Reciprocal rank fusionedit
This functionality is in technical preview and may be changed or removed in a future release. The syntax will likely change before GA. Elastic will work to fix any issues, but features in technical preview are not subject to the support SLA of official GA features.
Reciprocal rank fusion (RRF) is a method for combining multiple result sets with different relevance indicators into a single result set. RRF requires no tuning, and the different relevance indicators do not have to be related to each other to achieve high-quality results.
RRF uses the following formula to determine the score for ranking each document:
score = 0.0
for q in queries:
if d in result(q):
score += 1.0 / ( k + rank( result(q), d ) )
return score
# where
# k is a ranking constant
# q is a query in the set of queries
# d is a document in the result set of q
# result(q) is the result set of q
# rank( result(q), d ) is d's rank within the result(q) starting from 1
Reciprocal rank fusion APIedit
You can use RRF as part of a search to combine and rank documents using separate sets of top documents (result sets) from a combination of child retrievers using an RRF retriever. A minimum of two child retrievers is required for ranking.
An RRF retriever is an optional object defined as part of a search request’s retriever parameter. The RRF retriever object contains the following parameters:
-
retrievers -
(Required, array of retriever objects)
A list of child retrievers to specify which sets of returned top documents will have the RRF formula applied to them. Each child retriever carries an equal weight as part of the RRF formula. Two or more child retrievers are required.
-
rank_constant -
(Optional, integer)
This value determines how much influence documents in individual result sets per query have over the final ranked result set. A higher value indicates that lower ranked documents have more influence. This value must be greater than or equal to
1. Defaults to60. -
window_size -
(Optional, integer)
This value determines the size of the individual result sets per query. A higher value will improve result relevance at the cost of performance. The final ranked result set is pruned down to the search request’s size.
window_sizemust be greater than or equal tosizeand greater than or equal to1. Defaults to thesizeparameter.
An example request using RRF:
GET example-index/_search
{
"retriever": {
"rrf": {
"retrievers": [
{
"standard": {
"query": {
"term": {
"text": "shoes"
}
}
}
},
{
"knn": {
"field": "vector",
"query_vector": [1.25, 2, 3.5],
"k": 50,
"num_candidates": 100
}
}
],
"rank_window_size": 50,
"rank_constant": 20
}
}
}
In the above example, we execute the knn and standard retrievers
independently of each other. Then we use the rrf retriever to combine
the results.
|
First, we execute the kNN search specified by the |
|
|
Second, we execute the query specified by the |
|
|
Then, on a coordinating node, we combine the kNN search top documents with
the query top documents and rank them based on the RRF formula using parameters from
the |
Note that if k from a knn search is larger than rank_window_size, the results are
truncated to rank_window_size. If k is smaller than rank_window_size, the results are
k size.
Reciprocal rank fusion supported featuresedit
The rrf retriever supports:
The rrf retriever does not currently support:
Using unsupported features as part of a search with an rrf retriever results
in an exception.
Reciprocal rank fusion using multiple standard retrieversedit
The rrf retriever provides a way to combine and rank multiple
standard retrievers. A primary use case is combining top documents
from a traditional BM25 query and an ELSER
query to achieve improved relevance.
An example request using RRF with multiple standard retrievers:
GET example-index/_search
{
"retriever": {
"rrf": {
"retrievers": [
{
"standard": {
"query": {
"term": {
"text": "blue shoes sale"
}
}
}
},
{
"standard": {
"query": {
"text_expansion":{
"ml.tokens":{
"model_id":"my_elser_model",
"model_text":"What blue shoes are on sale?"
}
}
}
}
}
],
"rank_window_size": 50,
"rank_constant": 20
}
}
}
In the above example, we execute each of the two standard retrievers
independently of each other. Then we use the rrf retriever to combine
the results.
|
First we run the |
|
|
Next we run the |
|
|
The |
Not only does this remove the need to figure out what the appropriate weighting is using linear combination, but RRF is also shown to give improved relevance over either query individually.
Reciprocal rank fusion using sub searchesedit
RRF using sub searches is no longer supported. Use the retriever API instead. See using multiple standard retrievers for an example.
