print(f'Best precision for {l} is for cluster {best} with {(df.cluster==best).sum()} points, \
with precision {((labelled.cluster==best)&(labelled.label==l)).sum()/(labelled.cluster==best).sum():.2f} and recall {((labelled.cluster==best)&(labelled.label==l)).sum()/(labelled.label==l).sum():.2f}')