Hi, Thanks for your hint It just saved my day. Regards, Rajkumar
On Wed, Feb 21, 2018 at 4:28 PM, Christian Braune < christian.braun...@gmail.com> wrote: > Hi, > > if you have your original points stored in a numpy array, you can get all > points from a cluster i by doing the following: > > cluster_points = points[kmeans.labels_ == i] > > "kmeans.labels_" contains a list labels for each point. > "kmeans.labels_ == i" creates a mask that selects only those points that > belong to cluster i > and the whole line then gives you the points, finally. > > BTW: the fit method has the raw points as input parameter, not the > distance matrix. > > Regards, > Christian > > prince gosavi <princegosav...@gmail.com> schrieb am Mi., 21. Feb. 2018 um > 11:16 Uhr: > >> Hi, >> I have applied Kmeans clustering using the scikit library from >> >> kmeans=KMeans(max_iter=4,n_clusters=10,n_init=10).fit(euclidean_dist) >> >> After applying the algorithm.I would like to get the data points in the >> clusters so as to further use them to apply a model. >> >> Example: >> kmeans.cluster_centers_[1] >> >> gives me distance array of all the data points. >> >> Is there any way around this available in scikit so as to get the data >> points id/index. >> >> Regards >> _______________________________________________ >> scikit-learn mailing list >> scikit-learn@python.org >> https://mail.python.org/mailman/listinfo/scikit-learn >> > > _______________________________________________ > scikit-learn mailing list > scikit-learn@python.org > https://mail.python.org/mailman/listinfo/scikit-learn > > -- Regards
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