Matplotlib 如何在seaborn中绘制连接多个小提琴分布中间点的线？

我努力在每种叠加分布的中值（以及最小值和最大值）之间绘制一条点线。 我试着将小提琴图与seaborn.lineplot叠加，但失败了。我不确定用这种方法，我可以画点线，也可以链接同一类型分布的最小值和最大值。我还尝试使用seaborn.lineplot，但这里的挑战是在每个x轴值上绘制分布的最小值和最大值

下面是seaborn中小提琴情节的示例数据集和代码

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

x=[0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.2,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.4,0.6,0.6,0.6,0.6,0.6,0.6,0.6,0.6,0.6,0.6,0.6,0.6,0.8,0.8,0.8,0.8,0.8,0.8,0.8,0.8,0.8,0.8,0.8,0.8]
cate=['a','a','a','a','b','b','b','b','c','c','c','c','a','a','a','a','b','b','b','b','c','c','c','c','a','a','a','a','b','b','b','b','c','c','c','c','a','a','a','a','b','b','b','b','c','c','c','c']
y=[1.1,1.12,1.13,1.13,3.1,3.12,3.13,3.13,5.1,5.12,5.13,5.13,2.2,2.22,2.25,2.23,4.2,4.22,4.25,4.23,6.2,6.22,6.25,6.23,2.2,2.22,2.24,2.23,4.2,4.22,4.24,4.23,6.2,6.22,6.24,6.23,1.1,1.13,1.14,1.12,3.1,3.13,3.14,3.12,5.1,5.13,5.14,5.12]

my_pal =['red','green', 'purple']

df = pd.DataFrame({'x': x, 'Type': cate, 'y': y})
ax=sns.catplot(y='y', x='x',data=df, hue='Type', palette=my_pal,  kind="violin",dodge =False) 
sns.lineplot(y='y', x='x',data=df, hue='Type', palette=my_pal,  ci=100,legend=False)
plt.show()

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但它只在绘图左侧的缩小部分绘制线条。是否有技巧将线图与小提琴图重叠？

对于线图，“x”被视为数字。然而，对于小提琴图，“x”被认为是分类的（定位在0，1，2，…）

一个解决方案是将“x”转换为字符串，使两个图都认为它是分类的。< /P>

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

my_pal = ['red', 'green', 'purple']
N = 40
df = pd.DataFrame({'x': np.random.randint(1, 6, N*3) * 0.2,
                   'y': np.random.uniform(0, 1, N*3) + np.tile([2, 4, 6], N),
                   'Type': np.tile(list('abc'), N)})
df['x'] = [f'{x:.1f}' for x in df['x']]
ax = sns.violinplot(y='y', x='x', data=df, hue='Type', palette=my_pal, dodge=False)
ax = sns.lineplot(y='y', x='x', data=df, hue='Type', palette=my_pal, ci=100, legend=False, ax=ax)
ax.margins(0.15)  # slightly more padding for x and y axis
ax.legend(bbox_to_anchor=(1.01, 1), loc='upper left')
plt.tight_layout()
plt.show()