How to Sort a List of Dictionaries by Key in Python

A reusable helper function that sorts a list of dictionaries by a specified key, with optional descending order support.

Easy Python 3.9+ Aug 9, 2026 Data pipelines & processing 12 views 0 copies

sorting dictionaries data-pipelines lambda sorted

Python code

37 lines

Python 3.9+

from typing import List

def sort_records(records: List[dict], key: str, descending: bool = False) -> List[dict]:
    """Sort a list of dictionaries by a specified key."""
    return sorted(records, key=lambda record: record[key], reverse=descending)


def demonstrate_sorting() -> None:
    users = [
        {"name": "Alice", "age": 34, "score": 88.5},
        {"name": "Bob", "age": 27, "score": 92.0},
        {"name": "Charlie", "age": 29, "score": 76.3},
        {"name": "Diana", "age": 31, "score": 95.1},
    ]

    print("Original records:")
    for user in users:
        print(user)

    print("\nSorted by age (ascending):")
    sorted_by_age = sort_records(users, "age")
    for user in sorted_by_age:
        print(user)

    print("\nSorted by score (descending):")
    sorted_by_score = sort_records(users, "score", descending=True)
    for user in sorted_by_score:
        print(user)

    print("\nSorted by name (ascending):")
    sorted_by_name = sort_records(users, "name")
    for user in sorted_by_name:
        print(user)


if __name__ == "__main__":
    demonstrate_sorting()

Output

stdout

Original records:
{'name': 'Alice', 'age': 34, 'score': 88.5}
{'name': 'Bob', 'age': 27, 'score': 92.0}
{'name': 'Charlie', 'age': 29, 'score': 76.3}
{'name': 'Diana', 'age': 31, 'score': 95.1}

Sorted by age (ascending):
{'name': 'Bob', 'age': 27, 'score': 92.0}
{'name': 'Charlie', 'age': 29, 'score': 76.3}
{'name': 'Diana', 'age': 31, 'score': 95.1}
{'name': 'Alice', 'age': 34, 'score': 88.5}

Sorted by score (descending):
{'name': 'Diana', 'age': 31, 'score': 95.1}
{'name': 'Bob', 'age': 27, 'score': 92.0}
{'name': 'Alice', 'age': 34, 'score': 88.5}
{'name': 'Charlie', 'age': 29, 'score': 76.3}

Sorted by name (ascending):
{'name': 'Alice', 'age': 34, 'score': 88.5}
{'name': 'Bob', 'age': 27, 'score': 92.0}
{'name': 'Charlie', 'age': 29, 'score': 76.3}
{'name': 'Diana', 'age': 31, 'score': 95.1}

How it works

The sorted() function returns a new list and leaves the original untouched, which is ideal for data pipelines. The key=lambda record: record[key] tells Python to extract the sort value from each dictionary. The reverse parameter flips the order to descending when set to True. Type hints (List[dict], str, bool) make the helper self-documenting and friendly for static checkers.

Common mistakes

Using `.sort()` on the original list, which mutates source data and may break downstream pipeline steps
Forgetting to handle missing keys — `record[key]` raises `KeyError` when the key is absent
Assuming decimal and string values sort the same way without normalizing types first

Variations

Use `operator.itemgetter(key)` as a faster `key` argument instead of a lambda
Sort in-place with `records.sort(key=..., reverse=...)` if mutation is acceptable and memory matters

Real-world use cases

Ordering API response rows by a timestamp or priority field before writing them to a database table.
Ranking customer records by a computed score in an ETL batch before sending to a reporting dashboard.
Sorting log entries or event payloads by severity or time so downstream aggregations process them in order.

How to Sort a List of Dictionaries by Key in Python

Python code

Output

How it works

Common mistakes

Variations

Real-world use cases

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