osmnx：项目指向街道段

问题描述

我有一个由lat和lon给出的点，我想通过最小的欧几里得距离找到到该点的最近边缘。例如

// all imports here

const useStyles = makeStyles({
[...]
});

export default function OutlinedCard() {
  const [isEditable,setIsEditable] = useState(false);
  const [name,setName] = useState("");
  const classes = useStyles();

  const handleEdit = () => {
    console.log("I am here");
    setIsEditable(!isEditable);
  };

  return (
    <Card className={classes.root} variant="outlined">
      <CardContent>
        <Typography
          className={classes.title}
          color="textSecondary"
          gutterBottom
        >
          Word of the Day
        </Typography>
        <TextField
          id="standard-basic"
          label="Name:"
          disabled={!isEditable}
          value={name}
          onChange={(e) => setName(e.target.value)}
          onBlur={() => setIsEditable(false)}
        />
      </CardContent>
      <CardActions>
        <Button size="small" onClick={handleEdit}>
          Edit
        </Button>
      </CardActions>
    </Card>
  );
}

我明白了

import osmnx as ox
track = [(40.7052,-74.0069)]
fig,ax = ox.plot_graph(G,show=False,close=False)
for pairs in track:
    ax.scatter(pairs[1],pairs[0],c='red')
plt.show()

ox.distance.get_nearest_edge(G,track,return_geom=True,return_dist=True)

它输出边缘的顶点及其几何形状。点与最近的边缘之间的距离为162。但是如何找到我的点在该最近的边缘上的投影？

解决方法

这是一个完整的最小工作示例：

import osmnx as ox
from shapely.geometry import Point
ox.config(use_cache=True,log_console=True)

# create point tuple as (lat,lng)
point = (40.7052,-74.0069)
G = ox.graph_from_point(point,network_type='drive')
u,v,k,edge_geom,dist = ox.distance.get_nearest_edge(G,point,return_geom=True,return_dist=True)

# create shapely point geometry object as (x,y),that is (lng,lat)
point_geom = Point(reversed(point))

# use shapely to find the point along the edge that is closest to the reference point
nearest_point_on_edge = edge_geom.interpolate(edge_geom.project(point_geom))
nearest_point_on_edge.coords[0]

openstreetmap osmnx projection