如何在 Shapely 中找到 LineString 的第二个最近点

Question

给定某LineString和点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)])

离p最近的点的计算公式为：

n_p = nearest_points(linestring, p)[0]

结论是完全相同的点，这是正常的，因为线串中也有完全相同的值，但我需要知道最近的点，除了点本身。那么如何找到秒最近的点呢？

Answer 1

解决如下。

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)))

输出

第一个点是p，第二个点是线流中最接近p的点不等于p（所以第二个最接近的点）

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

Answer 2

在一般情况下，最简单的解决方案是从您的 LineString 构造一个新的几何对象，但没有最近点，然后用这个新几何获得最近点。:

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]

或者，如果您不想构造一个新对象，例如，由于内存限制，您可以运行在 LineString 中的所有点上使用 heapq.nsmallest:

import heapq

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

在你的具体情况下，当所有点共线时，你还可以计算与最近点的相邻点的距离：

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)

如何在 Shapely 中找到 LineString 的第二个最近点

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

python

point

shapely