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python - 频率和百分比不均匀组 sns barplot

转载 作者:太空狗 更新时间:2023-10-30 01:12:09 26 4
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我正在尝试按组显示相对百分比以及 sns 条形图中的总频率。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(“项目”)中具有与我的数据(“组”)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'

这是我编写的按组比较相对频率的函数。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')

result of running my function

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。

最佳答案

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 函数如下:

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

得到你的数字:

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

结果图如下所示:

the output

您可以根据需要更改此图例的外观,我只是添加括号作为示例。

关于python - 频率和百分比不均匀组 sns barplot,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44763643/

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