Reordering the high-level clusters from seaborn clustermap results

Question

Is there a way to get from a to b in the following figure with scripting? I am using seaborn.clustermap() to get to a (i.e. the order of the rows are preserved. However, columns order change only at second highest level).

I was wondering whether it is possible to use the seaborn.matrix.ClusterGrid that is returned by seaborn.clustermap(), modify it and plot the modified results. b P.S. The reason I am asking this is that the order has a meaning (first comes blue, next green, and finally red).

Update: Here is a small data set to generate the situation:

df = pd.DataFrame([[1, 1.1, 0.9, 1.9, 2, 2.1, 2.8, 3, 3.1], 
                   [1.8, 2, 2.1, 0.7, 1, 1.1, 2.7, 3, 3.3]],
              columns = ['d1', 'd2', 'd3', 
                         'l3', 'l2', 'l1', 
                         'b1', 'b2', 'b3'],
              index = ['p1', 'p2'])

cg = sns.clustermap(df); ## returns a ClusterGrid

The output is this:

We can think of columns starting with b as breakfast, l as lunch and d as dinner. Now, the order, is breakfast -> dinner -> lunch. I want to get to breakfast -> lunch -> dinner.

Can you create a small dataset with the code to generate the clustermap? — Scott Boston, May 22 '19 at 14:35
@ScottBoston: This is a rather big data set and as it is visible, I have truncated the rows. I am afraid that with a smaller number of rows I can not reproduce the separation that is shown at the moment. But I will do my best. I will update the question as soon as I have managed to create one. — Dataman, May 22 '19 at 14:40
@ScottBoston I now added some data. Hopefully, it is helpful. — Dataman, May 23 '19 at 06:52
do you want to keep the order `b1,b2,b3,l1,l2,l3,d1,d2,d3` or merely the general order `b,l,d` and each subclass in there is to be ordered by the algorithm? — Diziet Asahi, May 23 '19 at 11:45
@DizietAsahi I would like the latter i.e. "merely the general order b,l,d and each subclass in there is to be ordered by the algorithm" — Dataman, May 23 '19 at 12:10
@Dataman Yes, this small set of data is helpful. However, I am not able to determine a way to achieve your desired results. — Scott Boston, May 23 '19 at 17:27

Dataman · Answer 1 · 2020-08-23T08:55:44.417

This is how I solved my problem. It works but it might not be as elegant as one would hope for!

import pandas as pd
import seaborn as sns
import numpy as np
import matplotlib.pyplot as plt
from scipy.cluster.hierarchy import linkage, dendrogram

# set the desired order of groups eg: breakfast, lunch, dinner
groups = ['b', 'l', 'd'] 

# reorder indexes/indices besed on the desired order
new_order = []
for group in groups:
    indexes = cg.data2d.columns.str.startswith(group)
    indexes_locs = np.where(indexes)[0].tolist()
    new_order += indexes_locs
    
## reorder df based on the new order
ordered_df = cg.data2d.iloc[:, new_order]

## Run clustermap on the reordered dataframe by disabling 
## the clustering for both rows and columns
ocg = sns.clustermap(ordered_df, 
                     row_cluster=False, 
                     col_cluster=False,
                    );

## draw dendrogram x-axis
axx = ocg.ax_col_dendrogram.axes
axx.clear()

with plt.rc_context({'lines.linewidth': 0.5}):
    
    link = cg.dendrogram_col.linkage ## extract the linkage information

    ## manualy inspect the linkage and determine the new desired order
    link[[4, 2]] = link[[2, 4]]  ## swaping the two groups of higher hierarchy
    
    ## draw the the dendrogram on the x-axis
    dendrogram(link, 
           color_threshold=0, 
           ax=axx,
           truncate_mode='lastp',
           orientation='top',
           link_color_func=lambda x: 'k'
          );

axx.set_yticklabels(['']*len(axx.get_yticklabels()))
axx.tick_params(color='w')
    
## draw dendrogram y-axis (no chage here)
axy = ocg.ax_row_dendrogram.axes
axy.clear()

with plt.rc_context({'lines.linewidth': 0.5}):
    
    ## draw the the dendrogram on the y-axis
    dendrogram(cg.dendrogram_row.linkage, 
           color_threshold=0, 
           ax=axy,
           truncate_mode='lastp',
           orientation='left',
           link_color_func=lambda x: 'k',
          );

axy.set_xticklabels(['']*len(axy.get_yticklabels()))
axy.tick_params(color='w')
# axy.invert_yaxis() # we might need to invert y-axis

The output looks like this:

score 0 · Answer 2 · answered Apr 22 '22 at 06:31

This solution should be simpler and more elegant. The trick is to understand what is saved in the link. The key is that, if you have 9 samples, and you merge 0 & 1, then you gain a new "sample"/"group" 10. If you then merge 2 and 10, then you gain a new "sample"/"group" 11. You just need to find the group id and sway its position during merging.

from scipy.cluster.hierarchy import linkage, dendrogram

link = linkage(df.T)
link
_ = dendrogram(link)

array([[ 4.        ,  5.        ,  0.14142136,  2.        ],
       [ 0.        ,  1.        ,  0.2236068 ,  2.        ],
       [ 2.        , 10.        ,  0.2236068 ,  3.        ],
       [ 7.        ,  8.        ,  0.31622777,  2.        ],
       [ 3.        ,  9.        ,  0.31622777,  3.        ],
       [ 6.        , 12.        ,  0.36055513,  3.        ],
       [11.        , 13.        ,  1.28062485,  6.        ],
       [14.        , 15.        ,  1.74642492,  9.        ]])

    link[6][[0, 1]] = link[6][[1, 0]]
    link[7][[0, 1]] = link[7][[1, 0]]
    _ = dendrogram(link)

    cg = sns.clustermap(df, col_linkage=link)

Thank you for your time! :) The output figure in your solution is still `lunch -> dinner -> breakfast`! The desired output was: `breakfast -> lunch -> dinner`. Is something missing here? — Dataman, Apr 22 '22 at 10:17
The original output is breakfast->dinner->lunch, and this version is lunch->dinner->breakfast. I think this is what you want. If you want breakfast->lunch->dinner, just delete "link[7][[0, 1]] = link[7][[1, 0]]". — Shidan, Apr 22 '22 at 18:46

Reordering the high-level clusters from seaborn clustermap results

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