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  2. RTree: Count points in the neighbourhoods within each point of another set of points

RTree: Count points in the neighbourhoods within each point of another set of points

由 梦想的初衷 提交于 2019-12-05 04:23:18

Rather than answer your question directly, I'd argue that you're doing this wrong. After arguing this, I'll give a better answer.

Why you're doing it wrong

An r-tree is great for bounding-box queries in two or three Euclidean dimensions.

You are looking up longitude-latitude points on a two-dimensional surface curved in a three-dimensional space. The upshot is that your coordinate system will yield singularities and discontinuities: 180°W is the same as 180°E, 2°E by 90°N is close to 2°W by 90°N. The r-tree does not capture these sorts of things!

But, even if they were a good solution, your idea to take lat±r and lon±r yields a square region; rather, you probably want a circular region around your point.

How to do it right

  1. Rather than keeping the points in lon-lat format, convert them to xyz format using a spherical coordinate conversion. Now they are in a 3D Euclidean space and there are no singularities or discontinuities.

  2. Place the points in a three-dimensional kd-tree. This allows you to quickly, in O(log n) time, ask questions like "What are the k-nearest neighbours to this point?" and "What are all the points within a radius r of this points?" SciPy comes with an implementation.

  3. For your radius search, convert from a Great Circle radius to a chord: this makes the search in 3-space equivalent to a radius search on a circle wrapped to the surface of a sphere (in this case, the Earth).

Code for doing it right

I've implemented the foregoing in Python as a demonstration. Note that all spherical points are stored in (longitude,latitude)/(x-y) format using a lon=[-180,180], lat=[-90,90] scheme. All 3D points are stored in (x,y,z) format.

#/usr/bin/env python3

import numpy as np
import scipy as sp
import scipy.spatial

Rearth = 6371

#Generate uniformly-distributed lon-lat points on a sphere
#See: http://mathworld.wolfram.com/SpherePointPicking.html
def GenerateUniformSpherical(num):
  #Generate random variates
  pts      = np.random.uniform(low=0, high=1, size=(num,2))
  #Convert to sphere space
  pts[:,0] = 2*np.pi*pts[:,0]          #0-360 degrees
  pts[:,1] = np.arccos(2*pts[:,1]-1)   #0-180 degrees
  #Convert to degrees
  pts = np.degrees(pts)
  #Shift ranges to lon-lat
  pts[:,0] -= 180
  pts[:,1] -= 90
  return pts

def ConvertToXYZ(lonlat):
  theta  = np.radians(lonlat[:,0])+np.pi
  phi    = np.radians(lonlat[:,1])+np.pi/2
  x      = Rearth*np.cos(theta)*np.sin(phi)
  y      = Rearth*np.sin(theta)*np.sin(phi)
  z      = Rearth*np.cos(phi)
  return np.transpose(np.vstack((x,y,z)))

#Get all points which lie with `r_km` Great Circle kilometres of the query
#points `qpts`.
def GetNeighboursWithinR(qpts,kdtree,r_km):
  #We need to convert Great Circle kilometres into chord length kilometres in
  #order to use the kd-tree
  #See: http://mathworld.wolfram.com/CircularSegment.html
  angle        = r_km/Rearth
  chord_length = 2*Rearth*np.sin(angle/2)
  pts3d        = ConvertToXYZ(qpts)
  #See: https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.spatial.KDTree.query_ball_point.html#scipy.spatial.KDTree.query_ball_point
  #p=2 implies Euclidean distance, eps=0 implies no approximation (slower)
  return kdtree.query_ball_point(pts3d,chord_length,p=2,eps=0) 


##############################################################################
#WARNING! Do NOT alter pts3d or kdtree will malfunction and need to be rebuilt
##############################################################################

##############################
#Correctness tests on the North, South, East, and West poles, along with Kolkata
ptsll = np.array([[0,90],[0,-90],[0,0],[-180,0],[88.3639,22.5726]])
pts3d = ConvertToXYZ(ptsll)
kdtree = sp.spatial.KDTree(pts3d, leafsize=10) #Stick points in kd-tree for fast look-up

qptsll = np.array([[-3,88],[5,-85],[10,10],[-178,3],[175,4]])
GetNeighboursWithinR(qptsll, kdtree, 2000)

##############################
#Stress tests
ptsll = GenerateUniformSpherical(100000)    #Generate uniformly-distributed lon-lat points on a sphere
pts3d = ConvertToXYZ(ptsll)                 #Convert points to 3d
#See: https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.spatial.KDTree.html
kdtree = sp.spatial.KDTree(pts3d, leafsize=10) #Stick points in kd-tree for fast look-up

qptsll = GenerateUniformSpherical(100)      #We'll find neighbours near these points
GetNeighboursWithinR(qptsll, kdtree, 500)

I've attached code which should, with some minor modifications, do what you want.

