Python: Color pandas dataframe based on MultiIndex

Question

Consider this MultiIndex dataframe

i = pd.MultiIndex.from_tuples([(0, 0), (0, 1), (1, 0), (1, 1)], names=['level_0', 'level_1'])
df = pd.DataFrame(range(0, 4), index=i, columns=['foo'])

and this color dictionary

colors = {0: (0.6, 0.8, 0.8, 1), 1: (1, 0.9, 0.4, 1)}

where the first three items of the tuple are RGB values and the fourth item is the transparency alpha.

How can the df be colored based on the level_0 index?

This result would be nice to have:

But this one here I would consider fine art:

In the latter style, the lighter cells would have the same RGB settings but a transparency of 0.25.

jezrael · Accepted Answer · 2020-01-16 15:56:27Z

2

You can use Styler.set_table_styles for set styles only first level of MultiIndex:

i = pd.MultiIndex.from_tuples([(0, 0), (0, 1), (1, 0), (1, 1)], names=['level_0', 'level_1'])
df = pd.DataFrame(range(0, 4), index=i, columns=['foo']) 
#print (df)

import matplotlib.colors as col
colors = {0: (0.6, 0.8, 0.8, 1), 1: (1, 0.9, 0.4, 1)}

c = {k:col.rgb2hex(v) for k, v in colors.items()}
idx = df.index.get_level_values(0)

css = [{'selector': f'.row{i}.level0','props': [('background-color', c[v])]}
             for i,v in enumerate(idx)]
print (css)

df.style.set_table_styles(css)

Second is also possible, but more complicated:

colors = {0: (0.6, 0.8, 0.8, 1), 1: (1, 0.9, 0.4, 1)}
#converts grba to integers
c1 = {k: (int(r * 255),int(g * 255),int(b * 255), a) for k, (r,g,b,a) in colors.items()}
c2 = {k: (int(r * 255),int(g * 255),int(b * 255), 0.25) for k, (r,g,b,a) in colors.items()}

#get values of first level of MulitIndex
idx = df.index.get_level_values(0)
#set css for first level
css = [{'selector': f'.row{i}.level0', 
        'props': [('background-color', f'rgba{c1[j]}')]} for i,j in enumerate(idx)]
#counter per first level for pair and unpair coloring
zipped = zip(df.groupby(idx).cumcount(), enumerate(idx))

css1 = [{'selector': f'.row{i}', 'props': [('background-color', f'rgba{c1[j]}')]} 
       if v % 2 == 0 
       else {'selector': f'.row{i}', 'props': [('background-color', f'rgba{c2[j]}')]} 
       for v,(i, j) in zipped]


df.style.set_table_styles(css1 + css)

edited Jan 16, 2020 at 15:56

answered Jan 16, 2020 at 12:53

jezrael

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3 Comments

hyco Over a year ago

Excellent! Thanks, @jezrael, you've done the art. Note: To display, I only call df.style.set_table_styles(css1 + css).

jezrael Over a year ago

@hyco - yes, it was really interesting question, but really hard, it was challenge ;)

hyco Over a year ago

see below that cssisn't even necessary.

hyco · Accepted Answer · 2020-01-16 16:13:56Z

1

@jezrael has done the art!

import pandas as pd

i = pd.MultiIndex.from_tuples([(0, 'zero'), (0, 'one'), (0, 'two'), (1, 'zero'), (1, 'one')], names=['level_0', 'level_1'])
df = pd.DataFrame(range(0, 5), index=i, columns=['foo']) 

colors = {0: (0.6, 0.8, 0.8, 1), 1: (1, 0.9, 0.4, 1)}
#convert rgba to integers
c1 = {k: (int(r * 255),int(g * 255),int(b * 255), a) for k, (r,g,b,a) in colors.items()}
c2 = {k: (int(r * 255),int(g * 255),int(b * 255), 0.25) for k, (r,g,b,a) in colors.items()}

#get values of first level of MulitIndex
idx = df.index.get_level_values(0)
#counter per first level for pair and unpair coloring
zipped = zip(df.groupby(idx).cumcount(), enumerate(idx))

css = [{'selector': f'.row{i}', 'props': [('background-color', f'rgba{c1[j]}')]} 
       if v % 2 == 0 
       else {'selector': f'.row{i}', 'props': [('background-color', f'rgba{c2[j]}')]} 
       for v,(i, j) in zipped]

df.style.set_table_styles(css)

Compared to @jezrael's solution, the above is even shorter.

I also like that the solution doesn't depend on index labels being ordered integers. Note that I'm using strings (at least for the second index level).

answered Jan 16, 2020 at 16:13

hyco

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1 Comment

Haikal Yeo Over a year ago

How would I extend this to a longer dataframe so that I can alternate between the two colours based on the level 0 index?

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