count plot with stacked bars per hue

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You were basically there with your last part, using DataFrame.plot() with bar and stacked=True.

Instead of your aggregate function, you can accomplish what you want with a groupby + pivot.

df_plot = df.groupby(['class', 'survived']).size().reset_index().pivot(columns='class', index='survived', values=0)

class     First  Second  Third
survived                      
0            80      97    372
1           136      87    119

From here you can just plot it as a bar with the stacked=True argument

df_plot.plot(kind='bar', stacked=True)

enter image description here

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Marcello
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Marcello

I am system manager in the HW development team in Ericsson R&D Italy. Since 2017 I am learning Python for data analytics, dusting off my machine learning skills built during my university years... quite a lot of time ago. I am interested in supporting tech business development and data analytics competence build up in my local area, with some non-profiti initiatives: DataScienceSeed meetup TalenTour initiative

Updated on June 01, 2020

Comments

  • Marcello
    Marcello about 4 years

    I am looking for an efficient way of drawing a count plot with stacked bars according to "hue". Standard hue behavior is to split a count into parallel bars according to the value of a second column, what I am looking for is an efficient way to have the hue bars stacked in order to quickly compare totals.

    Let me explain with an example from the titanic dataset:

    import pandas as pd
    import numpy as np
    import seaborn as sns
    %matplotlib inline
    
    df = sns.load_dataset('titanic')
    sns.countplot(x='survived',hue='class',data=df)
    

    gives standard Seaborn behavior with countplot and hue Standard Seaborn behavior with countplot and hue

    what I am looking for is something like stacked bars per hue Stacked bars per hue

    to get the last image I used the following code

    def aggregate(rows,columns,df):
        column_keys = df[columns].unique()
        row_keys = df[rows].unique()
    
        agg = { key : [ len(df[(df[rows]==value) & (df[columns]==key)]) for value in row_keys]
                   for key in column_keys }
    
        aggdf = pd.DataFrame(agg,index = row_keys)
        aggdf.index.rename(rows,inplace=True)
    
        return aggdf
    
    aggregate('survived','class',df).plot(kind='bar',stacked=True)
    

    I am sure there is some more efficient way. I know seaborn is not very stacked bars friendly... so I tried to rearrange the dataset with my function and used matplotlib, but I guess there is a more clever way to do that as well.

    Thank you very much!

  • Wojciech Moszczyński
    Wojciech Moszczyński about 5 years
    doesn't work in 2019. You need use: df_plot.plot.bar(stacked=True)
  • ALollz
    ALollz about 5 years
    @WojciechMoszczyński. Thanks, it works with removing the x= argument too, which was causing the issue.
  • Sarvesh Dubey
    Sarvesh Dubey over 3 years
    @ALollz @Marcello With seaborn 0.11, you can achieve the same in much simple way: sns.displot(df, x='survived', hue='class', multiple='stack')
  • Jorge
    Jorge almost 2 years
    An improved version of this answer is simply pd.crosstab(df['survived'], df['class']).plot(kind='bar', stacked=True)