This rule raises an issue when a group iterator returned by itertools.groupby() is stored in a container that persists beyond the current loop iteration, instead of being consumed immediately.

Why is this an issue?

The itertools.groupby() function groups consecutive items from an iterable based on a key function. It returns pairs of (key, group_iterator) where each group_iterator is a sub-iterator that yields items for that specific group.

A critical characteristic of these group iterators is that they share the same underlying data source as the main groupby iterator. Every time the outer loop advances to the next group, the previous group iterator is immediately invalidated — regardless of whether it has been iterated or not. Storing a raw group iterator and consuming it after the loop has moved on will always yield an empty sequence.

What is the potential impact?

When group iterators are reused, they yield no items. Your code will process empty sequences instead of the actual grouped data, leading to incorrect results, silent failures, and data loss. These issues can be difficult to debug because the code does not raise exceptions but simply produces wrong output.

How to fix it

There are two correct approaches:

• Consume the group iterator immediately inside the loop body, never storing the raw iterator at all. This is the simplest pattern and avoids any risk of reuse.
• Materialize the group iterator into a concrete collection (list, tuple, set, etc.) before the loop advances. This allows the data to be stored and accessed later.

Code examples

Noncompliant code example

from itertools import groupby

data = [1, 1, 2, 2, 3]
groups = {}

for key, group in groupby(data):
    groups[key] = group  # Noncompliant

for key, group in groups.items():
    print(f"{key}: {list(group)}")  # Empty results

Compliant solution

from itertools import groupby

data = [1, 1, 2, 2, 3]
groups = {}

for key, group in groupby(data):
    groups[key] = list(group)  # Convert to list immediately

for key, group in groups.items():
    print(f"{key}: {group}")  # Correct results

Resources

Documentation

• Python Documentation - itertools.groupby