[FIXED] How to have clusters of stacked bars

Issue

So here is how my data set looks like :

In [1]: df1=pd.DataFrame(np.random.rand(4,2),index=["A","B","C","D"],columns=["I","J"])

In [2]: df2=pd.DataFrame(np.random.rand(4,2),index=["A","B","C","D"],columns=["I","J"])

In [3]: df1
Out[3]: 
          I         J
A  0.675616  0.177597
B  0.675693  0.598682
C  0.631376  0.598966
D  0.229858  0.378817

In [4]: df2
Out[4]: 
          I         J
A  0.939620  0.984616
B  0.314818  0.456252
C  0.630907  0.656341
D  0.020994  0.538303

I want to have stacked bar plot for each dataframe but since they have same index, I'd like to have 2 stacked bars per index.

I've tried to plot both on the same axes :

In [5]: ax = df1.plot(kind="bar", stacked=True)

In [5]: ax2 = df2.plot(kind="bar", stacked=True, ax = ax)

But it overlaps.

Then I tried to concat the two dataset first :

pd.concat(dict(df1 = df1, df2 = df2),axis = 1).plot(kind="bar", stacked=True)

but here everything is stacked

My best try is :

 pd.concat(dict(df1 = df1, df2 = df2),axis = 0).plot(kind="bar", stacked=True)

Which gives :

This is basically what I want, except that I want the bar ordered as

(df1,A) (df2,A) (df1,B) (df2,B) etc...

I guess there is a trick but I can't found it !

---

After @bgschiller's answer I got this :

Which is almost what I want. I would like the bar to be clustered by index, in order to have something visually clear.

Bonus : Having the x-label not redundant, something like :

df1 df2    df1 df2
_______    _______ ...
   A          B

Solution

I eventually found a trick (edit: see below for using seaborn and longform dataframe):

Solution with pandas and matplotlib

Here it is with a more complete example :

import pandas as pd
import matplotlib.cm as cm
import numpy as np
import matplotlib.pyplot as plt

def plot_clustered_stacked(dfall, labels=None, title="multiple stacked bar plot",  H="/", **kwargs):
    """Given a list of dataframes, with identical columns and index, create a clustered stacked bar plot. 
labels is a list of the names of the dataframe, used for the legend
title is a string for the title of the plot
H is the hatch used for identification of the different dataframe"""

    n_df = len(dfall)
    n_col = len(dfall[0].columns) 
    n_ind = len(dfall[0].index)
    axe = plt.subplot(111)

    for df in dfall : # for each data frame
        axe = df.plot(kind="bar",
                      linewidth=0,
                      stacked=True,
                      ax=axe,
                      legend=False,
                      grid=False,
                      **kwargs)  # make bar plots

    h,l = axe.get_legend_handles_labels() # get the handles we want to modify
    for i in range(0, n_df * n_col, n_col): # len(h) = n_col * n_df
        for j, pa in enumerate(h[i:i+n_col]):
            for rect in pa.patches: # for each index
                rect.set_x(rect.get_x() + 1 / float(n_df + 1) * i / float(n_col))
                rect.set_hatch(H * int(i / n_col)) #edited part     
                rect.set_width(1 / float(n_df + 1))

    axe.set_xticks((np.arange(0, 2 * n_ind, 2) + 1 / float(n_df + 1)) / 2.)
    axe.set_xticklabels(df.index, rotation = 0)
    axe.set_title(title)

    # Add invisible data to add another legend
    n=[]        
    for i in range(n_df):
        n.append(axe.bar(0, 0, color="gray", hatch=H * i))

    l1 = axe.legend(h[:n_col], l[:n_col], loc=[1.01, 0.5])
    if labels is not None:
        l2 = plt.legend(n, labels, loc=[1.01, 0.1]) 
    axe.add_artist(l1)
    return axe

# create fake dataframes
df1 = pd.DataFrame(np.random.rand(4, 5),
                   index=["A", "B", "C", "D"],
                   columns=["I", "J", "K", "L", "M"])
df2 = pd.DataFrame(np.random.rand(4, 5),
                   index=["A", "B", "C", "D"],
                   columns=["I", "J", "K", "L", "M"])
df3 = pd.DataFrame(np.random.rand(4, 5),
                   index=["A", "B", "C", "D"], 
                   columns=["I", "J", "K", "L", "M"])

# Then, just call :
plot_clustered_stacked([df1, df2, df3],["df1", "df2", "df3"])
    

And it gives that :

You can change the colors of the bar by passing a cmap argument:

plot_clustered_stacked([df1, df2, df3],
                       ["df1", "df2", "df3"],
                       cmap=plt.cm.viridis)

---

Solution with seaborn:

Given the same df1, df2, df3, below, I convert them in a long form:

df1["Name"] = "df1"
df2["Name"] = "df2"
df3["Name"] = "df3"
dfall = pd.concat([pd.melt(i.reset_index(),
                           id_vars=["Name", "index"]) # transform in tidy format each df
                   for i in [df1, df2, df3]],
                   ignore_index=True)

The problem with seaborn is that it doesn't stack bars natively, so the trick is to plot the cumulative sum of each bar on top of each other:

dfall.set_index(["Name", "index", "variable"], inplace=1)
dfall["vcs"] = dfall.groupby(level=["Name", "index"]).cumsum()
dfall.reset_index(inplace=True) 

>>> dfall.head(6)
  Name index variable     value       vcs
0  df1     A        I  0.717286  0.717286
1  df1     B        I  0.236867  0.236867
2  df1     C        I  0.952557  0.952557
3  df1     D        I  0.487995  0.487995
4  df1     A        J  0.174489  0.891775
5  df1     B        J  0.332001  0.568868

Then loop over each group of variable and plot the cumulative sum:

c = ["blue", "purple", "red", "green", "pink"]
for i, g in enumerate(dfall.groupby("variable")):
    ax = sns.barplot(data=g[1],
                     x="index",
                     y="vcs",
                     hue="Name",
                     color=c[i],
                     zorder=-i, # so first bars stay on top
                     edgecolor="k")
ax.legend_.remove() # remove the redundant legends 

It lacks the legend that can be added easily I think. The problem is that instead of hatches (which can be added easily) to differentiate the dataframes we have a gradient of lightness, and it's a bit too light for the first one, and I don't really know how to change that without changing each rectangle one by one (as in the first solution).

Tell me if you don't understand something in the code.

Feel free to re-use this code which is under CC0.

Answered By - jrjc