Plotting a dendrogram using Plotly Python

Question

I am trying to plot a dendrogram using Python, preferably using Plotly. I have a dataset containing a clustering of various objects. I can use this dataset to generate the required data or at least extrapolate. However, I don't understand what the input to the create_dendrogram actually is. The doc just says that it is a ndarray - Matrix of observations as array of arrays. I am familiar with Numpy ndarrays, but I would like to know what the array must contain.

More specifically, what is the significance of the value X[i][j]. It just seems to be a float between 0 and 1. I had looked at the Plotly API documentation for Python here - https://plot.ly/python/dendrogram/

import plotly.plotly as py
from plotly.tools import FigureFactory as FF

import numpy as np

X = np.random.rand(10, 10)
fig = FF.create_dendrogram(X, orientation='left', labels=names)
py.iplot(fig, filename='dendrogram_with_labels')

If there was an alternative and more intuitive way to get a dendrogram in Python I would also like to know that. I am new to this and any help would be appreciated. (Please let me know if I need to rephrase the question!)

Take a look at docs.scipy.org/doc/scipy/reference/generated/… — Warren Weckesser
Thanks Warren, but the ndarray used in scipy.cluster.hierarchy.dendrogram uses a 4 by (n-1) matrix (ndarray), whereas for Plotly it is a n x n ndarray. — Achyudh Ram
There is a way for doing this with R using ggplot2+dendextend+plotly — Tal Galili

biofunc1234 biofunc1234 · Accepted Answer · 2017-03-30T21:11:14

You can pass a linkage function to the create_dendrogram function. For example:

from scipy.cluster.hierarchy import linkage

...

figure = FF.create_dendrogram(
    data_array, orientation='bottom', labels=id_label_list,
    linkagefun=lambda x: linkage(data_array, 'ward', metric='euclidean')
)

Plotting a dendrogram using Plotly Python

1 Answers