Reciprocal rank fusion full exampleedit
We begin by creating a mapping for an index with a text field, a vector field, and an integer field along with indexing several documents. For this example we are going to use a vector with only a single dimension to make the ranking easier to explain.
PUT example-index
{
"mappings": {
"properties": {
"text" : {
"type" : "text"
},
"vector": {
"type": "dense_vector",
"dims": 1,
"index": true,
"similarity": "l2_norm"
},
"integer" : {
"type" : "integer"
}
}
}
}
PUT example-index/_doc/1
{
"text" : "rrf",
"vector" : [5],
"integer": 1
}
PUT example-index/_doc/2
{
"text" : "rrf rrf",
"vector" : [4],
"integer": 2
}
PUT example-index/_doc/3
{
"text" : "rrf rrf rrf",
"vector" : [3],
"integer": 1
}
PUT example-index/_doc/4
{
"text" : "rrf rrf rrf rrf",
"integer": 2
}
PUT example-index/_doc/5
{
"vector" : [0],
"integer": 1
}
POST example-index/_refresh
We now execute a search using an rrf retriever with a standard retriever
specifying a BM25 query, a knn retriever specifying a kNN search, and
a terms aggregation.
GET example-index/_search
{
"retriever": {
"rrf": {
"retrievers": [
{
"standard": {
"query": {
"term": {
"text": "rrf"
}
}
}
},
{
"knn": {
"field": "vector",
"query_vector": [3],
"k": 5,
"num_candidates": 5
}
}
],
"rank_window_size": 5,
"rank_constant": 1
}
},
"size": 3,
"aggs": {
"int_count": {
"terms": {
"field": "integer"
}
}
}
}
And we receive the response with ranked hits and the terms
aggregation result. Note that _score is null, and we instead
use _rank to show our top-ranked documents.
{
"took": ...,
"timed_out" : false,
"_shards" : {
"total" : 1,
"successful" : 1,
"skipped" : 0,
"failed" : 0
},
"hits" : {
"total" : {
"value" : 5,
"relation" : "eq"
},
"max_score" : null,
"hits" : [
{
"_index" : "example-index",
"_id" : "3",
"_score" : null,
"_rank" : 1,
"_source" : {
"integer" : 1,
"vector" : [
3
],
"text" : "rrf rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "2",
"_score" : null,
"_rank" : 2,
"_source" : {
"integer" : 2,
"vector" : [
4
],
"text" : "rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "4",
"_score" : null,
"_rank" : 3,
"_source" : {
"integer" : 2,
"text" : "rrf rrf rrf rrf"
}
}
]
},
"aggregations" : {
"int_count" : {
"doc_count_error_upper_bound" : 0,
"sum_other_doc_count" : 0,
"buckets" : [
{
"key" : 1,
"doc_count" : 3
},
{
"key" : 2,
"doc_count" : 2
}
]
}
}
}
Let’s break down how these hits were ranked. We
start by running the standard retriever specifying a query
and the knn retriever specifying a kNN search separately to
collect what their individual hits are.
First, we look at the hits for the query from the standard retriever.
"hits" : [
{
"_index" : "example-index",
"_id" : "4",
"_score" : 0.16152832,
"_source" : {
"integer" : 2,
"text" : "rrf rrf rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "3",
"_score" : 0.15876243,
"_source" : {
"integer" : 1,
"vector" : [3],
"text" : "rrf rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "2",
"_score" : 0.15350538,
"_source" : {
"integer" : 2,
"vector" : [4],
"text" : "rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "1",
"_score" : 0.13963442,
"_source" : {
"integer" : 1,
"vector" : [5],
"text" : "rrf"
}
}
]
Note that our first hit doesn’t have a value for the vector field. Now,
we look at the results for the kNN search from the knn retriever.
