scipy.cluster.hierarchy.to_tree(Z, rd=False)[source]

Convert 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)