how do I get the subtrees of dendrogram made by scipy.cluster.hierarchy
I had a confusion regarding this module (scipy.cluster.hierarchy) ... and still have some !
For example we have the following dendrogram:
My question is how can I extract the coloured subtrees (each one represent a cluster) in a nice format, say SIF format ? Now the code to get the plot above is:
import scipy
import scipy.cluster.hierarchy as sch
import matplotlib.pylab as plt
scipy.randn(100,2)
d = sch.distance.pdist(X)
Z= sch.linkage(d,method='complete')
P =sch.dendrogram(Z)
plt.savefig('plot_dendrogram.png')
T = sch.fcluster(Z, 0.5*d.max(), 'distance')
#array([4, 5, 3, 2, 2, 3, 5, 2, 2, 5, 2, 2, 2, 3, 2, 3, 2, 5, 4, 5, 2, 5, 2,
# 3, 3, 3, 1, 3, 4, 2, 2, 4, 2, 4, 3, 3, 2, 5, 5, 5, 3, 2, 2, 2, 5, 4,
# 2, 4, 2, 2, 5, 5, 1, 2, 3, 2, 2, 5, 4, 2, 5, 4, 3, 5, 4, 4, 2, 2, 2,
# 4, 2, 5, 2, 2, 3, 3, 2, 4, 5, 3, 4, 4, 2, 1, 5, 4, 2, 2, 5, 5, 2, 2,
# 5, 5, 5, 4, 3, 3, 2, 4], dtype=int32)
sch.leaders(Z,T)
# (array([190, 191, 182, 193, 194], dtype=int32),
# array([2, 3, 1, 4,5],dtype=int32))
So now, the output of fcluster() gives the clustering of the nodes (by their id's), and leaders() described here is supposed to return 2 arrays:
first one contains the leader nodes of the clusters generated by Z, here we can see we have 5 clusters, as well as in the plot
and the second one the id's of these clusters
So if this leaders() returns resp. L and M : L[2]=182 and M[2]=1, then cluster 1 is leaded by node id 182, which doesn't exist in the observations set X, the documentation says "... then it corresponds to a non-singleton cluster". But I can't get it ...
Also, I converted the Z to a tree by sch.to_tree(Z), that will return an easy-to-use tree object, which I want to visualize, but which tool should I use as a graphical platform that manipulate these kind of tree objects as inputs?
Answer
Answering the part of your question regarding tree manipulation...
As explained in aother answer, you can read the coordinates of the branches reading icoord and dcoord from the tree object. For each branch the coordinated are given from the left to the right.
If you want to manually plot the tree you can use something like:
def plot_tree(P, pos=None):
plt.clf()
icoord = scipy.array(P['icoord'])
dcoord = scipy.array(P['dcoord'])
color_list = scipy.array(P['color_list'])
xmin, xmax = icoord.min(), icoord.max()
ymin, ymax = dcoord.min(), dcoord.max()
if pos:
icoord = icoord[pos]
dcoord = dcoord[pos]
color_list = color_list[pos]
for xs, ys, color in zip(icoord, dcoord, color_list):
plt.plot(xs, ys, color)
plt.xlim(xmin-10, xmax + 0.1*abs(xmax))
plt.ylim(ymin, ymax + 0.1*abs(ymax))
plt.show()
Where, in your code, plot_tree(P) gives:
The function allows you to select just some branches:
plot_tree(P, range(10))
Now you have to know which branches to plot. Maybe the fcluster() output is a little obscure and another way to find which branches to plot based on a minimum and a maximum distance tolerance would be using the output of linkage() directly (Z in the OP's case):
dmin = 0.2
dmax = 0.3
pos = scipy.all( (Z[:,2] >= dmin, Z[:,2] <= dmax), axis=0 ).nonzero()
plot_tree( P, pos )
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