavaScript skill"""
    skills = item[1]["skills"]

    # Return default value that is equivalent to no skill
    return skills.get("python", 0) + skills.get("js", 0)

print(sorted(data.items(), key=get_relevant_skills, reverse=True))

In this example, you have a dictionary with numeric keys and a nested dictionary as a value. You want to sort by the combined Python and JavaScript skills, attributes found in the skills subdictionary.

Part of what makes sorting by the combined skill tricky is that the python and js keys aren’t present in the skills dictionary for all people. The skills dictionary is also nested. You use .get() to read the keys and provide 0 as a default value that’s used for missing skills.

You’ve also used the reverse argument because you want the top Python skills to appear first.

sorted(
    data.items(),
    key=lambda item: (
        item[1]["skills"].get("python", 0)
        + item[1]["skills"].get("js", 0)
    ),
    reverse=True,
)

You’ve successfully used a higher-order function as a sort key to sort a dictionary view by value. That was the hard part. Now there’s only one issue left to solve—converting the list that sorted() yields back into a dictionary.

Converting Back to a Dictionary

The only issue left to address with the default behavior of sorted() is that it returns a list, not a dictionary. There are a few ways to convert a list of tuples back into a dictionary.

You can iterate over the result with a for loop and populate a dictionary on each iteration:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}
>>> sorted_people = sorted(people.items(), key=lambda item: item[1])

>>> sorted_people_dict = {}
>>> for key, value in sorted_people:
...     sorted_people_dict[key] = value
...

>>> sorted_people_dict
{2: 'Jack', 4: 'Jane', 1: 'Jill', 3: 'Jim'}

This method gives you absolute control and flexibility in deciding how you want to construct your dictionary. This method can be quite lengthy to type out, though. If you don’t have any special requirements for constructing your dictionary, then you may want to go for a dictionary constructor instead:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}
>>> sorted_people = sorted(people.items(), key=lambda item: item[1])
>>> dict(sorted_people)
{2: 'Jack', 4: 'Jane', 1: 'Jill', 3: 'Jim'}

That’s nice and compact! You could also use a dictionary comprehension, but that only makes sense if you want to change the shape of the dictionary or swap the keys and values, for example. In the following comprehension, you swap the keys and values:

>>> {
...     value: key
...     for key, value in sorted(people.items(), key=lambda item: item[1])
... }
...
{'Jack': 2, 'Jane': 4, 'Jill': 1, 'Jim': 3}

Depending on how familiar you or your team are with comprehensions, this may be less readable than just using a normal for loop.

Congratulations, you’ve got your sorted dictionary! You can now sort it by any criteria that you’d like.

Now that you can sort your dictionary, you might be interested in knowing if there are any performance implications to using a sorted dictionary, or whether there are alternative data structures for key-value data.

Using Special Getter Functions to Increase Performance and Readability

You may have noticed that most of the sort key functions that you’ve used so far aren’t doing very much. All the function does is get a value from a tuple. Making a getter function is such a common pattern that Python has a special way to create special functions that get values more quickly than regular functions.

The itemgetter() function can produce highly efficient versions of getter functions.

You pass itemgetter() an argument, which is typically the key or index position that you want to select. The itemgetter() function will then return a getter object that you call like a function.

That’s right, it’s a function that returns a function. Using the itemgetter() function is another example of working with higher-order functions.

The getter object from itemgetter() will call the .__getitem__() method on the item that’s passed to it. When something makes a call to .__getitem__(), it needs to pass in the key or index of what to get. The argument that’s used for .__getitem__() is the same argument that you passed to itemgetter():

>>> item = ("name", "Guido")

>>> from operator import itemgetter

>>> getter = itemgetter(0)
>>> getter(item)
'name'
>>> getter = itemgetter(1)
>>> getter(item)
'Guido'

In the example, you start off with a tuple, similar to one that you might get as part of a dictionary view.

