问题
I have a dataframe with latitude and longitude pairs.
Here is my dataframe look like.
order_lat order_long
0 19.111841 72.910729
1 19.111342 72.908387
2 19.111342 72.908387
3 19.137815 72.914085
4 19.119677 72.905081
5 19.119677 72.905081
6 19.119677 72.905081
7 19.120217 72.907121
8 19.120217 72.907121
9 19.119677 72.905081
10 19.119677 72.905081
11 19.119677 72.905081
12 19.111860 72.911346
13 19.111860 72.911346
14 19.119677 72.905081
15 19.119677 72.905081
16 19.119677 72.905081
17 19.137815 72.914085
18 19.115380 72.909144
19 19.115380 72.909144
20 19.116168 72.909573
21 19.119677 72.905081
22 19.137815 72.914085
23 19.137815 72.914085
24 19.112955 72.910102
25 19.112955 72.910102
26 19.112955 72.910102
27 19.119677 72.905081
28 19.119677 72.905081
29 19.115380 72.909144
30 19.119677 72.905081
31 19.119677 72.905081
32 19.119677 72.905081
33 19.119677 72.905081
34 19.119677 72.905081
35 19.111860 72.911346
36 19.111841 72.910729
37 19.131674 72.918510
38 19.119677 72.905081
39 19.111860 72.911346
40 19.111860 72.911346
41 19.111841 72.910729
42 19.111841 72.910729
43 19.111841 72.910729
44 19.115380 72.909144
45 19.116625 72.909185
46 19.115671 72.908985
47 19.119677 72.905081
48 19.119677 72.905081
49 19.119677 72.905081
50 19.116183 72.909646
51 19.113827 72.893833
52 19.119677 72.905081
53 19.114100 72.894985
54 19.107491 72.901760
55 19.119677 72.905081
I want to cluster this points which are nearest to each other(200 meters distance) following is my distance matrix.
from scipy.spatial.distance import pdist, squareform
distance_matrix = squareform(pdist(X, (lambda u,v: haversine(u,v))))
array([[ 0. , 0.2522482 , 0.2522482 , ..., 1.67313071,
1.05925366, 1.05420922],
[ 0.2522482 , 0. , 0. , ..., 1.44111548,
0.81742536, 0.98978355],
[ 0.2522482 , 0. , 0. , ..., 1.44111548,
0.81742536, 0.98978355],
...,
[ 1.67313071, 1.44111548, 1.44111548, ..., 0. ,
1.02310118, 1.22871515],
[ 1.05925366, 0.81742536, 0.81742536, ..., 1.02310118,
0. , 1.39923529],
[ 1.05420922, 0.98978355, 0.98978355, ..., 1.22871515,
1.39923529, 0. ]])
Then I am applying DBSCAN clustering algorithm on distance matrix.
from sklearn.cluster import DBSCAN
db = DBSCAN(eps=2,min_samples=5)
y_db = db.fit_predict(distance_matrix)
I don't know how to choose eps & min_samples value. It clusters the points which are way too far, in one cluster.(approx 2 km in distance) Is it because it calculates euclidean distance while clustering? please help.
回答1:
DBSCAN is meant to be used on the raw data, with a spatial index for acceleration. The only tool I know with acceleration for geo distances is ELKI (Java) - scikit-learn unfortunately only supports this for a few distances like Euclidean distance (see sklearn.neighbors.NearestNeighbors).
But apparently, you can affort to precompute pairwise distances, so this is not (yet) an issue.
However, you did not read the documentation carefully enough, and your assumption that DBSCAN uses a distance matrix is wrong:
from sklearn.cluster import DBSCAN
db = DBSCAN(eps=2,min_samples=5)
db.fit_predict(distance_matrix)
uses Euclidean distance on the distance matrix rows, which obviously does not make any sense.
See the documentation of DBSCAN (emphasis added):
class sklearn.cluster.DBSCAN(eps=0.5, min_samples=5, metric='euclidean', algorithm='auto', leaf_size=30, p=None, random_state=None)
metric : string, or callable
The metric to use when calculating distance between instances in a feature array. If metric is a string or callable, it must be one of the options allowed by metrics.pairwise.calculate_distance for its metric parameter. If metric is “precomputed”, X is assumed to be a distance matrix and must be square. X may be a sparse matrix, in which case only “nonzero” elements may be considered neighbors for DBSCAN.
similar for fit_predict:
X : array or sparse (CSR) matrix of shape (n_samples, n_features), or array of shape (n_samples, n_samples)
A feature array, or array of distances between samples if metric='precomputed'.
