Select appropriate colors in stacked Seaborn barplot - pandas

Question

I want to create a stacked barplot using Seaborn with this MiltiIndex DataFrame
header = pd.MultiIndex.from_product([['#'],
['TE', 'SS', 'M', 'MR']])
dat = ([[100, 20, 21, 35], [100, 12, 5, 15]])
df = pd.DataFrame(dat, index=['JC', 'TTo'], columns=header)
df = df.stack()
df = df.sort_values('#', ascending=False).sort_index(level=0, sort_remaining=False)
The code I'm using for the plot is:
fontP = FontProperties()
fontP.set_size('medium')
colors = {'TE': 'green', 'SS': 'blue', 'M': 'yellow', 'MR': 'red'}
kwargs = {'alpha':0.5}
plt.figure(figsize=(12, 9))
sns.barplot(x=df2.index.get_level_values(0).unique(),
y=df2.loc[pd.IndexSlice[:, df2.index[0]], '#'],
color=colors[df2.index[0][1]], **kwargs)
sns.barplot(x=df2.index.get_level_values(0).unique(),
y=df2.loc[pd.IndexSlice[:, df2.index[1]], '#'],
color=colors[df2.index[1][1]], **kwargs)
sns.barplot(x=df2.index.get_level_values(0).unique(),
y=df2.loc[pd.IndexSlice[:, df2.index[2]], '#'],
color=colors[df2.index[2][1]], **kwargs)
bottom_plot = sns.barplot(x=df2.index.get_level_values(0).unique(),
y=df2.loc[pd.IndexSlice[:, df2.index[3]], '#'],
color=colors[df2.index[3][1]], **kwargs)
bar1 = plt.Rectangle((0, 0), 1, 1, fc='green', edgecolor="None")
bar2 = plt.Rectangle((0, 0), 0, 0, fc='yellow', edgecolor="None")
bar3 = plt.Rectangle((0, 0), 2, 2, fc='red', edgecolor="None")
bar4 = plt.Rectangle((0, 0), 3, 3, fc='blue', edgecolor="None")
l = plt.legend([bar1, bar2, bar3, bar4], [
"TE", "M",
'MR', 'SS'
],
bbox_to_anchor=(0.95, 1),
loc='upper left',
prop=fontP)
l.draw_frame(False)
sns.despine()
bottom_plot.set_ylabel("#")
axes = plt.gca()
axes.yaxis.grid()
And I get:
My problem is the order of the colors in the second bar ('TTo'), I want the colors to be automatically selected based on the level 1 index value (['TE', 'SS', 'M', 'MR']) so that they are ordered correctly. Further down the one with the highest value with its corresponding color, in front the next one with the next highest value and its color and so on, as the first bar shows ('JC).
Maybe there is a simpler way to do this in Seaborn than the one I'm using...

Answer 1

I'm not sure how to create such a plot with seaborn. Here is a way to create it with a loop through the rows and adding one matplotlib bar at each step:
import pandas as pd
import seaborn as sns
from matplotlib import pyplot as plt
sns.set()
header = pd.MultiIndex.from_product([['#'],
['TE', 'SS', 'M', 'MR']])
dat = ([[100, 20, 21, 35], [100, 12, 5, 15]])
df = pd.DataFrame(dat, index=['JC', 'TTo'], columns=header)
df = df.stack()
df = df.sort_values('#', ascending=False).sort_index(level=0, sort_remaining=False)
colors = {'TE': 'green', 'SS': 'blue', 'M': 'yellow', 'MR': 'red'}
prev_index0 = None
for (index0, index1), quantity in df.itertuples():
if index0 != prev_index0:
bottom = 0
plt.bar(index0, quantity, fc=colors[index1], ec='none', bottom=bottom, label=index1)
bottom += quantity
prev_index0 = index0
legend_handles = [plt.Rectangle((0, 0), 0, 0, color=colors[c], label=c) for c in colors]
plt.legend(handles=legend_handles)
plt.show()
To plot the bars back to front without stacking, the code can be simplified:
colors = {'TE': 'forestgreen', 'SS': 'cornflowerblue', 'M': 'gold', 'MR': 'crimson'}
for (index0, index1), quantity in df.itertuples():
plt.bar(index0, quantity, fc=colors[index1], ec='none', label=index1)
legend_handles = [plt.Rectangle((0, 0), 0, 0, color=colors[c], label=c, ec='black') for c in colors]
plt.legend(handles=legend_handles, bbox_to_anchor=(1.02, 1.02), loc='upper left')
plt.tight_layout()