如何计算 Pandas 数据框中每个坐标对（以米为单位）到最近邻居的欧氏距离？

Question

我有一个像这样的数据框

索引	地点 ID	var_lat_fact	var_lon_fact
0	167312091448	5.6679820000	-0.0144950000
1	167312091448	5.6686320000	-0.0157910000
2	167312091448	5.6653530000	-0.0181980000
3	167312091448	5.6700970000	-0.0191400000
4	167312091448	5.6689810000	-0.0104040000

对于每个坐标对（纬度，经度），我想计算到数据帧内最近邻居的欧几里德距离。因此，每个点都会在附加列中获得一个度量（例如，nearest_neighbour_dist），指示该距离（以米为单位）。

类似这样的事情

索引	地点 ID	var_lat_fact	var_lon_fact	最近邻居距离
0	167312091448	5.6679820000	-0.0144950000	123
1	167312091448	5.6686320000	-0.0157910000	342
2	167312091448	5.6653530000	-0.0181980000	312
3	167312091448	5.6700970000	-0.0191400000	42
4	167312091448	5.6689810000	-0.0104040000	23

我实在无法理解这个问题…… 任何帮助将不胜感激。

Answer 1

import pandas as pd
from io import StringIO
from scipy.spatial import KDTree

# Load test data
s = """
place_id,var_lat_fact,var_lon_fact
167312091448 5.6679820000 -0.0144950000
167312091448 5.6686320000 -0.0157910000
167312091448 5.6653530000 -0.0181980000
167312091448 5.6700970000 -0.0191400000
167312091448 5.6689810000 -0.0104040000
""".replace(' ', ',')

df = pd.read_csv(StringIO(s))

# Create kd Tree
points = df[['var_lat_fact', 'var_lon_fact']].values
kd = KDTree(points)

# Compute the closest two neighbors for each point
distances, indexes = kd.query(points, k=2)

# Discard the first 'neighbor' (the point itself, i.e. distance=0),
# And select the second.
df['nearest_neighbour_dist'] = distances[:, 1]

print(df)

       place_id  var_lat_fact  var_lon_fact  nearest_neighbour_dist
0  167312091448      5.667982     -0.014495                0.001450
1  167312091448      5.668632     -0.015791                0.001450
2  167312091448      5.665353     -0.018198                0.004068
3  167312091448      5.670097     -0.019140                0.003655
4  167312091448      5.668981     -0.010404                0.004211