I think your problem arose for one of two reasons:

  1. You were not correctly constructing the spatial index. Your responses to my comments suggested that you weren't wholly aware of how the spatial index was getting made.

  2. The bounding box for your spatial query was not built correctly.

I'll discuss both possibilities below.

Constructing the spatial index

As it turns out, the spatial index is constructed simply by typing:

sindex = gpd_df.sindex

Magic.

But from whence does gpd_df.sindex get its data? It assumes that the data is stored in a column called geometry in a shapely format. If you have not added data to such a column, it will raise a warning.

A correct initialization of the data frame would look like so:

#Generate random points throughout Oregon
x = np.random.uniform(low=oregon_xmin, high=oregon_xmax, size=10000)
y = np.random.uniform(low=oregon_ymin, high=oregon_ymax, size=10000)

#Turn the lat-long points into a geodataframe
gpd_df = gpd.GeoDataFrame(data={'x':x, 'y':y})
#Set up point geometries so that we can index the data frame
#Note that I am using x-y points!
gpd_df['geometry'] = gpd_df.apply(lambda row: shapely.geometry.Point((row['x'], row['y'])), axis=1)

#Automagically constructs a spatial index from the `geometry` column
gpd_df.sindex

Seeing the foregoing sort of example code in your question would have been helpful in diagnosing your problem and getting going on solving it.

Since you did not get the extremely obvious warning geopandas raises when a geometry column is missing:

AttributeError: No geometry data set yet (expected in column 'geometry'.

I think you've probably done this part right.

Constructing the bounding box

In your question, you form a bounding box like so:

nearest_index = list(sindex.intersection((pt_center.LATITUDE-r, pt_center.LONGITUDE-r, pt_center.LATITUDE+r, pt_center.LONGITUDE+r)))

As it turns out, bounding boxes have the form:

(West, South, East, North)

At least, they do for X-Y styled-points, e.g. shapely.geometry.Point(Lon,Lat)

In my code, I use the following:

bbox = (cpt.x-radius, cpt.y-radius, cpt.x+radius, cpt.y+radius)

Working example

Putting the above together leads me to this working example. Note that I also demonstrate how to sort points by distance, answering your second question.

#!/usr/bin/env python3

import numpy as np
import numpy.random
import geopandas as gpd
import shapely.geometry
import operator

oregon_xmin = -124.5664
oregon_xmax = -116.4633
oregon_ymin = 41.9920
oregon_ymax = 46.2938

def radius(gpd_df, cpt, radius):
  """
  :param gpd_df: Geopandas dataframe in which to search for points
  :param cpt:    Point about which to search for neighbouring points
  :param radius: Radius about which to search for neighbours
  :return:       List of point indices around the central point, sorted by
                 distance in ascending order
  """
  #Spatial index
  sindex = gpd_df.sindex
  #Bounding box of rtree search (West, South, East, North)
  bbox = (cpt.x-radius, cpt.y-radius, cpt.x+radius, cpt.y+radius)
  #Potential neighbours
  good = []
  for n in sindex.intersection(bbox):
    dist = cpt.distance(gpd_df['geometry'][n])
    if dist<radius:
      good.append((dist,n))
  #Sort list in ascending order by `dist`, then `n`
  good.sort() 
  #Return only the neighbour indices, sorted by distance in ascending order
  return [x[1] for x in good]

#Generate random points throughout Oregon
x = np.random.uniform(low=oregon_xmin, high=oregon_xmax, size=10000)
y = np.random.uniform(low=oregon_ymin, high=oregon_ymax, size=10000)

#Turn the lat-long points into a geodataframe
gpd_df = gpd.GeoDataFrame(data={'x':x, 'y':y})
#Set up point geometries so that we can index the data frame
gpd_df['geometry'] = gpd_df.apply(lambda row: shapely.geometry.Point((row['x'], row['y'])), axis=1)

#The 'x' and 'y' columns are now stored as part of the geometry, so we remove
#their columns in order to save space
del gpd_df['x']
del gpd_df['y']

for i, row in gpd_df.iterrows():
  neighbours = radius(gpd_df,row['geometry'],0.5)
  print(neighbours)
  #Use len(neighbours) here to construct a new row for the data frame

(What I had been requesting in the comments is code that looks like the foregoing, but which exemplifies your problem. Note the use of random to succinctly generate a dataset for experimentation.)

标签
gis
geospatial
spatial-index
r-tree
geopandas
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