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A library for efficiently storing and querying spatial data in the Go programming language.

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The R-tree is a popular data structure for efficiently storing and querying spatial objects; one common use is implementing geospatial indexes in database management systems. Both bounding-box queries and k-nearest-neighbor queries are supported.

R-trees are balanced, so maximum tree height is guaranteed to be logarithmic in the number of entries; however, good worst-case performance is not guaranteed. Instead, a number of rebalancing heuristics are applied that perform well in practice. For more details please refer to the references.

This implementation handles the general N-dimensional case; for a more efficient implementation for the 3-dimensional case, see Patrick Higgins' fork.

Getting Started

Get the source code from GitHub or, with Go 1 installed, run go get

Make sure you import in your Go source files.


Storing, updating, and deleting objects

To create a new tree, specify the number of spatial dimensions and the minimum and maximum branching factor:

rt := rtreego.NewTree(2, 25, 50)

You can also bulk-load the tree when creating it by passing the objects as a parameter.

rt := rtreego.NewTree(2, 25, 50, objects...)

Any type that implements the Spatial interface can be stored in the tree:

type Spatial interface {
  Bounds() *Rect

Rects are data structures for representing spatial objects, while Points represent spatial locations. Creating Points is easy--they're just slices of float64s:

p1 := rtreego.Point{0.4, 0.5}
p2 := rtreego.Point{6.2, -3.4}

To create a Rect, specify a location and the lengths of the sides:

r1, _ := rtreego.NewRect(p1, []float64{1, 2})
r2, _ := rtreego.NewRect(p2, []float64{1.7, 2.7})

To demonstrate, let's create and store some test data.

type Thing struct {
  where *Rect
  name string

func (t *Thing) Bounds() *Rect {
  return t.where

rt.Insert(&Thing{r1, "foo"})
rt.Insert(&Thing{r2, "bar"})

size := rt.Size() // returns 2

We can insert and delete objects from the tree in any order.

// do some stuff...

Note that Delete function does the equality comparison by comparing the memory addresses of the objects. If you do not have a pointer to the original object anymore, you can define a custom comparator.

type Comparator func(obj1, obj2 Spatial) (equal bool)

You can use a custom comparator with DeleteWithComparator function.

cmp := func(obj1, obj2 Spatial) bool {
  sp1 := obj1.(*IDRect)
  sp2 := obj2.(*IDRect)

  return sp1.ID == sp2.ID

rt.DeleteWithComparator(obj, cmp)

If you want to store points instead of rectangles, you can easily convert a point into a rectangle using the ToRect method:

var tol = 0.01

type Somewhere struct {
  location rtreego.Point
  name string
  wormhole chan int

func (s *Somewhere) Bounds() *Rect {
  // define the bounds of s to be a rectangle centered at s.location
  // with side lengths 2 * tol:
  return s.location.ToRect(tol)

rt.Insert(&Somewhere{rtreego.Point{0, 0}, "Someplace", nil})

If you want to update the location of an object, you must delete it, update it, and re-insert. Just modifying the object so that the *Rect returned by Location() changes, without deleting and re-inserting the object, will corrupt the tree.


Bounding-box and k-nearest-neighbors queries are supported.

Bounding-box queries require a search *Rect. It returns all objects that touch the search rectangle.

bb, _ := rtreego.NewRect(rtreego.Point{1.7, -3.4}, []float64{3.2, 1.9})

// Get a slice of the objects in rt that intersect bb:
results := rt.SearchIntersect(bb)


You can filter out values during searches by implementing Filter functions.

type Filter func(results []Spatial, object Spatial) (refuse, abort bool)

A filter for limiting results by result count is included in the package for backwards compatibility.

// maximum of three results will be returned
tree.SearchIntersect(bb, LimitFilter(3))

Nearest-neighbor queries find the objects in a tree closest to a specified query point.

q := rtreego.Point{6.5, -2.47}
k := 5

// Get a slice of the k objects in rt closest to q:
results = rt.NearestNeighbors(k, q)

More information

See GoDoc for full API documentation.



Written by Daniel Connelly (


rtreego is released under a BSD-style license, described in the LICENSE file.

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