seaborn plot_marginals多个kdeplots

Question

我希望能够在y轴边缘绘制多个重叠的kde图（不需要x轴边距图）。每个kde图将对应于颜色类别（有4个），因此我将有4个kde，每个描绘其中一个类别的分布。这是我得到的：

 boolean markDone(long id) {
    Transaction transaction = datastore.newTransaction();
    try {
      Entity task = transaction.get(keyFactory.newKey(id));
      if (task != null) {
        transaction.put(Entity.newBuilder(task).set("done", true).build());
      }
      transaction.commit();
      return task != null;
    } finally {
      if (transaction.isActive()) {
        transaction.rollback();
      }
    }

Answer 1

要绘制每个类别的分布，我认为最好的方法是首先将数据合并到pandas数据框中。然后，您可以通过过滤数据框来遍历每个唯一类别，并使用对sns.kdeplot的调用来绘制分布。

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


x = np.array([106405611, 107148674, 107151119, 107159869, 107183396, 107229405,
              107231917, 107236097, 107239994, 107259338, 107273842, 107275873,
              107281000, 107287770, 106452671, 106471246, 106478110, 106494135,
              106518400, 106539079])

y = np.array([9.09803208,   5.357552  ,   8.98868469,   6.84549005,
              8.17990909,  10.60640521,   9.89935692,   9.24079133,
              8.97441459,   9.09803208,  10.63753055,  11.82336724,
              7.93663794,   8.74819285,   8.07146236,   9.82336724,
              8.4429435 ,  10.53332973,   8.23361968,  10.30035256])

col = np.array([2, 4, 4, 1, 3, 4, 3, 3, 4, 1, 4, 3, 2, 4, 1, 1, 2, 2, 3, 1])

# Combine data into DataFrame
df = pd.DataFrame({'V': x, 'Distance': y, 'col': col})

# Define colormap and create corresponding color palette
cmap = sns.diverging_palette(20, 220, as_cmap=True)
colors = sns.diverging_palette(20, 220, n=4)

# Plot data onto seaborn JointGrid
g = sns.JointGrid('V', 'Distance', data=df, ratio=2)
g = g.plot_joint(plt.scatter, c=df['col'], edgecolor="black", cmap=cmap)

# Loop through unique categories and plot individual kdes
for c in df['col'].unique():
    sns.kdeplot(df['Distance'][df['col']==c], ax=g.ax_marg_y, vertical=True,
                color=colors[c-1], shade=True)
    sns.kdeplot(df['V'][df['col']==c], ax=g.ax_marg_x, vertical=False,
                color=colors[c-1], shade=True)

在我看来，这是一个比我原来的答案更好，更清晰的解决方案，我不必要地重新定义seaborn kdeplot，因为我没想过这样做。感谢mwaskom指出这一点。另请注意，已发布的解决方案中将删除图例标签，并使用

完成

g.ax_marg_x.legend_.remove()
g.ax_marg_y.legend_.remove()