seaborn 中的一半（不是分割！）小提琴图

Question

目前，seaborn 通过根据 split=True 变量设置

hue

来提供

 分割小提琴图的功能。  我想制作一个“半”小提琴图，即省略每把小提琴的一半的图。这样的图描绘了类似于每个连续变量的 pdf 的内容，仅绘制在每个分类变量的每条垂直线的一侧。 

我已经设法欺骗

seaborn

，用绘制的值范围之外的额外数据点和额外的虚拟色调来绘制此图，但我想知道这是否可以在不实际更改数据集的情况下完成，例如在 

sns.violinplot()

 参数内。

例如这张图：

由以下代码片段创建：

# imports
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt

# load dataset from seaborn
datalist = sns.get_dataset_names()
dataset_name = 'iris'
if dataset_name in datalist:
    df = sns.load_dataset(dataset_name)
else:
    print("Dataset with name: " + dataset_name + " was not found in the available datasets online by seaborn.")

# prepare data
df2 = df.append([-999,-999,-999,-999,'setosa'])
df2['huecol'] = 0.0
df2['huecol'].iloc[-1]= -999

# plot
fig = plt.figure(figsize=(6,6))
sns.violinplot(x='species',y="sepal_width",
            split=True, hue ='huecol', inner = 'quartile',
            palette="pastel", data=df2, legend=False)
plt.title('iris')

# remove hue legend
leg = plt.gca().legend()
leg.remove()
plt.ylim([1,5.0])
plt.show()

Answer 1

无需修改数据：

ax = sns.violinplot(
    data=tips,
    x="day", y="total_bill", hue=True,
    hue_order=[True, False], split=True,
)
ax.legend_ = None

Answer 2

我一直在寻找与此类似的解决方案，但没有找到令人满意的解决方案。我最终多次调用

seaborn.kdeplot，因为 violinplot 本质上是一个单侧核密度图。

示例

categorical_kde_plot

的函数定义如下

categorical_kde_plot(
    df,
    variable="tip",
    category="day",
    category_order=["Thur", "Fri", "Sat", "Sun"],
    horizontal=False,
)

使用

horizontal=True

，输出将如下所示：

代码

import seaborn as sns
from matplotlib import pyplot as plt


def categorical_kde_plot(
    df,
    variable,
    category,
    category_order=None,
    horizontal=False,
    rug=True,
    figsize=None,
):
    """Draw a categorical KDE plot

    Parameters
    ----------
    df: pd.DataFrame
        The data to plot
    variable: str
        The column in the `df` to plot (continuous variable)
    category: str
        The column in the `df` to use for grouping (categorical variable)
    horizontal: bool
        If True, draw density plots horizontally. Otherwise, draw them
        vertically.
    rug: bool
        If True, add also a sns.rugplot.
    figsize: tuple or None
        If None, use default figsize of (7, 1*len(categories))
        If tuple, use that figsize. Given to plt.subplots as an argument.
    """
    if category_order is None:
        categories = list(df[category].unique())
    else:
        categories = category_order[:]

    figsize = (7, 1.0 * len(categories))

    fig, axes = plt.subplots(
        nrows=len(categories) if horizontal else 1,
        ncols=1 if horizontal else len(categories),
        figsize=figsize[::-1] if not horizontal else figsize,
        sharex=horizontal,
        sharey=not horizontal,
    )

    for i, (cat, ax) in enumerate(zip(categories, axes)):
        sns.kdeplot(
            data=df[df[category] == cat],
            x=variable if horizontal else None,
            y=None if horizontal else variable,
            # kde kwargs
            bw_adjust=0.5,
            clip_on=False,
            fill=True,
            alpha=1,
            linewidth=1.5,
            ax=ax,
            color="lightslategray",
        )

        keep_variable_axis = (i == len(fig.axes) - 1) if horizontal else (i == 0)

        if rug:
            sns.rugplot(
                data=df[df[category] == cat],
                x=variable if horizontal else None,
                y=None if horizontal else variable,
                ax=ax,
                color="black",
                height=0.025 if keep_variable_axis else 0.04,
            )

        _format_axis(
            ax,
            cat,
            horizontal,
            keep_variable_axis=keep_variable_axis,
        )

    plt.tight_layout()
    plt.show()


def _format_axis(ax, category, horizontal=False, keep_variable_axis=True):

    # Remove the axis lines
    ax.spines["top"].set_visible(False)
    ax.spines["right"].set_visible(False)

    if horizontal:
        ax.set_ylabel(None)
        lim = ax.get_ylim()
        ax.set_yticks([(lim[0] + lim[1]) / 2])
        ax.set_yticklabels([category])
        if not keep_variable_axis:
            ax.get_xaxis().set_visible(False)
            ax.spines["bottom"].set_visible(False)
    else:
        ax.set_xlabel(None)
        lim = ax.get_xlim()
        ax.set_xticks([(lim[0] + lim[1]) / 2])
        ax.set_xticklabels([category])
        if not keep_variable_axis:
            ax.get_yaxis().set_visible(False)
            ax.spines["left"].set_visible(False)


if __name__ == "__main__":
    df = sns.load_dataset("tips")

    categorical_kde_plot(
        df,
        variable="tip",
        category="day",
        category_order=["Thur", "Fri", "Sat", "Sun"],
        horizontal=True,
    )

Answer 3

答案很简单，不，Seaborn 不可能不欺骗它认为存在

hue

 存在。 

这个答案展示了如何在matplotlib中做到这一点，原则上同样可以应用于seaborn小提琴图，即切掉一半的小提琴路径。

Answer 4

正如@

newbielp在最佳答案下的评论中提到的：

hue_order

 不适用于色调作为布尔参数。

如果您也有色调，但数据未配对（因此在一些全小提琴中会有半小提琴），请考虑使用您的色调类别而不是布尔参数。

    ax = sns.violinplot(
        data=df,
        x="day",
        y="total_bill",
        hue="sex",                    # when you have hue
        split=True,
        hue_order=["Male", "Female"], # set hue_order with cat instead of bool
        inner='quartile'
    )

生成上图的完整代码：

import seaborn as sns
import matplotlib.pyplot as plt

if __name__ == "__main__":
    df = sns.load_dataset("tips")
    # manually remove tip from man on Friday
    df = df.drop(df[(df["sex"] == "Male") & (df["day"] == "Fri")].index)
    print(df)
    ax = sns.violinplot(
        data=df,
        x="day",
        y="total_bill",
        hue="sex",
        split=True,
        scale_hue=False,
        saturation=0.75,
        hue_order=["Male", "Female"],
        dodge=True,
        inner='quartile'
    )
    plt.show(block=True)