In other words, you need to do
db = DBSCAN(eps=2, min_samples=5, metric="precomputed")
回答2:
You can cluster spatial latitude-longitude data with scikit-learn's DBSCAN without precomputing a distance matrix.
db = DBSCAN(eps=2/6371., min_samples=5, algorithm='ball_tree', metric='haversine').fit(np.radians(coordinates))
This comes from this tutorial on clustering spatial data with scikit-learn DBSCAN. In particular, notice that the eps value is still 2km, but it's divided by 6371 to convert it to radians. Also, notice that .fit() takes the coordinates in radian units for the haversine metric.
回答3:
I don't know what implementation of haversine you're using but it looks like it returns results in km so eps should be 0.2, not 2 for 200 m.
For the min_samples parameter, that depends on what your expected output is. Here are a couple of examples. My outputs are using an implementation of haversine based on this answer which gives a distance matrix similar, but not identical to yours.
This is with db = DBSCAN(eps=0.2, min_samples=5)
[ 0 -1 -1 -1 1 1 1 -1 -1 1 1 1 2 2 1 1 1 -1 -1 -1 -1 1 -1 -1 -1 -1 -1 1 1 -1 1 1 1 1 1 2 0 -1 1 2 2 0 0 0 -1 -1 -1 1 1 1 -1 -1 1 -1 -1 1]
This creates three clusters, 0, 1 and 2, and a lot of the samples don't fall into a cluster with at least 5 members and so are not assigned to a cluster (shown as -1).
Trying again with a smaller min_samples value:
db = DBSCAN(eps=0.2, min_samples=2)
[ 0 1 1 2 3 3 3 4 4 3 3 3 5 5 3 3 3 2 6 6 7 3 2 2 8 8 8 3 3 6 3 3 3 3 3 5 0 -1 3 5 5 0 0 0 6 -1 -1 3 3 3 7 -1 3 -1 -1 3]
Here most of the samples are within 200m of at least one other sample and so fall into one of eight clusters 0 to 7.
Edited to add
It looks like @Anony-Mousse is right, though I didn't see anything wrong in my results. For the sake of contributing something, here's the code I was using to see the clusters:
from math import radians, cos, sin, asin, sqrt
from scipy.spatial.distance import pdist, squareform
from sklearn.cluster import DBSCAN
import matplotlib.pyplot as plt
import pandas as pd
def haversine(lonlat1, lonlat2):
"""
Calculate the great circle distance between two points
on the earth (specified in decimal degrees)
"""
# convert decimal degrees to radians
lat1, lon1 = lonlat1
lat2, lon2 = lonlat2
lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
# haversine formula
dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * asin(sqrt(a))
r = 6371 # Radius of earth in kilometers. Use 3956 for miles
return c * r
X = pd.read_csv('dbscan_test.csv')
distance_matrix = squareform(pdist(X, (lambda u,v: haversine(u,v))))
db = DBSCAN(eps=0.2, min_samples=2, metric='precomputed') # using "precomputed" as recommended by @Anony-Mousse
y_db = db.fit_predict(distance_matrix)
X['cluster'] = y_db
plt.scatter(X['lat'], X['lng'], c=X['cluster'])
plt.show()
回答4:
@eos Gives the best answer I think - as well as making use of Haversine distance (the most relevant distance measure in this case), it avoids the need to generate a precomputed distance matrix. If you create a distance matrix then you need to calculate the pairwise distances for every combination of points (although you can obviously save a bit of time by taking advantage of the fact that your distance metric is symmetric).
If you just supply DBSCAN with a distance measure and use the ball_tree algorithm though, it can avoid the need to calculate every possible distance. This is because the ball tree algorithm can use the triangular inequality theorem to reduce the number of candidates that need to be checked to find the nearest neighbours of a data point (this is the biggest job in DBSCAN).
The triangular inequality theorem states:
|x+y| <= |x| + |y|
...so if a point p is distance x from its neighbour n, and another point q is a distance y from p, if x+y is greater than our nearest neighbour radius, we know that q must be too far away from n to be considered a neighbour, so we don't need to calculate its distance.
Read more about how ball trees work in the scikit-learn documentation
来源：https://stackoverflow.com/questions/34579213/dbscan-for-clustering-of-geographic-location-data