Specify Number of Nodes in Dendrogram Plot There are 100 data points in the original data set, X . Create a hierarchical binary cluster tree using linkage . Then, plot the dendrogram for the complete tree (100 leaf nodes) by setting the input argument P equal to 0 . Now, plot the dendrogram with only 25 leaf nodes.
The key to interpreting a dendrogram is to focus on the height at which any two objects are joined together. In the example above, we can see that E and F are most similar, as the height of the link that joins them together is the smallest. The next two most similar objects are A and B.
The input to linkage()
is either an n x m array, representing n points in
m-dimensional space, or a one-dimensional array containing the condensed distance matrix. In your example, mat
is 3 x 3, so you are clustering
three 3-d points. Clustering is based on the distance between these points.
Why does mat and 1-mat give identical clusterings here?
The arrays mat
and 1-mat
produce the same clustering because the clustering
is based on distances between the points, and neither a reflection (-mat
)
nor a translation (mat + offset
) of the entire data set change the relative
distances between the points.
How can I annotate the distance along each branch of the tree using dendrogram so that the distances between pairs of nodes can be compared?
In the code below, I
show how you can use the data returned by dendrogram to label the horizontal
segments of the diagram with the corresponding distance. The values associated
with the keys icoord
and dcoord
give the x and y coordinates of each
three-segment inverted-U of the figure. In augmented_dendrogram
this data
is used to add a label of the distance (i.e. y value) of each horizontal
line segment in dendrogram.
from scipy.cluster.hierarchy import dendrogram
import matplotlib.pyplot as plt
def augmented_dendrogram(*args, **kwargs):
ddata = dendrogram(*args, **kwargs)
if not kwargs.get('no_plot', False):
for i, d in zip(ddata['icoord'], ddata['dcoord']):
x = 0.5 * sum(i[1:3])
y = d[1]
plt.plot(x, y, 'ro')
plt.annotate("%.3g" % y, (x, y), xytext=(0, -8),
textcoords='offset points',
va='top', ha='center')
return ddata
For your mat
array, the augmented dendrogram is
So point 'a' and 'c' are 1.01 units apart, and point 'b' is 1.57 units from the cluster ['a', 'c'].
It seems that show_leaf_counts
flag is ignored, is there a way to turn it on
so that the number of objects in each class is shown?
The flag show_leaf_counts
only applies when not all the original data
points are shown as leaves. For example, when trunc_mode = "lastp"
,
only the last p
nodes are show.
Here's an example with 100 points:
import numpy as np
from scipy.cluster.hierarchy import linkage
import matplotlib.pyplot as plt
from augmented_dendrogram import augmented_dendrogram
# Generate a random sample of `n` points in 2-d.
np.random.seed(12312)
n = 100
x = np.random.multivariate_normal([0, 0], np.array([[4.0, 2.5], [2.5, 1.4]]),
size=(n,))
plt.figure(1, figsize=(6, 5))
plt.clf()
plt.scatter(x[:, 0], x[:, 1])
plt.axis('equal')
plt.grid(True)
linkage_matrix = linkage(x, "single")
plt.figure(2, figsize=(10, 4))
plt.clf()
plt.subplot(1, 2, 1)
show_leaf_counts = False
ddata = augmented_dendrogram(linkage_matrix,
color_threshold=1,
p=6,
truncate_mode='lastp',
show_leaf_counts=show_leaf_counts,
)
plt.title("show_leaf_counts = %s" % show_leaf_counts)
plt.subplot(1, 2, 2)
show_leaf_counts = True
ddata = augmented_dendrogram(linkage_matrix,
color_threshold=1,
p=6,
truncate_mode='lastp',
show_leaf_counts=show_leaf_counts,
)
plt.title("show_leaf_counts = %s" % show_leaf_counts)
plt.show()
These are the points in the data set:
With p=6
and trunc_mode="lastp"
, dendrogram
only shows the "top"
of the dendrogram. The following shows the effect of show_leaf_counts
.
I think there's a couple misunderstandings as to the use of the functions that you are trying to use. Here's a fully working code snippet to illustrate my points:
import matplotlib.pyplot as plt
from scipy.cluster.hierarchy import dendrogram, linkage
from numpy import array
import numpy as np
mat = array([184, 222, 177, 216, 231,
45, 123, 128, 200,
129, 121, 203,
46, 83,
83])
dist_mat = mat
linkage_matrix = linkage(dist_mat, 'single')
print linkage_matrix
plt.figure(101)
plt.subplot(1, 2, 1)
plt.title("ascending")
dendrogram(linkage_matrix,
color_threshold=1,
truncate_mode='lastp',
labels=array(['a', 'b', 'c', 'd', 'e', 'f']),
distance_sort='ascending')
plt.subplot(1, 2, 2)
plt.title("descending")
dendrogram(linkage_matrix,
color_threshold=1,
truncate_mode='lastp',
labels=array(['a', 'b', 'c', 'd', 'e', 'f']),
distance_sort='descending')
def make_fake_data():
amp = 1000.
x = []
y = []
for i in range(0, 10):
s = 20
x.append(np.random.normal(30, s))
y.append(np.random.normal(30, s))
for i in range(0, 20):
s = 2
x.append(np.random.normal(150, s))
y.append(np.random.normal(150, s))
for i in range(0, 10):
s = 5
x.append(np.random.normal(-20, s))
y.append(np.random.normal(50, s))
plt.figure(1)
plt.title('fake data')
plt.scatter(x, y)
d = []
for i in range(len(x) - 1):
for j in range(i+1, len(x) - 1):
d.append(np.sqrt(((x[i]-x[j])**2 + (y[i]-y[j])**2)))
return d
mat = make_fake_data()
plt.figure(102)
plt.title("Three Clusters")
linkage_matrix = linkage(mat, 'single')
print "three clusters"
print linkage_matrix
dendrogram(linkage_matrix,
truncate_mode='lastp',
color_threshold=1,
show_leaf_counts=True)
plt.show()
First of all, the computation m -> m - 1 didn't really change your result since the distance matrix, which basically describes the relative distances between all unique pairs, didn't change in your specific case. (In my example code above, all distances are Euclidean so all are positive and consistent from points on a 2d plane.)
For your second question, you probably need to roll out your own annotation routine to do what you want, since I don't think dendromgram natively supports it...
For the last question, show_leaf_counts seems to work only when you try to display non-singleton leaf nodes with truncate_mode='lastp' option. Basically a leaves are bunched up so close together that they are not easy to see. So you have an option of just displaying a leaf but have an option of showing (in parenthesis) how many are bunched up in that leaf.
Hope this helps.
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