How to plot percentage with seaborn distplot / histplot / displot

Question

Is there a way to plot the percentage instead of the count on a distplot?

ax = sns.FacetGrid(telcom, hue='Churn', palette=["teal", "crimson"], size=5, aspect=1)
ax = ax.map(sns.distplot, "tenure",  hist=True, kde=False)
ax.fig.suptitle('Tenure distribution in customer churn', y=1, fontsize=16, fontweight='bold');
plt.legend();

Answer 1

• As of seaborn 0.11.2
  • seaborn.distplot is replaced with the Figure level seaborn.displot and Axes level seaborn.histplot , which have a stat parameter. Use stat='percent' .
• For both types of plots, experiment with common_bins and common_norm .
  • For example, common_norm=True will show the percent as a part of the entire population, whereas False will show the percent relative to the group.
• The implementation shown in this answer shows how to add annotations.

import seaborn as sns
import matplotlib.pyplot as ply

# data
data = sns.load_dataset('titanic')

Figure Level

p = sns.displot(data=data, x='age', stat='percent', hue='sex', height=3)
plt.show()

p = sns.displot(data=data, x='age', stat='percent', col='sex', height=3)
plt.show()

• Type annotations ( := ) used in labels requires python >= 3.8 . This can be implemented with a for-loop , without using := .

fg = sns.displot(data=data, x='age', stat='percent', col='sex', height=3.5, aspect=1.25)

for ax in fg.axes.ravel():
    
    # add annotations
    for c in ax.containers:

        # custom label calculates percent and add an empty string so 0 value bars don't have a number
        labels = [f'{w:0.1f}%' if (w := v.get_height()) > 0 else '' for v in c]

        ax.bar_label(c, labels=labels, label_type='edge', fontsize=8, rotation=90, padding=2)
    
    ax.margins(y=0.2)

plt.show()

Axes Level

fig = plt.figure(figsize=(4, 3))
p = sns.histplot(data=data, x='age', stat='percent', hue='sex')
plt.show()

Answer 2

You could use norm_hist = True .

From the documentation :

norm_hist : bool, optional

If True, the histogram height shows a density rather than a count. This is implied if a KDE or fitted density is plotted.

Answer 3

You could chose a barplot, and set an estimator defining the normalization in percentages:

import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt

df = pd.DataFrame(dict(x=np.random.poisson(10, 1_000)))
ax = sns.barplot(x="x",
                 y="x",
                 data=df,
                 palette=["teal", "crimson"],
                 estimator=lambda x: len(x) / len(df) * 100
                 )
ax.set(xlabel="tenure")
ax.set(ylabel="Percent")

plt.show()

Giving: