Quick Tips · Tip 05 · 0:21
Speed up recursive Python functions with one decorator: functools.cache
A plain fib(30) makes 2,692,537 calls because it repeats the same work. With functools.cache every answer is computed once and the count drops to 31 (Python 3.9+).
Key moments
What you’ll needPython 3.9+
What you’ll need
- python.org/downloads →
Python 3.9 or newerRuns the code. Check yours with python3 --version. - code.visualstudio.com →
VS Code or CursorEither works; so does any editor you like. - git-scm.com/downloads →
gitOptional for a tip: to keep your own copy of the code. - No packages needed: it uses only the Python standard library.
Set it up
- 1Make a virtual environmentIt keeps this project's packages separate from the rest of your computer.
python3 -m venv .venv source .venv/bin/activate
- 2Run itSave the snippet below as fib.py.
python3 fib.py
fib.pyOpen on GitHub →
from functools import cache # Python 3.9+
def fib(n):
global calls; calls += 1
return n if n < 2 else fib(n-1) + fib(n-2)
calls = 0; fib(30); print(calls)
fib = cache(fib); calls = 0; fib(30); print(calls)Transcript
This function repeats the same work millions of times. Each call splits in two, and counts itself. Run it plain, then wrapped in functools cache. Each answer computed once, then remembered.
2.7 million calls, down to 31. Repeated work? Cache it.
Read the lesson on GitHub →- #python
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