How to pipe pandas groupby parts to seaborn distplot?

user_5200329 0
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I was learning using .pipe method in pandas and wondering if we can use it to plot the distplot for each group of groupby.

MWE


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

# data
np.random.seed(100)
data = {'year': np.random.choice([2016, 2018, 2020], size=400),
        'item': np.random.choice(['Apple', 'Banana', 'Carrot'], size=400),
        'price': np.random.random(size=400)}

df = pd.DataFrame(data)

# distplots
for year in df['year'].unique():
    x = df['price'][df['year'] == year]
    sns.distplot(x, hist=False, rug=True)

Question

Can we get the same plot using pandas groupby and without using for loop?

My attempt:

df.groupby('year').pipe(lambda dfx: sns.distplot(dfx['price']))
# TypeError: cannot convert the series to <class 'float'>

# df[['year','price']].groupby('year').pipe(sns.distplot)
# TypeError: float() argument must be a string or a number, not 'DataFrame'

Required output

Same output as for-loop but using pandas pipe. enter image description here

pandas seaborn python
2 Answers

If you also want labels, you can do following:

import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt

iris = sns.load_dataset('iris')

#iris.groupby('species')['sepal_length'].apply(lambda x:sns.distplot(x,hist=False, rug=False,label = x.name))

iris.groupby('species')['sepal_length']\
    .apply(lambda x: sns.kdeplot(x, label = x.name))

plt.xlabel('sepal_length')
plt.ylabel('kde')
plt.show()

enter image description here

user_52003290

Not quite pipe, but you can use apply:

df.groupby('year')['price'].apply(sns.distplot, hist=False, rug=True);

Output (which is the same):

enter image description here

user_42384080
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