在单个图中，所有列的箱线图由“标签"分开.柱子 [英] In a single figure, boxplot of all columns split by a "label" column

问题描述

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除了我希望它被包含特定标签的列分隔，类似于在以下示例中通过 hue 参数实现的功能:

  ax = sns.boxplot(x ="day"，y ="total_bill"，hue ="smoker"，data = tips，调色板="Set3") 

此示例仅在定义 x 和 y 时有效，但是对于我想要实现的目标，我希望 x 本质上是每一列在我的数据框中(显然不包括标签)，以 y 作为频率，类似于第一个示例中所示.如果无法通过seaborn实现，我愿意尝试其他一些针对python的可视化库.

解决方案

您需要拆栈"或融化"数据，使所有内容都是值，而不是变量(长格式而不是宽格式).

这是下面的样子:

  iris_xtab = seaborn.load_dataset("iris")iris_long = pandas.melt(iris，id_vars ='species')seaborn.boxplot(x ='species'，y ='value'，hue ='variable'，data = iris_long) 

或者将种类值忽略为x(您必须按照前面建议的那样分配一个虚拟值

  ax = seaborn.boxplot(x ='pos'，y ='value'，hue ='variable'，数据= iris_long.assign(pos = 1)) 

Looking at the boxplot API page, I want something that looks like a combination of this:

>>> iris = sns.load_dataset("iris")
>>> ax = sns.boxplot(data=iris, orient="h", palette="Set2")

Except I want it split by a column that holds a certain label, similarly to what is achieved by the hue argument in an example that follows:

ax = sns.boxplot(x="day", y="total_bill", hue="smoker", data=tips, palette="Set3")

This example only works when x and y are defined, but for what I want to achieve I want x to essentially be every column in my dataframe (except the label, obviously) and y to be the frequency, similarly to what is shown in the first example. If not possible to achieve with seaborn, I am willing to try some other visualization library for python.

解决方案

You need to "unstack" or "melt" the data such everything is value, not a variable (long format instead of wide format).

Here's what that looks like:

iris_xtab = seaborn.load_dataset("iris")
iris_long = pandas.melt(iris, id_vars='species')
seaborn.boxplot(x='species', y='value', hue='variable', data=iris_long)

Or leaving out the species value as x (you have to assign a dummy value as suggested earlier

ax = seaborn.boxplot(x='pos', y='value', hue='variable', 
                     data=iris_long.assign(pos=1))

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