"hits" : [
{
"_index" : "example-index",
"_id" : "3",
"_score" : 1.0,
"_source" : {
"integer" : 1,
"vector" : [3],
"text" : "rrf rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "2",
"_score" : 0.5,
"_source" : {
"integer" : 2,
"vector" : [4],
"text" : "rrf rrf"
}
},
{
"_index" : "example-index",
"_id" : "1",
"_score" : 0.2,
"_source" : {
"integer" : 1,
"vector" : [5],
"text" : "rrf"
}
},
{
"_index" : "example-index",
"_id" : "5",
"_score" : 0.1,
"_source" : {
"integer" : 1,
"vector" : [0]
}
}
]
We can now take the two individually ranked result sets and apply the
RRF formula to them using parameters from the rrf retriever to get
our final ranking.
# doc | query | knn | score _id: 1 = 1.0/(1+4) + 1.0/(1+3) = 0.4500 _id: 2 = 1.0/(1+3) + 1.0/(1+2) = 0.5833 _id: 3 = 1.0/(1+2) + 1.0/(1+1) = 0.8333 _id: 4 = 1.0/(1+1) = 0.5000 _id: 5 = 1.0/(1+4) = 0.2000
We rank the documents based on the RRF formula with a rank_window_size of 5
truncating the bottom 2 docs in our RRF result set with a size of 3.
We end with _id: 3 as _rank: 1, _id: 2 as _rank: 2, and
_id: 4 as _rank: 3. This ranking matches the result set from the
original RRF search as expected.
Pagination in RRFedit
When using rrf you can paginate through the results using the from parameter.
As the final ranking is solely dependent on the original query ranks, to ensure
consistency when paginating, we have to make sure that while from changes, the order
of what we have already seen remains intact. To that end, we’re using a fixed rank_window_size
as the whole available result set upon which we can paginate.
This essentially means that if:
-
from + size≤rank_window_size: we could getresults[from: from+size]documents back from the finalrrfranked result set -
from + size>rank_window_size: we would get 0 results back, as the request would fall outside the availablerank_window_size-sized result set.
An important thing to note here is that since rank_window_size is all the results that we’ll get to see
from the individual query components, pagination guarantees consistency, i.e. no documents are skipped
or duplicated in multiple pages, iff rank_window_size remains the same. If rank_window_size changes, then the order
of the results might change as well, even for the same ranks.
To illustrate all of the above, let’s consider the following simplified example where we have
two queries, queryA and queryB and their ranked documents:
| queryA | queryB | _id: | 1 | 5 | _id: | 2 | 4 | _id: | 3 | 3 | _id: | 4 | 1 | _id: | | 2 |
For rank_window_size=5 we would get to see all documents from both queryA and queryB.
Assuming a rank_constant=1, the rrf scores would be:
# doc | queryA | queryB | score _id: 1 = 1.0/(1+1) + 1.0/(1+4) = 0.7 _id: 2 = 1.0/(1+2) + 1.0/(1+5) = 0.5 _id: 3 = 1.0/(1+3) + 1.0/(1+3) = 0.5 _id: 4 = 1.0/(1+4) + 1.0/(1+2) = 0.533 _id: 5 = 0 + 1.0/(1+1) = 0.5
So the final ranked result set would be [1, 4, 2, 3, 5] and we would paginate over that, since
rank_window_size == len(results). In this scenario, we would have:
-
from=0, size=2would return documents [1,4] with ranks[1, 2] -
from=2, size=2would return documents [2,3] with ranks[3, 4] -
from=4, size=2would return document [5] with rank[5] -
from=6, size=2would return an empty result set as it there are no more results to iterate over
Now, if we had a rank_window_size=2, we would only get to see [1, 2] and [5, 4] documents
for queries queryA and queryB respectively. Working out the math, we would see that the results would now
be slightly different, because we would have no knowledge of the documents in positions [3: end] for either query.
# doc | queryA | queryB | score _id: 1 = 1.0/(1+1) + 0 = 0.5 _id: 2 = 1.0/(1+2) + 0 = 0.33 _id: 4 = 0 + 1.0/(1+2) = 0.33 _id: 5 = 0 + 1.0/(1+1) = 0.5
The final ranked result set would be [1, 5, 2, 4], and we would be able to paginate
on the top rank_window_size results, i.e. [1, 5]. So for the same params as above, we would now have:
-
from=0, size=2would return [1,5] with ranks[1, 2] -
from=2, size=2would return an empty result set as it would fall outside the availablerank_window_sizeresults.