- scipy.cluster.hierarchy.to_tree(Z, rd=False)¶
Converts a linkage matrix into an easy-to-use tree object.
The reference to the root ClusterNode object is returned (by default).
Each ClusterNode object has a left, right, dist, id, and count attribute. The left and right attributes point to ClusterNode objects that were combined to generate the cluster. If both are None then the ClusterNode object is a leaf node, its count must be 1, and its distance is meaningless but set to 0.
Note: This function is provided for the convenience of the library user. ClusterNodes are not used as input to any of the functions in this library.
Z : ndarray
The linkage matrix in proper form (see the linkage function documentation).
rd : bool, optional
When False (default), a reference to the root ClusterNode object is returned. Otherwise, a tuple (r, d) is returned. r is a reference to the root node while d is a list of ClusterNode objects - one per original entry in the linkage matrix plus entries for all clustering steps. If a cluster id is less than the number of samples n in the data that the linkage matrix describes, then it corresponds to a singleton cluster (leaf node). See linkage for more information on the assignment of cluster ids to clusters.
tree : ClusterNode or tuple (ClusterNode, list of ClusterNode)
If rd is False, a ClusterNode. If rd is True, a list of length 2*n - 1, with n the number of samples. See the description of rd above for more details.
>>> from scipy.cluster import hierarchy >>> x = np.random.rand(10).reshape(5, 2) >>> Z = hierarchy.linkage(x) >>> hierarchy.to_tree(Z) <scipy.cluster.hierarchy.ClusterNode object at ... >>> rootnode, nodelist = hierarchy.to_tree(Z, rd=True) >>> rootnode <scipy.cluster.hierarchy.ClusterNode object at ... >>> len(nodelist) 9