sklearn.metrics.normalized_mutual_info_score¶
- sklearn.metrics.normalized_mutual_info_score(labels_true, labels_pred, max_n_classes=5000)[source]¶
Normalized Mutual Information between two clusterings
Normalized Mutual Information (NMI) is an normalization of the Mutual Information (MI) score to scale the results between 0 (no mutual information) and 1 (perfect correlation). In this function, mutual information is normalized by sqrt(H(labels_true) * H(labels_pred))
This measure is not adjusted for chance. Therefore adjusted_mustual_info_score might be preferred.
This metric is independent of the absolute values of the labels: a permutation of the class or cluster label values won’t change the score value in any way.
This metric is furthermore symmetric: switching label_true with label_pred will return the same score value. This can be useful to measure the agreement of two independent label assignments strategies on the same dataset when the real ground truth is not known.
Read more in the User Guide.
Parameters: labels_true : int array, shape = [n_samples]
A clustering of the data into disjoint subsets.
labels_pred : array, shape = [n_samples]
A clustering of the data into disjoint subsets.
max_n_classes: int, optional (default=5000) :
Maximal number of classes handled by the adjusted_rand_score metric. Setting it too high can lead to MemoryError or OS freeze
Returns: nmi: float :
score between 0.0 and 1.0. 1.0 stands for perfectly complete labeling
See also
- adjusted_rand_score
- Adjusted Rand Index
- adjusted_mutual_info_score
- Adjusted Mutual Information (adjusted against chance)
Examples
Perfect labelings are both homogeneous and complete, hence have score 1.0:
>>> from sklearn.metrics.cluster import normalized_mutual_info_score >>> normalized_mutual_info_score([0, 0, 1, 1], [0, 0, 1, 1]) 1.0 >>> normalized_mutual_info_score([0, 0, 1, 1], [1, 1, 0, 0]) 1.0
If classes members are completely split across different clusters, the assignment is totally in-complete, hence the NMI is null:
>>> normalized_mutual_info_score([0, 0, 0, 0], [0, 1, 2, 3]) 0.0