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How can I run hierarchical clustering on a correlation matrix in scipy/numpy? I have a matrix of 100 rows by 9 columns, and I'd like to hierarchically cluster by correlations of each entry across the 9 conditions. I'd like to use 1-pearson correlation as the distances for clustering. Assuming I have a numpy array X that contains the 100 x 9 matrix, how can I do this?

I tried using hcluster, based on this example:

Y=pdist(X, 'seuclidean')
Z=linkage(Y, 'single')
dendrogram(Z, color_threshold=0)

However, pdist is not what I want, since that's a euclidean distance. Any ideas?

thanks.

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2 Answers 2

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Just change the metric to correlation so that the first line becomes:

Y=pdist(X, 'correlation')

However, I believe that the code can be simplified to just:

Z=linkage(X, 'single', 'correlation')
dendrogram(Z, color_threshold=0)

because linkage will take care of the pdist for you.

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4 Comments

Does 'correlation' here mean Pearson or Spearman? Also, shouldn't it be 1 - pearson in order to be a valid distance metric that can be used for pdist? Does pdist do that automatically? thanks.
It looks like it is 1 - pearson to me. You can look at it yourself in site-packages/scipy/spatial/distance.py
It's fairly rare for "correlation" mentioned alone to mean Spearman correlation. Usually if it's Spearman people will say so, otherwise assume Pearson.
I you want the correlations between the columns, you need to pass the transpose of X: Z=linkage(X.T, 'single', 'correlation')
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I find it helpful to perform and visualize the hierarchical clustering using the seaborn clustermap (which uses underneath scipy for the clustering), after having used 'correlation' as a metric for pdist:

import seaborn as sns
from scipy.cluster.hierarchy import dendrogram
from scipy.spatial.distance import pdist, squareform

D = squareform(pdist(X.T, 'correlation'))
h = sns.clustermap(D, cmap='Reds')

You can also recover the corresponding linkage matrix and plot the dendrogram

Z = h.dendrogram_col.linkage    
dendrogram(Z, color_threshold=0)
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