What is the difference between a `dict` and a `defaultdict` in Python?

Q: What is the difference between a `dict` and a `defaultdict` in Python?

Learn the answer to "What is the difference between a `dict` and a `defaultdict` in Python?" with detailed explanations, code examples, and best practices on DeployU.

The Scenario

You are a backend engineer at a social media company. You are writing a new service that needs to count the number of times each word appears in a large text file.

You are considering using either a dict or a defaultdict to store the word counts.

The Challenge

Explain the difference between a dict and a defaultdict in Python. What are the pros and cons of each approach, and which one would you choose for this use case?

Wrong Approach

A junior engineer might just use a `dict` and write a lot of boilerplate code to check if a key exists before incrementing its value. They might not be aware of the `defaultdict` class, which is a much more elegant solution.

Addresses symptoms, not root cause

Right Approach

A senior engineer would know that a `defaultdict` is the perfect tool for this job. They would be able to explain the difference between a `dict` and a `defaultdict`, and they would be able to write a concise and elegant solution using a `defaultdict`.

Step 1: Understand the Key Differences

Feature	`dict`	`defaultdict`
Missing Keys	Raises a `KeyError` if you try to access a key that does not exist.	Returns a default value if you try to access a key that does not exist.
Syntax	`{}`	`defaultdict(default_factory)`
Use Cases	When you want to handle missing keys explicitly.	When you want to provide a default value for missing keys.

Step 2: Choose the Right Tool for the Job

For our use case, a defaultdict is the best choice. It allows us to provide a default value of 0 for missing keys, which makes it easy to increment the count for a new word.

Step 3: Code Examples

Here are some code examples that show the difference between the two approaches:

dict:

word_counts = {}
for word in words:
  if word in word_counts:
    word_counts[word] += 1
  else:
    word_counts[word] = 1

defaultdict:

from collections import defaultdict

word_counts = defaultdict(int)
for word in words:
  word_counts[word] += 1

As you can see, the defaultdict solution is much more concise and elegant.

Systematic, production-ready debugging

Practice Question

You want to create a dictionary that groups a list of words by their first letter. Which of the following would be the most elegant way to do this?

Questions