O.rka - 8 months ago 170

Python Question

Usually when I do dendrograms and heatmaps, I use a distance matrix and do a bunch of

`SciPy`

`Seaborn`

`Seaborn`

I essentially want to use

`seaborn`

Here's the usage:

`seaborn.clustermapÂ¶`

seaborn.clustermap(data, pivot_kws=None, method='average', metric='euclidean',

z_score=None, standard_scale=None, figsize=None, cbar_kws=None, row_cluster=True,

col_cluster=True, row_linkage=None, col_linkage=None, row_colors=None,

col_colors=None, mask=None, **kwargs)

My code below:

`from sklearn.datasets import load_iris`

iris = load_iris()

X, y = iris.data, iris.target

DF = pd.DataFrame(X, index = ["iris_%d" % (i) for i in range(X.shape[0])], columns = iris.feature_names)

I don't think my method is correct below because I'm giving it a precomputed distance matrix and NOT a rectangular data matrix as it requests. There's no examples of how to use a correlation/distance matrix with

`clustermap`

`sns.heatmap`

`DF_corr = DF.T.corr()`

DF_dism = 1 - DF_corr

sns.clustermap(DF_dism)

Answer

You can pass the precomputed distance matrix as linkage to `clustermap()`

:

```
import pandas as pd, seaborn as sns
import scipy.spatial as sp, scipy.cluster.hierarchy as hc
from sklearn.datasets import load_iris
sns.set(font="monospace")
iris = load_iris()
X, y = iris.data, iris.target
DF = pd.DataFrame(X, index = ["iris_%d" % (i) for i in range(X.shape[0])], columns = iris.feature_names)
DF_corr = DF.T.corr()
DF_dism = 1 - DF_corr # distance matrix
linkage = hc.linkage(sp.distance.squareform(DF_dism), method='average')
sns.clustermap(DF_dism, row_linkage=linkage, col_linkage=linkage)
```

For `clustermap(distance_matrix)`

(i.e., without linkage passed), the linkage is calculated internally based on pairwise distances of the rows and columns in the distance matrix (see note below for full details) instead of using the elements of the distance matrix directly (the correct solution). As a result, the output is somewhat different from the one in the question:

Note: if no `row_linkage`

is passed to `clustermap()`

, the row linkage is determined internally by considering each row a "point" (observation) and calculating the pairwise distances between the points. So the row dendrogram reflects row similarity. Analogous for `col_linkage`

, where each column is considered a point. This explanation should likely be added to the docs. Here the docs's first example modified to make the internal linkage calculation explicit:

```
import seaborn as sns; sns.set()
import scipy.spatial as sp, scipy.cluster.hierarchy as hc
flights = sns.load_dataset("flights")
flights = flights.pivot("month", "year", "passengers")
row_linkage, col_linkage = (hc.linkage(sp.distance.pdist(x), method='average')
for x in (flights.values, flights.values.T))
g = sns.clustermap(flights, row_linkage=row_linkage, col_linkage=col_linkage)
# note: this produces the same plot as "sns.clustermap(flights)", where
# clustermap() calculates the row and column linkages internally
```