下面是我对这个问题的计算部分的看法：

• 为了比较kde，需要用相同的带宽来计算它们。（默认带宽取决于x值的个数，这两个集合的x值可以不同。）
• 两条正曲线的交点正好是它们的最小值。
• 曲线的面积可以通过梯形法则来近似：np.trapz .

下面是这些想法转换成的一些示例代码和说明情节：

import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
from scipy.stats import gaussian_kde

df = sns.load_dataset('titanic')

x0 = df.loc[df['survived'] == 0, 'fare']
x1 = df.loc[df['survived'] == 1, 'fare']

kde0 = gaussian_kde(x0, bw_method=0.3)
kde1 = gaussian_kde(x1, bw_method=0.3)

xmin = min(x0.min(), x1.min())
xmax = max(x0.max(), x1.max())
dx = 0.2 * (xmax - xmin) # add a 20% margin, as the kde is wider than the data
xmin -= dx
xmax += dx

x = np.linspace(xmin, xmax, 500)
kde0_x = kde0(x)
kde1_x = kde1(x)
inters_x = np.minimum(kde0_x, kde1_x)

plt.plot(x, kde0_x, color='b', label='No')
plt.fill_between(x, kde0_x, 0, color='b', alpha=0.2)
plt.plot(x, kde1_x, color='orange', label='Yes')
plt.fill_between(x, kde1_x, 0, color='orange', alpha=0.2)
plt.plot(x, inters_x, color='r')
plt.fill_between(x, inters_x, 0, facecolor='none', edgecolor='r', hatch='xx', label='intersection')

area_inters_x = np.trapz(inters_x, x)

handles, labels = plt.gca().get_legend_handles_labels()
labels[2] += f': {area_inters_x * 100:.1f} %'
plt.legend(handles, labels, title='Survived?')
plt.title('Fare vs Survived')
plt.tight_layout()
plt.show()