Edit seaborn legend

Question

Using a data frame and this code in Python, I was able to create a plot:

g = sns.lmplot(\'credibility\', \'percentWatched\', data=data, hue = \'millennial\',         


        

Answer 1

If legend_out is set to True then legend is available thought g._legend property and it is a part of a figure. Seaborn legend is standard matplotlib legend object. Therefore you may change legend texts like:

import seaborn as sns

tips = sns.load_dataset("tips")
g = sns.lmplot(x="total_bill", y="tip", hue="smoker",
 data=tips, markers=["o", "x"], legend_out = True)

# title
new_title = 'My title'
g._legend.set_title(new_title)
# replace labels
new_labels = ['label 1', 'label 2']
for t, l in zip(g._legend.texts, new_labels): t.set_text(l)

sns.plt.show()

Another situation if legend_out is set to False. You have to define which axes has a legend (in below example this is axis number 0):

import seaborn as sns

tips = sns.load_dataset("tips")
g = sns.lmplot(x="total_bill", y="tip", hue="smoker",
 data=tips, markers=["o", "x"], legend_out = False)

# check axes and find which is have legend
leg = g.axes.flat[0].get_legend()
new_title = 'My title'
leg.set_title(new_title)
new_labels = ['label 1', 'label 2']
for t, l in zip(leg.texts, new_labels): t.set_text(l)
sns.plt.show()

Moreover you may combine both situations and use this code:

import seaborn as sns

tips = sns.load_dataset("tips")
g = sns.lmplot(x="total_bill", y="tip", hue="smoker",
 data=tips, markers=["o", "x"], legend_out = True)

# check axes and find which is have legend
for ax in g.axes.flat:
    leg = g.axes.flat[0].get_legend()
    if not leg is None: break
# or legend may be on a figure
if leg is None: leg = g._legend

# change legend texts
new_title = 'My title'
leg.set_title(new_title)
new_labels = ['label 1', 'label 2']
for t, l in zip(leg.texts, new_labels): t.set_text(l)

sns.plt.show()

This code works for any seaborn plot which is based on Grid class.