seaborn distplot / displot with multiple distributions

Question

I am using seaborn to plot a distribution plot. I would like to plot multiple distributions on the same plot in different colors:

Here's how I start the distribution plot:

import numpy as np
import pandas as pd
from sklearn.datasets import load_iris
iris = load_iris()
iris = pd.DataFrame(data= np.c_[iris['data'], iris['target']],columns= iris['feature_names'] + ['target'])

   sepal length (cm)  sepal width (cm)  petal length (cm)  petal width (cm)  target
0                5.1               3.5                1.4               0.2     0.0
1                4.9               3.0                1.4               0.2     0.0
2                4.7               3.2                1.3               0.2     0.0
3                4.6               3.1                1.5               0.2     0.0
4                5.0               3.6                1.4               0.2     0.0

sns.distplot(iris[['sepal length (cm)']], hist=False, rug=True);

The 'target' column contains 3 values: 0, 1, 2.

I would like to see one distribution plot for sepal length, where target ==0, target ==1, and target ==2, for a total of 3 plots.

From [`seaborn v0.11.0`](https://seaborn.pydata.org/whatsnew.html#v0-11-0-september-2020), see newer answers using [`sns.displot`](https://seaborn.pydata.org/generated/seaborn.displot.html), which replaces [`sns.distplot`](https://seaborn.pydata.org/generated/seaborn.distplot.html) — Trenton McKinney, Aug 30 '21 at 17:52

score 55 · Accepted Answer · edited Aug 30 '21 at 16:52

The important thing is to sort the dataframe by values where target is 0, 1, or 2.

import numpy as np
import pandas as pd
from sklearn.datasets import load_iris
import seaborn as sns

iris = load_iris()
iris = pd.DataFrame(data=np.c_[iris['data'], iris['target']],
                    columns=iris['feature_names'] + ['target'])

# Sort the dataframe by target
target_0 = iris.loc[iris['target'] == 0]
target_1 = iris.loc[iris['target'] == 1]
target_2 = iris.loc[iris['target'] == 2]

sns.distplot(target_0[['sepal length (cm)']], hist=False, rug=True)
sns.distplot(target_1[['sepal length (cm)']], hist=False, rug=True)
sns.distplot(target_2[['sepal length (cm)']], hist=False, rug=True)

plt.show()

The output looks like:

If you don't know how many values target may have, find the unique values in the target column, then slice the dataframe and add to the plot appropriately.

import numpy as np
import pandas as pd
from sklearn.datasets import load_iris
import seaborn as sns

iris = load_iris()
iris = pd.DataFrame(data=np.c_[iris['data'], iris['target']],
                    columns=iris['feature_names'] + ['target'])

unique_vals = iris['target'].unique()  # [0, 1, 2]

# Sort the dataframe by target
# Use a list comprehension to create list of sliced dataframes
targets = [iris.loc[iris['target'] == val] for val in unique_vals]

# Iterate through list and plot the sliced dataframe
for target in targets:
    sns.distplot(target[['sepal length (cm)']], hist=False, rug=True)

This creates separate plots for me (in 2021), maybe it has changed since — Ricardo Guerreiro, Sep 15 '21 at 09:43

score 50 · Answer 2 · answered Feb 10 '18 at 21:43

A more common approach for this type of problems is to recast your data into long format using melt, and then let map do the rest.

import numpy as np
import pandas as pd
from sklearn.datasets import load_iris
import seaborn as sns

iris = load_iris()
iris = pd.DataFrame(data=np.c_[iris['data'], iris['target']], 
                    columns=iris['feature_names'] + ['target'])

# recast into long format 
df = iris.melt(['target'], var_name='cols',  value_name='vals')

df.head()

   target               cols  vals
0     0.0  sepal length (cm)   5.1
1     0.0  sepal length (cm)   4.9
2     0.0  sepal length (cm)   4.7
3     0.0  sepal length (cm)   4.6
4     0.0  sepal length (cm)   5.0

You can now plot simply by creating a FacetGrid and using map:

g = sns.FacetGrid(df, col='cols', hue="target", palette="Set1")
g = (g.map(sns.distplot, "vals", hist=False, rug=True))

Amit Amola · Answer 3 · 2021-10-04T08:48:34.593

Anyone trying to build the same plot using the new 0.11.0 version, Seaborn has or is deprecating distplot and replacing it with displot.

So the new version wise the code would be:

import numpy as np
import pandas as pd
from sklearn.datasets import load_iris
import seaborn as sns

iris = load_iris()
iris = pd.DataFrame(data=np.c_[iris['data'], iris['target']],
                    columns=iris['feature_names'] + ['target'])

sns.displot(data=iris, x='sepal length (cm)', hue='target', kind='kde', fill=True, palette=sns.color_palette('bright')[:3], height=5, aspect=1.5)

Edit

As asked by Raghav in the comment section, can we change the labels in the chart without changing the dataframe itself. Yes we absolutely can. So we start by assigning the plot to a variable called chart and then do the following:

chart = sns.displot(data=iris, x='sepal length (cm)', hue='target', kind='kde', fill=True, palette=sns.color_palette('bright')[:3], height=5, aspect=1.5)

## Changing title
new_title = 'This is a NEW title'
chart._legend.set_title(new_title)

# Replacing labels
new_labels = ['label 1', 'label 2', 'label 3']
for t, l in zip(chart._legend.texts, new_labels):
    t.set_text(l)

And the final chart looks like as below:

Hope this helps Raghav.

Is there a way to change the values of legend from 0,1,2, to some other value without changing the name in the dataframe itself — Raghav Arora, Oct 04 '21 at 03:27
Oh it is surely possible, let me answer that in the answer itself. — Amit Amola, Oct 04 '21 at 08:42

score 9 · Answer 4 · answered Apr 19 '20 at 23:17

I have found a simpler solution using FacetGrid on https://github.com/mwaskom/seaborn/issues/861 by citynorman:

import numpy as np
import pandas as pd
from sklearn.datasets import load_iris
iris = load_iris()
iris = pd.DataFrame(data= np.c_[iris['data'], iris['target']],columns= iris['feature_names'] + ['target'])

g = sns.FacetGrid(iris, hue="target")
g = g.map(sns.distplot, "sepal length (cm)",  hist=False, rug=True)

score 9 · Answer 5 · answered Sep 23 '20 at 12:10

9

A more recent and simpler option:

sns.displot(data=iris, x='sepal length (cm)', hue='target', kind='kde')

answered Sep 23 '20 at 12:10

Rafael Toledo

974
13
19

score 2 · Answer 6 · edited Aug 30 '21 at 17:48

If anyone is looking to get a facetgrid of sns.distplot, its been replaced with a figure-level option, sns.displot, and an axes-level function, sns.histplot

This makes it quite easy to convert the data from a wide format (as shown in the OP) to long format, by using pandas.DataFrame.melt

import pandas as pd
import seaborn as sns

iris = sns.load_dataset('iris')

# convert the dataframe from wide to long form
iris_melt = iris.melt(id_vars='species')

iris_melt.head()

  species      variable  value
0  setosa  sepal_length    5.1
1  setosa  sepal_length    4.9
2  setosa  sepal_length    4.7
3  setosa  sepal_length    4.6
4  setosa  sepal_length    5.0

sns.displot(
    data=iris_melt, 
    x='value', 
    hue='species', 
    kind='kde', 
    fill=True,
    col='variable'
)

The image is small here, but if you right click on the image and open it in a new tab or window, you can see the details better.

seaborn distplot / displot with multiple distributions

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