You make the first getter by passing 0 as an argument to itemgetter(). When the resultant getter receives the tuple, it returns the first item in the tuple—the value at index 0. If you call itemgetter() with an argument of 1, then it gets the value at index position 1.

You can use this itemgetter as a key for the sorted() function:

>>> from operator import itemgetter

>>> fruit_inventory = [
...     ("banana", 5), ("orange", 15), ("apple", 3), ("kiwi", 0)
... ]

>>> # Sort by key
>>> sorted(fruit_inventory, key=itemgetter(0))
[('apple', 3), ('banana', 5), ('kiwi', 0), ('orange', 15)]

>>> # Sort by value
>>> sorted(fruit_inventory, key=itemgetter(1))
[('kiwi', 0), ('apple', 3), ('banana', 5), ('orange', 15)]

>>> sorted(fruit_inventory, key=itemgetter(2))
Traceback (most recent call last):
  File "<input>", line 1, in <module>
    sorted(fruit_inventory, key=itemgetter(2))
IndexError: tuple index out of range

In this example, you start by using itemgetter() with 0 as an argument. Since it’s operating on each tuple from the fruit_inventory variable, it gets the first element from each tuple. Then the example demonstrates initializing an itemgetter with 1 as an argument, which selects the second item in the tuple.

Finally, the example shows what would happen if you used itemgetter() with 2 as an argument. Since these tuples only have two index positions, trying to get the third element, with index 2, results in a IndexError.

You can use the function produced by itemgetter() in place of the getter functions that you’ve been using up until now:

>>> people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}

>>> from operator import itemgetter
>>> sorted(people.items(), key=itemgetter(1))
[(2, 'Jack'), (4, 'Jane'), (1, 'Jill'), (3, 'Jim')]

The itemgetter() function produces a function that has exactly the same effect as the value_getter() function from previous sections. The main reason you’d want to use the function from itemgetter() is because it’s more efficient. In the next section, you’ll start to put some numbers on just how much more efficient it is.

Measuring Performance When Using itemgetter()

So, you end up with a function that behaves like the original value_getter() from the previous sections, except that the version returned from itemgetter() is more efficient. You can use the timeit module to compare their performance:

# compare_lambda_vs_getter.py

from timeit import timeit

dict_to_order = {
    1: "requests",
    2: "pip",
    3: "jinja",
    4: "setuptools",
    5: "pandas",
    6: "numpy",
    7: "black",
    8: "pillow",
    9: "pyparsing",
    10: "boto3",
    11: "botocore",
    12: "urllib3",
    13: "s3transfer",
    14: "six",
    15: "python-dateutil",
    16: "pyyaml",
    17: "idna",
    18: "certifi",
    19: "typing-extensions",
    20: "charset-normalizer",
    21: "awscli",
    22: "wheel",
    23: "rsa",
}

sorted_with_lambda = "sorted(dict_to_order.items(), key=lambda item: item[1])"
sorted_with_itemgetter = "sorted(dict_to_order.items(), key=itemgetter(1))"

sorted_with_lambda_time = timeit(stmt=sorted_with_lambda, globals=globals())
sorted_with_itemgetter_time = timeit(
    stmt=sorted_with_itemgetter,
    setup="from operator import itemgetter",
    globals=globals(),
)

print(
    f"""\
{sorted_with_lambda_time=:.2f} seconds
{sorted_with_itemgetter_time=:.2f} seconds
itemgetter is {(
    sorted_with_lambda_time / sorted_with_itemgetter_time
):.2f} times faster"""
)

This code uses the timeit module to compare the sorting processes of the function from itemgetter() and a lambda function.

Running this script from the shell should give you similar results to what’s below:

$ python compare_lambda_vs_getter.py
sorted_with_lambda_time=1.81 seconds
sorted_with_itemgetter_time=1.29 seconds
itemgetter is 1.41 times faster

A savings of around 40 percent is significant!

