如何找到离点最近的LINESTRING？

Question

如何为某个点附近最近的 LINESTRING 提供资金？

首先，我有一个 LINESTRING 和点值的列表。 我如何获得最接近点 (5.41 3.9) 的 LINESTRING 以及距离？

from shapely.geometry import Point, LineString

line_string = [LINESTRING (-1.15.12 9.9, -1.15.13 9.93), LINESTRING (-2.15.12 8.9, -2.15.13 8.93)]
point = POINT (5.41 3.9)

#distance 
line_string [0].distance(point)

到目前为止，我认为我通过为第一个 LINESTRING 执行 line_string [0].distance(point) 获得了距离值，但我只是想确保我以正确的方式进行。

Answer 1

这是一个 function ，它接受一个LineString s 和一个point的列表，并返回最接近该point的LineString以及距离。

from shapely.geometry import Point, LineString

# set up lines and points
line_string_list = [LineString([(-1,1),(1,.5)]), LineString([(-1,-.5),(.5,-1)]), LineString([(-1,0),(.5,-.5)])]
point = Point(.25,-.75)

def closest_line(lines, point):
    # get distances
    distance_list = [line.distance(point) for line in line_string_list]
    shortest_distance = min(distance_list) # find the line closest to the point
    return(lines[distance_list.index(shortest_distance)], # return the closest line
           shortest_distance) # return the distance to that line
    
print(closest_line(line_string_list, point))

Answer 2

• 您的示例几何图形对线串无效，已修改
• 使用sjoin_nearest()很容易实现

import geopandas as gpd
import shapely.wkt
import shapely.geometry

line_string = ["LINESTRING (-1.15.12 9.9, -1.15.13 9.93)", "LINESTRING (-2.15.12 8.9, -2.15.13 8.93)"]
# fix invalid wkt string...
line_string = ["LINESTRING (-1.15 9.9, -1.15 9.93)", "LINESTRING (-2.15 8.9, -2.15 8.93)"]
point = "POINT (5.41 3.9)"

gdf_p = gpd.GeoDataFrame(geometry=[shapely.wkt.loads(point)])
gdf_l = gpd.GeoDataFrame(geometry=pd.Series(line_string).apply(shapely.wkt.loads))

gpd.sjoin_nearest(gdf_p, gdf_l).merge(gdf_l, left_on="index_right", right_index=True)

	几何_x	index_right	几何_y
0	点（5.41 3.9）	0	线串（-1.15 9.9，-1.15 9.93）