Define which neighbours a Ball tree should return
Question
I have a dataframe with several locations. I want to find each locations nearest neighbours.
To do this, I am using a Ball tree. However, the output seems to be comparing all of the locations with each other, including the original location. For example, I have locations A,B,C..... the output will list A as a neighbour for A.
Also, I have a column for time that I want to use in my analysis. I have set the time column to the index before fitting the Ball tree. But the output will return A at time 1, A at time 2, A at time 3 as neighbours of A.
I created a smaller dataframe with fake data to mirror my own (displayed below) and using this smaller dataset, I can run the tree with a larger number of neighbours that I would otherwise use and remove the 'wrong' neighbours from the output.
However, this method is too computationally expensive to use with my real data.
Is there a way to defining which type of neighbours to return in the Ball Tree?
sample code:
from sklearn.neighbors import BallTree
import numpy as np
import pandas as pd
test_data = pd.DataFrame({'latitude':[51.51, 51.52,61.53,61.54,71.55, 71.56,
51.51, 51.52,61.53,61.54,71.55, 71.56,
51.51, 51.52,61.53,61.54,71.55, 71.56],
'longitude':[-0.13,-0.13,-0.13,-0.14,-0.13,-0.13,
-0.13,-0.13,-0.13,-0.14,-0.13,-0.13,
-0.13,-0.13,-0.13,-0.14,-0.13,-0.13],
'id':['A','B','C','D','E','F',
'A','B','C','D','E','F',
'A','B','C','D','E','F'],
'target':[35,410,1,100,114,78,
14,254,101,278,3578,435,
254,254,37,47,38,101],
'time':['2019-03-10 11:00:00','2019-03-10 11:00:00','2019-03-10 11:00:00','2019-03-10 11:00:00','2019-03-10 11:00:00','2019-03-10 11:00:00',
'2019-03-10 11:10:00','2019-03-10 11:10:00','2019-03-10 11:10:00','2019-03-10 11:10:00','2019-03-10 11:10:00','2019-03-10 11:10:00',
'2019-03-10 11:20:00','2019-03-10 11:20:00','2019-03-10 11:20:00','2019-03-10 11:20:00','2019-03-10 11:20:00','2019-03-10 11:20:00',
]})
# --- STEP 1) Prepairing the data
test_data=test_data.reset_index()
# Convert latitude and longitude to radions
for column in test_data[['latitude','longitude']]:
rad = np.deg2rad(test_data[column].values)
test_data[f'{column}'] = rad
# Creating a duplicate of the time column, one will be set as the index
test_data['time2']=test_data['time']
# Convert time to datetime
test_data['time']=pd.to_datetime(test_data['time'])
test_data = test_data.set_index('time').astype('str')
# --- STEP 2) FITTING THE BALL TREE
locations_a = test_data
locations_b = test_data
col_name = 'ss_id'
latitude = "latitude"
longitude = "longitude"
# make ball tree
ball = BallTree(locations_a[[latitude, longitude]].values, metric='haversine')
# The amount of neighbors to return
k = 6
# Calculating distances
distances, indices = ball.query(locations_b[[latitude, longitude]].values, k = k)
# --- STEP 3) Merging Results into dataframe
dists = pd.DataFrame(distances).stack()
rel = pd.DataFrame(indices).stack()
# Create dataframe
neighbor_info_df = pd.merge(dists.rename('distance'), rel.rename('neighbor_idx'), right_index=True, left_index=True)
# Resetting and renaming indexes
neighbor_info_df = neighbor_info_df.reset_index(level=1).rename({'level_1': 'neighbor_number'}, axis=1)
neighbor_info_df = neighbor_info_df.reset_index().rename({'index': 'id_index_no'}, axis=1)
neighbor_info_df.head(10)
1 answers
solution1
1 ACCPTED 2023-01-27 08:44:11
As I suggested, you have to remove the first match to remove the point itself from the k neighbors (distance=0). However, you need to group by time before:
# Setup: let the default numeric index (0->N), don't use time as index
# test_data = pd.DataFrame(...)
# test_data['time']=pd.to_datetime(test_data['time'])
def find_k_neighbors(df, k=6):
"""Find k neighbors exclude current point itself."""
