CODEWITHSUNDEEP
pythonpointshapely
Wout VC
Wout VC

Reputation: 377

How to find the 2nd nearest point of a LineString in Shapely

Given a certain LineString and point p:

from shapely.ops import nearest_points
from shapely.geometry import Point

p = Point(51.21745162000732, 4.41871738126533)
linestring = LineString([(51.2176008, 4.4177154), (51.21758, 4.4178548), (51.2175729, 4.4179023), (51.21745162000732, 4.41871738126533)])

The nearest point to p is calculated by:

n_p = nearest_points(linestring, p)[0]

Conclusion it's the exact same point, which is normal since the exact same value is also in the linestring, but I need to know the nearest point, apart from the point itself. So how can I find the second nearest point?

Upvotes: 1

Views: 1936

Answers (2)

Georgy
Georgy

Reputation: 13687

In the general case, the simplest solution would be to construct a new geometric object from your LineString but without the nearest point, and then get the nearest point with this new geometry.:

from shapely.geometry import LineString, MultiPoint, Point
from shapely.ops import nearest_points

point = Point(51.21745162000732, 4.41871738126533)
line = LineString([(51.2176008, 4.4177154), (51.21758, 4.4178548), 
                   (51.2175729, 4.4179023), (51.21745162000732, 4.41871738126533)])

nearest_point = nearest_points(line, point)[0]
line_points_except_nearest = MultiPoint([point for point in linestring.coords 
                                         if point != (nearest_point.x, nearest_point.y)])
second_nearest = nearest_points(line_points_except_nearest, point)[0]

Alternatively, if you don't want to construct a new object because of, for example, memory constraints, you could run over all the points in the LineString with heapq.nsmallest:

import heapq

line_points = map(Point, line.coords)
nearest, second_nearest = heapq.nsmallest(2, line_points, key=point.distance)

In your specific case, when all the points are collinear, you can also calculate distances with the neighboring points of the nearest point:

index = list(line.coords).index((point.x, point.y))
if index == 0:
    second_nearest = Point(line.coords[1])
elif index == len(line.coords) - 1:
    second_nearest = Point(line.coords[-2])
else:
    second_nearest = min(Point(line.coords[index - 1]),
                         Point(line.coords[index + 1]),
                         key=point.distance)

Upvotes: 1

DarrylG
DarrylG

Reputation: 17156

Solve as folllows.

from shapely.ops import nearest_points
from shapely.geometry import Point
from shapely.geometry import LineString

def second_nearest(p, linestring):
  """ Finds nearest point of p in linestring
      if p in linestring, finds second nearest"""
  # coordinates of p and linestring
  p_coords = list(p.coords)[0]
  linestring_coords = list(linestring.coords)

  if p_coords in linestring_coords:
    # form a new linestring if p is in linestring
    linestring_coords.remove(p_coords)
    linestring = LineString(linestring_coords)

  return nearest_points(p, linestring)

p = Point(51.21745162000732, 4.41871738126533)

linestring = LineString([(51.2176008,4.4177154), (51.21758,4.4178548), (51.2175729,4.4179023), (51.21745162000732,4.41871738126533)])

n_p = second_nearest(p, linestring)

print(list(map(str, n_p)))

Output

First point is p, Second point is closest point to p in linestream not equal to p (so second closest point)

['POINT (51.21745162000732 4.41871738126533)', 
 'POINT (51.2175729 4.4179023)']

Upvotes: 1

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