How do I add a title to this Seaborne plot? Let's give it a title 'I AM A TITLE'.
tips = sns.load_dataset("tips")
g = sns.FacetGrid(tips, col="sex", row="smoker", margin_titles=True)
g.map(sns.plt.scatter, "total_bill", "tip")
If you create the FacetGrid directly, as in the original example, it automatically adds column and row labels instead of individual subplot titles. We can still add a title to the whole thing: from matplotlib. pyplot import scatter as plt_scatter g = sns.
To show the title for the diagram for Seaborn pairplot(), we can use pp. fig. suptitle() method.
To change the position of a legend in a seaborn plot, you can use the plt. legend() command. The default location is “best” – which is where Matplotlib automatically finds a location for the legend based on where it avoids covering any data points.
Updating slightly, with seaborn 0.11.1:
Seaborn's relplot
function creates a FacetGrid and gives each subplot its own explanatory title. You can add a title over the whole thing:
import seaborn as sns
tips = sns.load_dataset('tips')
rp = sns.relplot(data=tips, x='total_bill', y='tip',
col='sex', row='smoker',
kind='scatter')
# rp is a FacetGrid;
# relplot is a nice organized way to use it
rp.fig.subplots_adjust(top=0.9) # adjust the Figure in rp
rp.fig.suptitle('ONE TITLE FOR ALL')
If you create the FacetGrid directly, as in the original example, it automatically adds column and row labels instead of individual subplot titles. We can still add a title to the whole thing:
from matplotlib.pyplot import scatter as plt_scatter
g = sns.FacetGrid(tips, col='sex', row='smoker',
margin_titles=True)
g.map(plt_scatter, 'total_bill', 'tip')
g.fig.subplots_adjust(top=0.9)
g.fig.suptitle('TITLE!')
The FacetGrid objects are built with matplotlib Figure objects, so we can use subplots_adjust
, suptitle
that may be familiar from matplotlib in general.
g.fig.subplots_adjust(top=0.9)
g.fig.suptitle('Title', fontsize=16)
More info here: http://matplotlib.org/api/figure_api.html
In ipython notebook, this worked for me!
sns.plt.title('YOUR TITLE HERE')
What worked for me was:
sns.plt.suptitle('YOUR TITLE HERE')
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