python - 计算 Pandas 数据框中最近邻居的平均距离

Question

我有一组对象及其随时间的位置。我想获得每辆车与其最近的邻居之间的距离，并计算每个时间点的平均值。示例数据框如下:

 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})
 df

         x       y      car
 time
  0     216     13       1
  0     218     12       2
  0     217     12       3
  1     280     110      1
  1     290     109      3
  2     130     3        4
  2     132     56       5

对于每个时间点，我想知道每辆车最近的邻居。示例:

df2

          car    nearest_neighbour    euclidean_distance  
 time
  0       1            3                    1.41
  0       2            3                    1.00
  0       3            1                    1.41
  1       1            3                    10.05
  1       3            1                    10.05
  2       4            5                    53.04
  2       5            4                    53.04

我知道我可以从 How to apply euclidean distance function to a groupby object in pandas dataframe? 计算汽车之间的成对距离但是我如何为每辆车找到最近的邻居呢？

之后，使用 groupby 获得每帧的平均距离似乎很简单，但真正让我失望的是第二步。帮助表示赞赏!

Answer 1

这可能有点矫枉过正，但您可以使用 nearest neighbors from scikit

一个例子:

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:

      car  euclidean_distance  nearest_neighbour
time                                            
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

(请注意，在时间 0，汽车 3 的最近邻居是汽车 2。sqrt((217-216)**2 + 1) 大约是 1.4142135623730951 而 sqrt((218-217)**2 + 0) = 1)