Skip to content

Use nth_element for median computation in IndexLSH #4653

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
wants to merge 2 commits into
base: main
Choose a base branch
from
+8 −6
Open

Use nth_element for median computation in IndexLSH #4653

Changes from 1 commit
Commits
Show all changes
2 commits
Select commit Hold shift + click to select a range
29b2891
Use nth_element for median computation in IndexLSH
Ray0907 Nov 2, 2025
e88a52d
Refactor median computation based on review feedback.
Ray0907 Nov 8, 2025
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
14 changes: 9 additions & 5 deletions faiss/IndexLSH.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -86,12 +86,16 @@ void IndexLSH::train(idx_t n, const float* x) {

for (idx_t i = 0; i < nbits; i++) {
float* xi = transposed_x.get() + i * n;
// std::nth_element
std::sort(xi, xi + n);
if (n % 2 == 1)
// Use nth_element (O(n)) instead of sort (O(n log n)) for median
if (n % 2 == 1) {
std::nth_element(xi, xi + n / 2, xi + n);
thresholds[i] = xi[n / 2];
else
thresholds[i] = (xi[n / 2 - 1] + xi[n / 2]) / 2;
} else {
std::nth_element(xi, xi + n / 2, xi + n);
float median_high = xi[n / 2];
std::nth_element(xi, xi + n / 2 - 1, xi + n);
Copy link
Contributor

@alexanderguzhva alexanderguzhva Nov 24, 2025

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

looks good, unless running 2x std::nth_element() is too costly.

thresholds[i] = (xi[n / 2 - 1] + median_high) / 2;
}
Copy link

@subhadeepkaran subhadeepkaran Nov 7, 2025

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

maybe we can do like this?

// Use nth_element (O(n)) instead of sort (O(n log n)) for median
std::nth_element(xi, xi + n / 2, xi + n);
float median = xi[n / 2];
if (n % 2 == 0) {
    std::nth_element(xi, xi + n / 2 - 1, xi + n);
    median = (median + xi[n / 2 - 1]) / 2;
}
thresholds[i] = median;

Copy link
Contributor Author

@Ray0907 Ray0907 Nov 8, 2025

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Good point! Your version is cleaner!

}
}
is_trained = true;
Expand Down
Loading