# Prepare data and get neighbors
coords = np.radians(df[['latitude', 'longitude']])
tree = BallTree(coords, metric='haversine')
distances, indices = tree.query(coords, k=k+1)
# Flat all arrays then build dataframe
distances = np.ravel(distances[:, 1:])
indices = np.ravel(indices[:, 1:])
return (pd.DataFrame({'id_index_no': df.index.repeat(k),
'neighbor_number': np.tile(range(1, k+1), len(df)),
'distance': distances,
'neighbor_idx': df.iloc[indices].index}))
# For demonstration purpose, I limit k=3
neighbor_info_df = test_data.groupby('time').apply(find_k_neighbors, k=3).droplevel(1)
Output:
>>> neighbor_info_df
id_index_no neighbor_number distance neighbor_idx
time
2019-03-10 11:00:00 0 1 0.000175 1
2019-03-10 11:00:00 0 2 0.174882 2
2019-03-10 11:00:00 0 3 0.175057 3
2019-03-10 11:00:00 1 1 0.000175 0
2019-03-10 11:00:00 1 2 0.174707 2
2019-03-10 11:00:00 1 3 0.174882 3
2019-03-10 11:00:00 2 1 0.000193 3
2019-03-10 11:00:00 2 2 0.174707 1
2019-03-10 11:00:00 2 3 0.174882 0
2019-03-10 11:00:00 3 1 0.000193 2
2019-03-10 11:00:00 3 2 0.174707 4
2019-03-10 11:00:00 3 3 0.174882 5
2019-03-10 11:00:00 4 1 0.000175 5
2019-03-10 11:00:00 4 2 0.174707 3
2019-03-10 11:00:00 4 3 0.174882 2
2019-03-10 11:00:00 5 1 0.000175 4
2019-03-10 11:00:00 5 2 0.174882 3
2019-03-10 11:00:00 5 3 0.175057 2
2019-03-10 11:10:00 6 1 0.000175 7
2019-03-10 11:10:00 6 2 0.174882 8
2019-03-10 11:10:00 6 3 0.175057 9
2019-03-10 11:10:00 7 1 0.000175 6
2019-03-10 11:10:00 7 2 0.174707 8
2019-03-10 11:10:00 7 3 0.174882 9
2019-03-10 11:10:00 8 1 0.000193 9
2019-03-10 11:10:00 8 2 0.174707 7
2019-03-10 11:10:00 8 3 0.174882 6
2019-03-10 11:10:00 9 1 0.000193 8
2019-03-10 11:10:00 9 2 0.174707 10
2019-03-10 11:10:00 9 3 0.174882 11
2019-03-10 11:10:00 10 1 0.000175 11
2019-03-10 11:10:00 10 2 0.174707 9
2019-03-10 11:10:00 10 3 0.174882 8
2019-03-10 11:10:00 11 1 0.000175 10
2019-03-10 11:10:00 11 2 0.174882 9
2019-03-10 11:10:00 11 3 0.175057 8
2019-03-10 11:20:00 12 1 0.000175 13
2019-03-10 11:20:00 12 2 0.174882 14
2019-03-10 11:20:00 12 3 0.175057 15
2019-03-10 11:20:00 13 1 0.000175 12
2019-03-10 11:20:00 13 2 0.174707 14
2019-03-10 11:20:00 13 3 0.174882 15
2019-03-10 11:20:00 14 1 0.000193 15
2019-03-10 11:20:00 14 2 0.174707 13
2019-03-10 11:20:00 14 3 0.174882 12
2019-03-10 11:20:00 15 1 0.000193 14
2019-03-10 11:20:00 15 2 0.174707 16
2019-03-10 11:20:00 15 3 0.174882 17
2019-03-10 11:20:00 16 1 0.000175 17
2019-03-10 11:20:00 16 2 0.174707 15
2019-03-10 11:20:00 16 3 0.174882 14
2019-03-10 11:20:00 17 1 0.000175 16
2019-03-10 11:20:00 17 2 0.174882 15
2019-03-10 11:20:00 17 3 0.175057 14
Input dataframe:
>>> test_data
latitude longitude id target time
0 51.51 -0.13 A 35 2019-03-10 11:00:00
1 51.52 -0.13 B 410 2019-03-10 11:00:00
2 61.53 -0.13 C 1 2019-03-10 11:00:00
3 61.54 -0.14 D 100 2019-03-10 11:00:00
4 71.55 -0.13 E 114 2019-03-10 11:00:00
5 71.56 -0.13 F 78 2019-03-10 11:00:00
6 51.51 -0.13 A 14 2019-03-10 11:10:00
7 51.52 -0.13 B 254 2019-03-10 11:10:00
8 61.53 -0.13 C 101 2019-03-10 11:10:00
9 61.54 -0.14 D 278 2019-03-10 11:10:00
10 71.55 -0.13 E 3578 2019-03-10 11:10:00
11 71.56 -0.13 F 435 2019-03-10 11:10:00
12 51.51 -0.13 A 254 2019-03-10 11:20:00
13 51.52 -0.13 B 254 2019-03-10 11:20:00
14 61.53 -0.13 C 37 2019-03-10 11:20:00
15 61.54 -0.14 D 47 2019-03-10 11:20:00
16 71.55 -0.13 E 38 2019-03-10 11:20:00
17 71.56 -0.13 F 101 2019-03-10 11:20:00
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