频率和百分比不均匀组 sns barplot

Question

我试图在 sns 条形图中按组显示相对百分比以及总频率。我比较的两组在大小上有很大不同，这就是为什么我在下面的函数中按组显示百分比。

这是我创建的示例数据框的语法，它在目标分类变量 ('item') 中具有与我的数据 ('groups') 相似的相对组大小。 'rand' 只是我用来制作 df 的一个变量。

# import pandas and seaborn
import pandas as pd
import seaborn as sns
import numpy as np

# create dataframe
foobar = pd.DataFrame(np.random.randn(100, 3), columns=('groups', 'item', 'rand'))

# get relative groupsizes
for row, val in enumerate(foobar.rand) :
    if  val > -1.2 :
        foobar.loc[row, 'groups'] = 'A'
    else: 
        foobar.loc[row, 'groups'] = 'B'

    # assign categories that I am comparing graphically
    if row < 20:
        foobar.loc[row, 'item'] = 'Z'
    elif row < 40:
        foobar.loc[row, 'item'] = 'Y'
    elif row < 60:
        foobar.loc[row, 'item'] = 'X'
    elif row < 80:
        foobar.loc[row, 'item'] = 'W'
    else:
        foobar.loc[row, 'item'] = 'V'

这是我写的函数，它按组比较相对频率。它有一些默认变量，但我已经为这个问题重新分配了它们。

def percent_categorical(item, df=IA, grouper='Active Status') :
    # plot categorical responses to an item ('column name')
    # by percent by group ('diff column name w categorical data')
    # select a data frame (default is IA)
    # 'Active Status' is default grouper

    # create df of item grouped by status
    grouped = (df.groupby(grouper)[item]
    # convert to percentage by group rather than total count
                .value_counts(normalize=True)
                # rename column 
                .rename('percentage')
                # multiple by 100 for easier interpretation
                .mul(100)
                # change order from value to name
                .reset_index()
            .sort_values(item))

    # create plot
    PercPlot = sns.barplot(x=item,
                         y='percentage',
                         hue=grouper,
                         data=grouped,
                         palette='RdBu'
                         ).set_xticklabels(
                                 labels = grouped[item
                                      ].value_counts().index.tolist(), rotation=90)
    #show plot
    return PercPlot

函数和结果图如下：

percent_categorical('item', df=foobar, grouper='groups')

这很好，因为它允许我按组显示相对百分比。但是，我还想显示每个组的绝对数字，最好在图例中显示。在这种情况下，我希望它显示 A 组共有 89 名成员，B 组共有 11 名成员。

提前感谢您的帮助。

Answer 1

我通过拆分 groupby 操作解决了这个问题：一个用于获取百分比，一个用于计算对象的数量。

我调整了你的percent_catergorical功能如下：

def percent_categorical(item, df=IA, grouper='Active Status') :
    # plot categorical responses to an item ('column name')
    # by percent by group ('diff column name w categorical data')
    # select a data frame (default is IA)
    # 'Active Status' is default grouper

    # create groupby of item grouped by status
    groupbase = df.groupby(grouper)[item]
    # count the number of occurences
    groupcount = groupbase.count()       
    # convert to percentage by group rather than total count           
    groupper = (groupbase.value_counts(normalize=True)
                # rename column 
                .rename('percentage')
                # multiple by 100 for easier interpretation
                .mul(100)
                # change order from value to name
                .reset_index()
                .sort_values(item))

    # create plot
    fig, ax = plt.subplots()
    brplt = sns.barplot(x=item,
                         y='percentage',
                         hue=groupper,
                         data=groupper,
                         palette='RdBu',
                         ax=ax).set_xticklabels(
                                 labels = grouper[item
                                      ].value_counts().index.tolist(), rotation=90)
    # get the handles and the labels of the legend
    # these are the bars and the corresponding text in the legend
    thehandles, thelabels = ax.get_legend_handles_labels()
    # for each label, add the total number of occurences
    # you can get this from groupcount as the labels in the figure have
    # the same name as in the values in column of your df
    for counter, label in enumerate(thelabels):
        # the new label looks like this (dummy name and value)
        # 'XYZ (42)'
        thelabels[counter] = label + ' ({})'.format(groupcount[label])
    # add the new legend to the figure
    ax.legend(thehandles, thelabels)
    #show plot
    return fig, ax, brplt

得到你的身材：

fig, ax, brplt = percent_categorical('item', df=foobar, grouper='groups')

结果图如下所示：

您可以随意更改此图例的外观，我只是添加括号作为示例。

频率和百分比不均匀组 sns barplot

frequency and percentage uneven groups sns barplot

python

matplotlib

python-3.x

pandas

seaborn