Bear in mind that when timing code execution, times can vary significantly between systems. That said, in this case, the ratio should be relatively stable across systems.

From the results of this test, you can see that using itemgetter() is preferable from a performance standpoint. Plus, it’s part of the Python standard library, so there’s no cost to using it.

Now you can squeeze a bit more performance out of your dictionary sorting, but it’s worth taking a step back and considering whether using a sorted dictionary as your preferred data structure is the best choice. A sorted dictionary isn’t a very common pattern, after all.

Coming up, you’ll be asking yourself some questions about what you what you want to do with your sorted dictionary and whether it’s the best data structure for your use case.

Judging Whether You Want to Use a Sorted Dictionary

If you’re considering making a sorted key-value data structure, then there are a few things you might want to take into consideration.

If you’re going to be adding data to a dictionary, and you want it to stay sorted, then you might be better off using a structure like a list of tuples or a list of dictionaries:

# Dictionary
people = {3: "Jim", 2: "Jack", 4: "Jane", 1: "Jill"}

# List of tuples
people = [
    (3, "Jim"),
    (2, "Jack"),
    (4, "Jane"),
    (1, "Jill")
]

# List of dictionaries
people = [
    {"id": 3, "name": "Jim"},
    {"id": 2, "name": "Jack"},
    {"id": 4, "name": "Jane"},
    {"id": 1, "name": "Jill"},
]

A list of dictionaries is the most widespread pattern because of its cross-language compatibility, known as language interoperability.

Language interoperability is especially relevant if you create an HTTP REST API, for instance. Making your data available over the Internet will likely mean serializing it in JSON.

If someone using JavaScript were to consume JSON data from a REST API, then the equivalent data structure would be an object. The kicker is that JavaScript objects are not ordered, so the order would end up scrambled!

This scrambling behavior would be true for many languages, and objects are even defined in the JSON specification as an unordered data structure. So, if you took care to order your dictionary before serializing to JSON, it wouldn’t matter by the time it got into most other environments.

Another option is to simply not worry about ordering the data if you don’t need to. Including id, priority, or other equivalent attributes for each object can be enough to express order. If the ordering gets mixed up for any reason, then there’ll always be an unambiguous way to sort it:

people = {
    3: {"priority": 2, "name": "Jim"},
    2: {"priority": 4, "name": "Jack"},
    4: {"priority": 1, "name": "Jane"},
    1: {"priority": 2, "name": "Jill"}
}

With a priority attribute, for instance, it’s clear that Jane should be first in line. Being clear about your intended ordering is nicely in agreement with the old Python adage of explicit is better than implicit, from the Zen of Python.

What are the performance trade-offs with using a list of dictionaries versus a dictionary of dictionaries, though? In the next section, you’ll start to get some data on just that very question.

Comparing the Performance of Different Data Structures

If performance is a consideration—maybe you’ll be working with large datasets, for example—then you should carefully consider what you’ll be doing with the dictionary.

The two main questions you’ll seek to answer in the next few sections are:

1. Will you be sorting once and then making lots of lookups?
2. Will you be sorting many times and making very few lookups?

Once you’ve decided what usage patterns you’ll be subjecting your data structure to, then you can use the timeit module to test the performance. These measurements can vary a lot with the exact shape and size of the data being tested.

In this example, you’ll be pitting a dictionary of dictionaries against a list of dictionaries to see how they differ in terms of performance. You’ll be timing sorting operations and lookup operations with the following sample data:

