Merge rankings with Reciprocal Rank Fusion

What you'll build

In this tutorial, you build the Reciprocal Rank Fusion function used by this repository's hybrid retriever. You combine lexical and vector rankings without comparing their incompatible raw scores, promote chunks found by more than one retriever, and return one deterministic result list.

Time: About 30 minutes.

Before you begin

You need:

• Python 3.12 or later.
• uv.
• A local clone with dependencies installed.
• Implement Okapi BM25 from scratch.
• Build and search a vector index.

You do not need an API key. Every command uses small in-memory rankings.

See it work

Run the focused RRF tests and read the behaviors they confirm:

uv run pytest tests/test_rag_rrf.py -v

test_rrf_promotes_chunks_found_by_multiple_retrievers PASSED
test_rrf_deduplicates_within_one_ranking PASSED
test_rrf_uses_chunk_id_for_deterministic_ties PASSED
test_rrf_validates_configuration PASSED
4 passed

The tests prove that shared chunks gain influence from multiple rankings, duplicate results do not count twice, ties remain stable, and invalid fusion settings fail before scoring begins.

How Reciprocal Rank Fusion works

A rank fusion method combines multiple ordered result lists into one. The input lists may come from different retrieval systems, such as BM25 and cosine similarity, whose raw scores are not comparable.

Reciprocal Rank Fusion uses a result's position instead of its source score. A result at position $r$ contributes:

weight / (k + r)

The constant k reduces the difference between nearby positions. This project uses k=60 by default. A chunk found by multiple retrievers receives one contribution from each ranking, so repeated independent evidence can move it above chunks found by only one retriever.

The input scores remain incomparable score spaces: BM25 scores depend on term rarity and corpus statistics, while cosine similarity stays between -1 and 1. RRF never adds those source scores. Its output score is a new reciprocal-rank score used only for the fused ordering.

Preserve the reference implementation

The finished implementation already exists so that you can run it before rebuilding it. Work on a disposable branch or temporary copy of the repository.

In PowerShell, preserve the file and clear the original:

Copy-Item app/rag/rrf.py app/rag/rrf_reference.txt
Clear-Content app/rag/rrf.py

In Bash, run the equivalent commands:

cp app/rag/rrf.py app/rag/rrf_reference.txt
: > app/rag/rrf.py

Validate fusion settings

1. In app/rag/rrf.py, add the module imports:

   """Reciprocal Rank Fusion for combining independent retrieval rankings."""
   
   from __future__ import annotations
   
   from collections.abc import Sequence
   
   from app.rag.models import Chunk, SearchResult
   

2. Add the function signature and validate k:

   def reciprocal_rank_fusion(
       rankings: Sequence[Sequence[SearchResult]],
       *,
       k: int = 60,
       weights: Sequence[float] | None = None,
   ) -> list[SearchResult]:
       """Fuse rankings by chunk ID without comparing source scores."""
       if k < 1:
           raise ValueError("k must be greater than zero")
   

3. Create default weights and require one weight per ranking:

       if weights is None:
           weights = [1.0] * len(rankings)
       if len(weights) != len(rankings):
           raise ValueError("weights must match the number of rankings")
       if any(weight < 0 for weight in weights):
           raise ValueError("weights cannot be negative")
   

   A weight changes one retriever's influence without making its source scores comparable to another retriever's scores.

Accumulate reciprocal ranks

1. Create dictionaries for source chunks, fused scores, and best positions:

       chunks: dict[str, Chunk] = {}
       scores: dict[str, float] = {}
       best_ranks: dict[str, int] = {}
   

2. Iterate through each ranking and its weight:

       for results, weight in zip(rankings, weights, strict=True):
           seen_in_ranking: set[str] = set()
           for position, result in enumerate(results, start=1):
               chunk_id = result.chunk.chunk_id
   

3. Ignore a duplicate chunk within the same ranking:

               if chunk_id in seen_in_ranking:
                   continue
               seen_in_ranking.add(chunk_id)
   

   One retriever may return a duplicate by mistake. Counting it twice would imitate independent evidence that does not exist.

4. Store the chunk and add its reciprocal-rank contribution:

               chunks[chunk_id] = result.chunk
               scores[chunk_id] = scores.get(
                   chunk_id, 0.0
               ) + weight / (k + position)
               best_ranks[chunk_id] = min(
                   best_ranks.get(chunk_id, position), position
               )
   

   A chunk present in two rankings receives two contributions. best_ranks records its strongest source position for deterministic tie-breaking.

Build the fused ranking

1. Sort chunk IDs by descending fused score, best source position, and chunk ID:

       ordered_ids = sorted(
           scores,
           key=lambda chunk_id: (
               -scores[chunk_id],
               best_ranks[chunk_id],
               chunk_id,
           ),
       )
   

2. Convert the ordered IDs into new SearchResult values:

       return [
           SearchResult(
               chunk=chunks[chunk_id],
               score=scores[chunk_id],
               rank=rank,
           )
           for rank, chunk_id in enumerate(ordered_ids, start=1)
       ]
   

   The returned score belongs to the fused ranking. It does not preserve or reinterpret any source retriever's score.

Verify your work

Run the focused tests:

uv run pytest tests/test_rag_rrf.py -q

4 passed

Check the implementation and tests with Ruff:

uv run ruff check app/rag/rrf.py tests/test_rag_rrf.py

After your implementation matches the reference behavior, remove the temporary copy:

Remove-Item app/rag/rrf_reference.txt

Break it on purpose

Run the test that deliberately passes incompatible configuration:

uv run pytest tests/test_rag_rrf.py::test_rrf_validates_configuration -v

test_rrf_validates_configuration PASSED
1 passed

The test supplies one ranking but no matching weight and asserts that the function raises ValueError. It also passes k=0, which would make the fusion constant invalid, and asserts the second error. These failures keep configuration mistakes from producing a plausible but misleading fused order.

RRF still has an important limit: it can promote agreement only among the rankings it receives. If every retriever misses the relevant chunk, fusion cannot recover it. Retrieval evaluation measures that end-to-end behavior.

Troubleshooting

Symptom	Resolution	Source
ValueError: weights must match the number of rankings	Pass one weight for each ranking, or omit weights for equal weighting.	Observed
ValueError: k must be greater than zero	Use a positive fusion constant.	Observed
One duplicate chunk receives too much score	Keep seen_in_ranking inside the outer ranking loop.	Predicted
Equal scores change order between runs	Include best rank and chunk ID in the sort key.	Predicted
A strong source score has no effect	Expected: RRF uses rank positions, not source scores.	Predicted

Next steps

Continue with Benchmark retrieval quality to measure whether fused or individual rankings return relevant chunks at a useful cutoff. A later hybrid-retriever tutorial wires BM25, vector search, and RRF into one query path.

For the original method, read Reciprocal Rank Fusion outperforms Condorcet and individual Rank Learning Methods.

Tested against

Item	Value
Python	3.13.14; the repository requires 3.12 or later
pytest	9.1.1
Verified	2026-08-17