I am working on clustering the customer base of a business-to-business company. I have data on customers that consists of both numerical (e.g. # of purchases made, avg. spend per purchase) and categorical (e.g. industry code) data.

Additionally, I have latitude and longitude information for each customer, which I would like to include in the clustering. Normal categorical and numerical data can be clustered using e.g. PAM / K-Prototypes / Hierarchical Clustering (anything where a distance matrix has to be computed, since there are distance functions that can differentiate between both types).

However, I do not know how to go about including latitude and longitude values. Latitude and longitude are in decimal degrees, therefore metrics like Euclidean distance cannot be used. Some possible approaches I have considered are:

  • calculating x, y, z points on a sphere from lat / lon coordinates using

    $x = \cos(lat) \times cos(lon)$

    $y = cos(lat) \times sin(lon)$

    $z = sin(lat)$

    which could then be treated as 3 numeric attributes using Euclidean distance.

  • somehow implement haversine distance in the calculation of the distance matrix. So create a distance function that calculates numeric differences using Euclidean, categorical (after one-hot encoding) using e.g. Jaccard, and lat-long dissimilarity using Haversine. How could I potentially go about implementing something like this? Is it possible, or am I overlooking something?

  • creating regions, such as "EMEA" (Europe, Middle East, Africa), "APAC (Asia Pacific), "NA" (North America) from the lat-lon values, thereby creating more categorical attributes.

Can someone comment on what a suitable approach might be?

  • $\begingroup$ Are you thinking here that because your data is geographically global, then if you project it onto a usual sort of 2-D rectangular map, some places that are close together in the world inevitably become very far apart on the map? $\endgroup$
    – DavidH
    Aug 29, 2022 at 8:03
  • $\begingroup$ Yes, that was what I was thinking. For some lat, lon pairs, the pairwise distance on a 2D map does not correspond to the actual shortest distance between locations, such as for eastern Russia and Alaska. They are very close, but if they were projected onto a 2D map, the distance between them would be large. Or have I made a mistake in my reasoning? $\endgroup$
    – Ultralite
    Aug 29, 2022 at 10:02
  • $\begingroup$ If that is what you are trying to solve you could embed lat-longs in 3-D space - that is, assign them a point in 3-D space that corresponds to their position on the globe. Euclidean distance between those points is then a reasonable proxy for actual (great circle) distance between points, even if it's not quite the same. There's guidance on how to do this at stackoverflow.com/questions/10473852/…. (I see now I re-read you've already considered this. seems a decent approach to me) $\endgroup$
    – DavidH
    Aug 29, 2022 at 10:36

1 Answer 1


I don't have your specific data, because you didn't share anything, but I think this generic example should suffice.

import statsmodels.api as sm
import numpy as np
import pandas as pd

df = your_df

from numpy import unique
from numpy import where
from sklearn.datasets import make_classification
from sklearn.cluster import KMeans
from matplotlib import pyplot

# define dataset
X = df[['lat','lon']]

# define the model
model = KMeans(n_clusters=8)
# fit the model

# assign a cluster to each example
yhat = model.predict(X)


pyplot.scatter(X['lat'], X['lon'], c=X['kmeans'], cmap='some_variable', s=50, alpha=0.8)

Your specific outcome will certainly be different, but you should see something essentially like this.

enter image description here

The key thing to focus on is this:

X = df[['lat','lon']]

Add more variables as you desire, and check the outcome each time you make a change. Ultimately, you need to figure out what makes sense, because clustering is non-supervised, so there is no R^2 metric, or accuracy metric, are any kind of supervised metric.

You can find more detail here.



  • $\begingroup$ I don't see how this solves the problem of latitude and longitude representing points in three-dimensional space, resulting in inaccuracies when calculating the distance between these points. I believe this method would only work if locations were confined to a geographically similar space (e.g. all points in Germany). Even then, Euclidean distance would only provide an approximation and not the true distance between locations. $\endgroup$
    – Ultralite
    Aug 29, 2022 at 10:04
  • $\begingroup$ Sorry. I just posted two updated links. $\endgroup$
    – ASH
    Aug 29, 2022 at 13:25

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