blocks|key|3185953|text|从matplotlib+v1.4.0rc4开始，图例对象中添加了一个remove方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3185954|用法：|3185955|ax.get_legend().remove()|code-block|syntax|javascript|3185956|或|3185957|legend+=+ax.legend(...)
...
legend.remove()|3185958|请参阅here，了解引入此功能的提交。|3185959|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/cimarronm/matplotlib/releases/tag/v1.4.0rc4|1|https://github.com/matplotlib/matplotlib/pull/2798^0|C|9|Y|6|1|K|0|0|0|0|0|0|3|4|1|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|6|7|1A|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|1B|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1C|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1D|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|S|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]|10|$5|V|W|X|E|$Y|11]]]]

As of <a href="https://github.com/cimarronm/matplotlib/releases/tag/v1.4.0rc4" rel="noreferrer">matplotlib <code>v1.4.0rc4</code></a>, a <code>remove</code> method has been added to the legend object.

Usage:

<pre><code>ax.get_legend().remove()
</code></pre>

or

<pre><code>legend = ax.legend(...)
...
legend.remove()
</code></pre>

See <a href="https://github.com/matplotlib/matplotlib/pull/2798" rel="noreferrer">here</a> for the commit where this was introduced.

blocks|key|2972249|text|如果要绘制Pandas数据帧并删除图例，请将legend=None作为参数添加到plot命令。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2972250|import+pandas+as+pd
import+numpy+as+np
import+matplotlib.pyplot+as+plt

df2+=+pd.DataFrame(np.random.randn(10,+5))
df2.plot(legend=None)
plt.show()|code-block|syntax|javascript|2972251|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to plot a Pandas dataframe and want to remove the legend, add legend=None as parameter to the plot command.

<pre><code>import pandas as pd
import numpy as np
import matplotlib.pyplot as plt

df2 = pd.DataFrame(np.random.randn(10, 5))
df2.plot(legend=None)
plt.show()
</code></pre>

blocks|key|273038|text|您可以使用图例的set_visible方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|273039|ax.legend().set_visible(False)
draw()|code-block|syntax|javascript|273040|这是基于一段时间前我在here上遇到的一个类似问题的回答|273041|(感谢你的回答，Jouni+-对不起，我无法将问题标记为已回答...也许有权限的人可以帮我做这件事？)|273042|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1349202/show-hide-a-plots-legend^0|8|B|0|0|B|4|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

You could use the legend's <code>set_visible</code> method:

<pre><code>ax.legend().set_visible(False)
draw()
</code></pre>

This is based on a answer provided to me in response to a similar question I had some time ago <a href="https://stackoverflow.com/questions/1349202/show-hide-a-plots-legend">here</a>

(Thanks for that answer Jouni - I'm sorry I was unable to mark the question as answered... perhaps someone who has the authority can do so for me?)

blocks|key|269452|text|您必须添加以下代码行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|269453|ax+=+gca()
ax.legend_+=+None
draw()|code-block|syntax|javascript|269454|gca()返回当前的轴句柄，并具有该属性legend_|269455|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

you have to add the following lines of code:

<pre><code>ax = gca()
ax.legend_ = None
draw()
</code></pre>

gca() returns the current axes handle, and has that property legend_

blocks|key|3186105|text|如果以plt的形式调用pyplot|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3186106|frameon=False将移除图例周围的边框|3186107|和'‘传递了图例中不应包含任何变量的信息|3186108|import+matplotlib.pyplot+as+plt
plt.legend('',frameon=False)|code-block|syntax|javascript|3186109|entityMap^0|3|3|B|6|0|0|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Y|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|Z|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

if you call <code>pyplot</code> as <code>plt</code>

<code>frameon=False</code> is to remove the border around the legend

and '' is passing the information that no variable should be in the legend 

<pre><code>import matplotlib.pyplot as plt
plt.legend('',frameon=False)
</code></pre>

blocks|key|273102|text|如果您没有使用fig和ax绘图对象，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|273103|import+matplotlib.pyplot+as+plt

#+do+plot+specifics
plt.legend('')
plt.show()|code-block|syntax|javascript|273104|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you are not using fig and ax plot objects you can do it like so:

