Python：基于 MultiIndex 的彩色熊猫数据框

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.

Answer 1

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)

Answer 2

@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).