python之计算 Pandas 数据框中最近邻居的平均距离

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!

请您参考如下方法:

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)