使用 GeoPandas 的两个数据帧中的 K 最近点

Question

GeoPandas 在引擎盖下使用匀称。为了获得最近的邻居，我看到了nearest_points from shapely 的使用。但是，这种方法不包括 k-最近点。

我需要计算从到 GeoDataFrames 到最近点的距离，并将距离插入到包含“从该点开始”数据的 GeoDataFrame 中。

这是我使用GeoSeries.distance() 而不使用其他包或库的方法。请注意，当k == 1 时，返回值基本上显示了到最近点的距离。 There is also a GeoPandas-only solution for nearest point by @cd98 which inspired my approach.

这对我的数据很有效，但我想知道是否有更好或更快的方法或其他好处可以使用 shapely 或 sklearn.neighbors？

import pandas as pd
import geopandas as gp

gdf1 > GeoDataFrame with point type geometry column - distance from this point
gdf2 > GeoDataFrame with point type geometry column - distance to this point

def knearest(from_points, to_points, k):
    distlist = to_points.distance(from_points)
    distlist.sort_values(ascending=True, inplace=True) # To have the closest ones first
    return distlist[:k].mean()

# looping through a list of nearest points
for Ks in [1, 2, 3, 4, 5, 10]:
    name = 'dist_to_closest_' + str(Ks) # to set column name
    gdf1[name] = gdf1.geometry.apply(knearest, args=(gdf2, closest_x))

Answer 1

如果您的数据在网格坐标中，则该方法会更精简一些，但有一个关键问题。

以sutan's answer 为基础，精简来自 Uni Helsinki 的街区...

要获得多个邻居，您编辑 k_neighbors 参数....并且还必须在函数体中硬编码变量（请参阅我在“最近”和“最近距离”下面添加的内容）并将它们添加到返回语句中。

因此，如果你想要两个最近的点，它看起来像：

from sklearn.neighbors import BallTree
import numpy as np

def get_nearest(src_points, candidates, k_neighbors=2):
    """
    Find nearest neighbors for all source points from a set of candidate points
    modified from: https://automating-gis-processes.github.io/site/notebooks/L3/nearest-neighbor-faster.html
    """
    

    # Create tree from the candidate points
    tree = BallTree(candidates, leaf_size=15, metric='euclidean')

    # Find closest points and distances
    distances, indices = tree.query(src_points, k=k_neighbors)

    # Transpose to get distances and indices into arrays
    distances = distances.transpose()
    indices = indices.transpose()

    # Get closest indices and distances (i.e. array at index 0)
    # note: for the second closest points, you would take index 1, etc.
    closest = indices[0]
    closest_dist = distances[0]
    closest_second = indices[1] # *manually add per comment above*
    closest_second_dist = distances[1] # *manually add per comment above*

    # Return indices and distances
    return (closest, closest_dist, closest_sec, closest_sec_dist)

输入是 (x,y) 元组的列表。因此，由于（按问题标题）您的数据位于 GeoDataframe 中：

# easier to read
in_pts = [(row.geometry.x, row.geometry.y) for idx, row in gdf1.iterrows()]
qry_pts = [(row.geometry.x, row.geometry.y) for idx, row in gdf2.iterrows()]

# faster (by about 7X)
in_pts = [(x,y) for x,y in zip(gdf1.geometry.x , gdf1.geometry.y)]
qry_pts =  [(x,y) for x,y in zip(gdf2.geometry.x , gdf2.geometry.y)]

我对距离不感兴趣，所以我没有注释掉函数，而是运行：

idx_nearest, _, idx_2ndnearest, _ = get_nearest(in_pts, qry_pts)

并获得两个长度相同的 in_pts 数组，它们分别包含 qry_pts 的原始地理数据帧中最近和次近点的索引值。

Answer 2

是的，但首先，我必须感谢来自automating GIS process 的赫尔辛基大学，这里是the source code。方法如下
首先，读取数据，例如，为每个建筑物查找最近的公交车站。

# Filepaths
stops = gpd.read_file('data/pt_stops_helsinki.gpkg')
buildings = read_gdf_from_zip('data/building_points_helsinki.zip')

定义函数，这里可以调整k_neighbors

from sklearn.neighbors import BallTree
import numpy as np

def get_nearest(src_points, candidates, k_neighbors=1):
    """Find nearest neighbors for all source points from a set of candidate points"""

    # Create tree from the candidate points
    tree = BallTree(candidates, leaf_size=15, metric='haversine')

    # Find closest points and distances
    distances, indices = tree.query(src_points, k=k_neighbors)

    # Transpose to get distances and indices into arrays
    distances = distances.transpose()
    indices = indices.transpose()

    # Get closest indices and distances (i.e. array at index 0)
    # note: for the second closest points, you would take index 1, etc.
    closest = indices[0]
    closest_dist = distances[0]

    # Return indices and distances
    return (closest, closest_dist)


def nearest_neighbor(left_gdf, right_gdf, return_dist=False):
    """
    For each point in left_gdf, find closest point in right GeoDataFrame and return them.

    NOTICE: Assumes that the input Points are in WGS84 projection (lat/lon).
    """

    left_geom_col = left_gdf.geometry.name
    right_geom_col = right_gdf.geometry.name

    # Ensure that index in right gdf is formed of sequential numbers
    right = right_gdf.copy().reset_index(drop=True)

    # Parse coordinates from points and insert them into a numpy array as RADIANS
    left_radians = np.array(left_gdf[left_geom_col].apply(lambda geom: (geom.x * np.pi / 180, geom.y * np.pi / 180)).to_list())
    right_radians = np.array(right[right_geom_col].apply(lambda geom: (geom.x * np.pi / 180, geom.y * np.pi / 180)).to_list())

    # Find the nearest points
    # -----------------------
    # closest ==> index in right_gdf that corresponds to the closest point
    # dist ==> distance between the nearest neighbors (in meters)

    closest, dist = get_nearest(src_points=left_radians, candidates=right_radians)

    # Return points from right GeoDataFrame that are closest to points in left GeoDataFrame
    closest_points = right.loc[closest]

    # Ensure that the index corresponds the one in left_gdf
    closest_points = closest_points.reset_index(drop=True)

    # Add distance if requested
    if return_dist:
        # Convert to meters from radians
        earth_radius = 6371000  # meters
        closest_points['distance'] = dist * earth_radius

    return closest_points

做最近邻分析

# Find closest public transport stop for each building and get also the distance based on haversine distance
# Note: haversine distance which is implemented here is a bit slower than using e.g. 'euclidean' metric
# but useful as we get the distance between points in meters
closest_stops = nearest_neighbor(buildings, stops, return_dist=True)

现在加入 from 和 to 数据框

# Rename the geometry of closest stops gdf so that we can easily identify it
closest_stops = closest_stops.rename(columns={'geometry': 'closest_stop_geom'})

# Merge the datasets by index (for this, it is good to use '.join()' -function)
buildings = buildings.join(closest_stops)

Answer 3

上面使用自动化 GIS 进程的答案非常好，但是将点转换为 RADIANS 的 numpy 数组时出现错误。经纬度颠倒了。

left_radians = np.array(left_gdf[left_geom_col].apply(lambda geom: (geom.y * np.pi / 180, geom.x * np.pi / 180)).to_list())

确实点是用 (lat, lon) 给出的，但经度对应于平面或球体的 x 轴，纬度对应于 y 轴。