# seaborn distplot / displot with multiple distributions

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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
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.

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
import seaborn as sns

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
import seaborn as sns

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)
``````

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
import seaborn as sns

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")

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))
``````

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
import seaborn as sns

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.

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
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)
``````

A more recent and simpler option:

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

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

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

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.

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