# samples.py

dictionary_of_dictionaries = {
    1: {"first_name": "Dorthea", "last_name": "Emmanuele", "age": 29},
    2: {"first_name": "Evelina", "last_name": "Ferras", "age": 91},
    3: {"first_name": "Frederica", "last_name": "Livesay", "age": 99},
    4: {"first_name": "Murray", "last_name": "Linning", "age": 36},
    5: {"first_name": "Annette", "last_name": "Garioch", "age": 93},
    6: {"first_name": "Rozamond", "last_name": "Todd", "age": 36},
    7: {"first_name": "Tiffi", "last_name": "Varian", "age": 28},
    8: {"first_name": "Noland", "last_name": "Cowterd", "age": 51},
    9: {"first_name": "Dyana", "last_name": "Fallows", "age": 100},
    10: {"first_name": "Diahann", "last_name": "Cutchey", "age": 44},
    11: {"first_name": "Georgianne", "last_name": "Steinor", "age": 32},
    12: {"first_name": "Sabina", "last_name": "Lourens", "age": 31},
    13: {"first_name": "Lynde", "last_name": "Colbeck", "age": 35},
    14: {"first_name": "Abdul", "last_name": "Crisall", "age": 84},
    15: {"first_name": "Quintus", "last_name": "Brando", "age": 95},
    16: {"first_name": "Rowena", "last_name": "Geraud", "age": 21},
    17: {"first_name": "Maurice", "last_name": "MacAindreis", "age": 83},
    18: {"first_name": "Pall", "last_name": "O'Cullinane", "age": 79},
    19: {"first_name": "Kermie", "last_name": "Willshere", "age": 20},
    20: {"first_name": "Holli", "last_name": "Tattoo", "age": 88}
}

list_of_dictionaries = [
    {"id": 1, "first_name": "Dorthea", "last_name": "Emmanuele", "age": 29},
    {"id": 2, "first_name": "Evelina", "last_name": "Ferras", "age": 91},
    {"id": 3, "first_name": "Frederica", "last_name": "Livesay", "age": 99},
    {"id": 4, "first_name": "Murray", "last_name": "Linning", "age": 36},
    {"id": 5, "first_name": "Annette", "last_name": "Garioch", "age": 93},
    {"id": 6, "first_name": "Rozamond", "last_name": "Todd", "age": 36},
    {"id": 7, "first_name": "Tiffi", "last_name": "Varian", "age": 28},
    {"id": 8, "first_name": "Noland", "last_name": "Cowterd", "age": 51},
    {"id": 9, "first_name": "Dyana", "last_name": "Fallows", "age": 100},
    {"id": 10, "first_name": "Diahann", "last_name": "Cutchey", "age": 44},
    {"id": 11, "first_name": "Georgianne", "last_name": "Steinor", "age": 32},
    {"id": 12, "first_name": "Sabina", "last_name": "Lourens", "age": 31},
    {"id": 13, "first_name": "Lynde", "last_name": "Colbeck", "age": 35},
    {"id": 14, "first_name": "Abdul", "last_name": "Crisall", "age": 84},
    {"id": 15, "first_name": "Quintus", "last_name": "Brando", "age": 95},
    {"id": 16, "first_name": "Rowena", "last_name": "Geraud", "age": 21},
    {"id": 17, "first_name": "Maurice", "last_name": "MacAindreis", "age": 83},
    {"id": 18, "first_name": "Pall", "last_name": "O'Cullinane", "age": 79},
    {"id": 19, "first_name": "Kermie", "last_name": "Willshere", "age": 20},
    {"id": 20, "first_name": "Holli", "last_name": "Tattoo", "age": 88}
]

Each data structure has the same information, except one is structured as a dictionary of dictionaries, and the other is a list of dictionaries. First up, you’ll be getting some metrics on the performance of sorting these two data structures.

