如何将 pipe pandas 分组到 seaborn 分布图？

Question

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.

Answer 1

If you also want labels, you can do following:

iris = sns.load_dataset('iris')
iris.groupby('species')['sepal_length'].apply(lambda x: sns.distplot(x,
        hist=False, rug=False,label = x.name));
plt.xlabel('sepal_length')
plt.ylabel('kde')

Answer 2

Not quite pipe , but you can use apply :

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

Output (which is the same):