Python 编辑seaborn图例
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Edit seaborn legend
提问by Oliver G
Using a data frame and this code in Python, I was able to create a plot:
在 Python 中使用数据框和此代码，我能够创建一个图：
g = sns.lmplot('credibility', 'percentWatched', data=data, hue = 'millennial', markers = ["+", "."], x_jitter = True, y_jitter = True, size=5)
g.set(xlabel = 'Credibility Ranking\n ← Low High →', ylabel = 'Percent of Video Watched [%]')
However having the legend say "+ 0" and ". 1" isn't very helpful to readers. How can I edit the legend's labels? Ideally instead of saying 'millennial' it would say 'Generation' and "+ Millennial" ". Older Generations"
然而，让图例说“+ 0”和“. 1”对读者不是很有帮助。如何编辑图例的标签？理想情况下，不是说“千禧一代”，而是说“一代”和“+千禧一代”“。老一代”
采纳答案by Serenity
If legend_outis set to Truethen legend is available thought g._legendproperty and it is a part of a figure. Seaborn legend is standard matplotlib legend object. Therefore you may change legend texts like:
如果legend_out设置为，True则图例是可用的思想g._legend属性，并且它是图形的一部分。Seaborn 图例是标准的 matplotlib 图例对象。因此，您可以更改图例文本，例如：
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_outis set to False. You have to define which axes has a legend (in below example this is axis number 0):
另一种情况 iflegend_out设置为False。您必须定义哪些轴具有图例（在下面的示例中，这是轴编号 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:
此外，您可以结合这两种情况并使用以下代码：
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 Gridclass.
此代码适用于任何基于Gridclass 的seaborn 图。
回答by Glen Thompson
Took me a while to read through the above. This was the answer for me:
我花了一些时间来阅读上面的内容。这是我的答案：
import seaborn as sns
import matplotlib.pyplot as plt
tips = sns.load_dataset("tips")
g = sns.lmplot(
x="total_bill",
y="tip",
hue="smoker",
data=tips,
legend=False
)
plt.legend(title='Smoker', loc='upper left', labels=['Hell Yeh', 'Nah Bruh'])
plt.show(g)
Reference this for more arguments: matplotlib.pyplot.legend
更多参数请参考：matplotlib.pyplot.legend
回答by Batuhan Tüter
If you just want to change the legend title, you can do the following:
如果只想更改图例标题，可以执行以下操作：
import seaborn as sns
import matplotlib.pyplot as plt
tips = sns.load_dataset("tips")
g = sns.lmplot(
x="total_bill",
y="tip",
hue="smoker",
data=tips,
legend=True
)
g._legend.set_title("New Title")
