如何调整此代码以返回第二个和第三个“最近的邻居”？

Question

根据calculating average distance of nearest neighbours in pandas dataframe 中的这段代码，我该如何调整它以便将第二个和第三个最近邻居返回到新列中？

（或者创建一个可调整的参数来定义返回多少个邻居）：

示例代码：

import numpy as np 
from sklearn.neighbors import NearestNeighbors
import pandas as pd

def nn(x):
    nbrs = NearestNeighbors(
        n_neighbors=2, 
        algorithm='auto', 
        metric='euclidean'
    ).fit(x)
    distances, indices = nbrs.kneighbors(x)
    return distances, indices

time = [0, 0, 0, 1, 1, 2, 2]
x = [216, 218, 217, 280, 290, 130, 132]
y = [13, 12, 12, 110, 109, 3, 56] 
car = [1, 2, 3, 1, 3, 4, 5]
df = pd.DataFrame({'time': time, 'x': x, 'y': y, 'car': car})

#This has the index of the nearest neighbor in the group, as well as the distance
nns = df.drop('car', 1).groupby('time').apply(lambda x: nn(x.as_matrix()))

groups = df.groupby('time')
nn_rows = []
for i, nn_set in enumerate(nns):
    group = groups.get_group(i)
    for j, tup in enumerate(zip(nn_set[0], nn_set[1])):
        nn_rows.append({'time': i,
                    'car': group.iloc[j]['car'],
                    'nearest_neighbour': group.iloc[tup[1][1]]['car'],
                    'euclidean_distance': tup[0][1]})

nn_df = pd.DataFrame(nn_rows).set_index('time')

结果数据框：

>>> nn_df
time car euclidean_distance nearest_neighbour           
0    1   1.414214           3
0    2   1.000000           3
0    3   1.000000           2
1    1   10.049876          3
1    3   10.049876          1
2    4   53.037722          5
2    5   53.037722          4

如何获得最近邻 2 和 3 以及 N 的输出并将它们插入新列？

Answer 1

这是NearestNeighbors 方法的文档。

我认为您的问题可以使用n_neighbors 参数解决。该参数指定要返回的最近邻数的indices and distances。

通常使用的值是2，当我们的目标是找到除了点本身之外的单个最近邻。最近的邻居总是它自己，因为距离为 0。

要找到第二个和第三个最近的邻居，n_neighbors 应该设置为 4。这将返回点本身，然后是下一个 N-1 个最近的邻居

# Argument
n_neighbor = 4

# Indices
[point_itself, neighbor_1, neighbor_2, neighbor_3]

# Distances
[ 0, distance_1, distance_2, distance_3]