Fix bugs of bluetooth feature extraction when number of unique bt_address is 2
parent
8414977331
commit
68e12a2563
|
@ -73,18 +73,19 @@ def ownership_based_on_clustering(bt_frequency):
|
|||
diff_k2 = [(X_array[xi] - centers_k2[labels_k2[xi]])**2 for xi in range(0, len(X_array))]
|
||||
sum_dist_k2 = sum(diff_k2)
|
||||
|
||||
# By default, model with K = 2 is chosen
|
||||
labels = labels_k2
|
||||
centers = centers_k2
|
||||
numclust = 2
|
||||
if len(X_array) > 2:
|
||||
# K = 3, devices I own VS devices my partner/roommate owns (can also be other devices I own though) VS devices other people own
|
||||
kmeans_k3 = KMeans(n_clusters=3, init=initial_k3, n_init = 1).fit(X)
|
||||
labels_k3 = kmeans_k3.labels_
|
||||
centers_k3 = [c[0] for c in kmeans_k3.cluster_centers_]
|
||||
diff_k3 = [(X_array[xi] - centers_k3[labels_k3[xi]])**2 for xi in range(0, len(X_array))]
|
||||
sum_dist_k3 = sum(diff_k3)
|
||||
|
||||
if sum_dist_k2 < sum_dist_k3: # K = 2 is better
|
||||
labels = labels_k2
|
||||
centers = centers_k2
|
||||
numclust = 2
|
||||
else:
|
||||
# Model with K = 3 is chosen if sum of squared distances between clustered points and cluster centers is smaller or equal to what we get with K = 2
|
||||
if sum_dist_k3 <= sum_dist_k2:
|
||||
labels = labels_k3
|
||||
centers = centers_k3
|
||||
numclust = 3
|
||||
|
|
Loading…
Reference in New Issue