Comparing the Performance of Sorting

In the following code, you’ll be using timeit to compare the time it takes to sort the two data structures by the age attribute:

# compare_sorting_dict_vs_list.py

from timeit import timeit
from samples import dictionary_of_dictionaries, list_of_dictionaries

sorting_list = "sorted(list_of_dictionaries, key=lambda item:item['age'])"
sorting_dict = """
dict(
    sorted(
        dictionary_of_dictionaries.items(), key=lambda item: item[1]['age']
    )
)
"""

sorting_list_time = timeit(stmt=sorting_list, globals=globals())
sorting_dict_time = timeit(stmt=sorting_dict, globals=globals())

print(
    f"""\
{sorting_list_time=:.2f} seconds
{sorting_dict_time=:.2f} seconds
list is {(sorting_dict_time/sorting_list_time):.2f} times faster"""
)

This code imports the sample data structures for sorting on the age attribute. It may seem like you aren’t using the imports from samples, but it’s necessary for these samples to be in the global namespace so that the timeit context has access to them.

Running the code for this test on the command line should provide you with some interesting results:

$ python compare_sorting_dict_vs_list.py
sorting_list_time=1.15 seconds
sorting_dict_time=2.26 seconds
list is 1.95 times faster

Sorting a list can be almost twice as fast as the process required to sort a dictionary view and then create a new sorted dictionary. So, if you plan on sorting your data very regularly, then a list of tuples might be better than a dictionary for you.

One of the main overheads when sorting a dictionary, as opposed to a list, is reconstructing the dictionary after sorting it. If you were to take out the outer dict() constructor, then you’d significantly cut the execution time.

In the next section, you’ll be looking at the time it takes to look up values in a dictionary of dictionaries versus in a list of dictionaries.

Comparing the Performance of Lookups

However, if you plan to use the dictionary to sort your data once and use that dictionary mainly for lookups, then a dictionary will definitely make more sense than a list:

# compare_lookup_dict_vs_list.py

from timeit import timeit
from samples import dictionary_of_dictionaries, list_of_dictionaries

lookups = [15, 18, 19, 16, 6, 12, 5, 3, 9, 20, 2, 10, 13, 17, 4, 14, 11, 7, 8]

list_setup = """
def get_key_from_list(key):
    for item in list_of_dictionaries:
        if item["id"] == key:
            return item
"""

lookup_list = """
for key in lookups:
    get_key_from_list(key)
"""

lookup_dict = """
for key in lookups:
    dictionary_of_dictionaries[key]
"""

lookup_list_time = timeit(stmt=lookup_list, setup=list_setup, globals=globals())
lookup_dict_time = timeit(stmt=lookup_dict, globals=globals())

print(
    f"""\
{lookup_list_time=:.2f} seconds
{lookup_dict_time=:.2f} seconds
dict is {(lookup_list_time / lookup_dict_time):.2f} times faster"""
)

This code makes a series of lookups to both the list and the dictionary. You’ll note that with the list, you have to write a special function to make a lookup. The function to make the list lookup involves going through all the list elements one by one until you find the target element, which isn’t ideal.

Running this comparison script from the command line should yield a result showing that dictionary lookups are significantly faster:

$ python compare_lookup_dict_vs_list.py
lookup_list_time=6.73 seconds
lookup_dict_time=0.38 seconds
dict is 17.83 times faster

Nearly eighteen times faster! That’s a whole bunch. So, you certainly want to weigh the blazing speed of dictionary lookups against the data structure’s slower sorting. Bear in mind that this ratio can vary significantly from system to system, not to mention the variation that might come from differently sized dictionaries or lists.

Dictionary lookups are certainly faster, though, no matter how you slice it. That said, if you’re just doing lookups, then you could just as easily do that with a regular unsorted dictionary. Why would you need a sorted dictionary in that case? Leave your use case in the comments!

Now you should have a relatively good idea of some trade-offs between two ways to store your key-value data. The conclusion that you can reach is that, most of the time, if you want a sorted data structure, then you should probably steer clear of the dictionary, mainly for language interoperability reasons.

That said, give Grant Jenks’ aforementioned sorted dictionary a try. It uses some ingenious strategies to get around typical performance drawbacks.

Do you have any interesting or performant implementations of a sorted key-value data structure? Share them in the comments, along with your use cases for a sorted dictionary!