<pre><code>import matplotlib.pyplot as plt

# do plot specifics
plt.legend('')
plt.show()
</code></pre>

blocks|key|2972390|text|根据@naitsirhc的信息，我想找到官方的API文档。这是我的发现和一些示例代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2972391|通过seaborn.scatterplot().|offset|length|style|CODE|2972392|The函数I+created+a+matplotlib.Axes+object+|unordered-list-item|2972393|matplotlib.Axes+ax.get_legend()将返回您调用的matplotlib.legned.Legend+.remove()函数以从绘图中删除图例。|ordered-list-item|2972394|2972395|2972396|ax+=+sns.scatterplot(......)
_lg+=+ax.get_legend()
_lg.remove()|code-block|syntax|javascript|2972397|如果查看matplotlib.legned.Legend接口文档，您将看不到.remove()函数。|2972398|原因是matplotlib.legned.Legend继承了matplotlib.artist.Artist。因此，当您调用ax.get_legend().remove()时，基本上就调用了matplotlib.artist.Artist.remove()。|2972399|最后，您甚至可以将代码简化为两行。|2972400|ax+=+sns.scatterplot(......)
ax.get_legend().remove()|2972401|entityMap|0|LINK|mutability|MUTABLE|url|https://seaborn.pydata.org/generated/seaborn.scatterplot.html|1|https://matplotlib.org/3.1.1/api/legend_api.html#matplotlib.legend.Legend|2|https://matplotlib.org/3.1.1/api/artist_api.html#matplotlib.artist.Artist|3|https://matplotlib.org/3.1.1/api/artist_api.html#figure-and-axes^0|0|2|L|2|L|0|0|H|F|0|0|F|G|F|12|O|1R|9|0|0|0|0|4|O|12|9|4|O|1|0|3|O|U|O|1Q|O|2N|X|U|O|2|2N|X|3|0|0|0^^$0|@$1|2|3|4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1H|8|@$D|1I|E|1J|F|G]]|9|@$D|1K|E|1L|1|1M]]|A|$]]|$1|H|3|I|5|J|7|1N|8|@$D|1O|E|1P|F|G]]|9|@]|A|$]]|$1|K|3|L|5|M|7|1Q|8|@$D|1R|E|1S|F|G]|$D|1T|E|1U|F|G]|$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|20|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|21|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|22|8|@$D|23|E|24|F|G]|$D|25|E|26|F|G]]|9|@$D|27|E|28|1|29]]|A|$]]|$1|W|3|X|5|6|7|2A|8|@$D|2B|E|2C|F|G]|$D|2D|E|2E|F|G]|$D|2F|E|2G|F|G]|$D|2H|E|2I|F|G]]|9|@$D|2J|E|2K|1|2L]|$D|2M|E|2N|1|2O]]|A|$]]|$1|Y|3|Z|5|6|7|2P|8|@]|9|@]|A|$]]|$1|10|3|11|5|R|7|2Q|8|@]|9|@]|A|$S|T]]|$1|12|3|-4|5|6|7|2R|8|@]|9|@]|A|$]]]|13|$14|$5|15|16|17|A|$18|19]]|1A|$5|15|16|17|A|$18|1B]]|1C|$5|15|16|17|A|$18|1D]]|1E|$5|15|16|17|A|$18|1F]]]]

According to the information from @naitsirhc, I wanted to find the official API documentation. Here are my finding and some sample code.
<ol>
<li>I created a <code>matplotlib.Axes</code> object by <a href="https://seaborn.pydata.org/generated/seaborn.scatterplot.html" rel="noreferrer"><code>seaborn.scatterplot()</code></a>.</li>
<li>The <code>ax.get_legend()</code> will return a <code>matplotlib.legned.Legend</code> instance.</li>
<li>Finally, you call <code>.remove()</code> function to remove the legend from your plot.</li>
</ol>
<pre class="lang-py prettyprint-override"><code>ax = sns.scatterplot(......)
_lg = ax.get_legend()
_lg.remove()
</code></pre>
If you check the <a href="https://matplotlib.org/3.1.1/api/legend_api.html#matplotlib.legend.Legend" rel="noreferrer"><code>matplotlib.legned.Legend</code></a> API document, you won't see the <code>.remove()</code> function.
The reason is that the <code>matplotlib.legned.Legend</code> inherited the <a href="https://matplotlib.org/3.1.1/api/artist_api.html#matplotlib.artist.Artist" rel="noreferrer"><code>matplotlib.artist.Artist</code></a>. Therefore, when you call <code>ax.get_legend().remove()</code> that basically call <a href="https://matplotlib.org/3.1.1/api/artist_api.html#figure-and-axes" rel="noreferrer"><code>matplotlib.artist.Artist.remove()</code></a>.
In the end, you could even simplify the code into two lines.
<pre class="lang-py prettyprint-override"><code>ax = sns.scatterplot(......)
ax.get_legend().remove()
</code></pre>

blocks|key|275009|text|我创建了一个图例，将其添加到图中，而不是添加到轴(matplotlib+2.2.2)。为了删除它，我将图形的legends属性设置为一个空列表：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|275010|import+matplotlib.pyplot+as+plt

fig+=+plt.figure()
ax1+=+fig.add_subplot(111)
ax2+=+ax1.twinx()

ax1.plot(range(10),+range(10,+20),+label='line+1')
ax2.plot(range(10),+range(30,+20,+-1),+label='line+2')

fig.legend()

fig.legends+=+[]

plt.show()|code-block|syntax|javascript|275011|entityMap^0|1I|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I made a legend by adding it to the figure, not to an axis (matplotlib 2.2.2). To remove it, I set the <code>legends</code> attribute of the figure to an empty list:

<pre><code>import matplotlib.pyplot as plt

fig = plt.figure()
ax1 = fig.add_subplot(111)
ax2 = ax1.twinx()

ax1.plot(range(10), range(10, 20), label='line 1')
ax2.plot(range(10), range(30, 20, -1), label='line 2')

fig.legend()

fig.legends = []

plt.show()
</code></pre>

blocks|key|2972412|text|如果您使用的是seaborn，则可以使用参数legend。即使您在同一张图中多次绘制。使用一些df的示例|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2972413|import+seaborn+as+sns

#+Will+display+legend
ax1+=+sns.lineplot(x='cars',+y='miles',+hue='brand',+data=df)

#+No+legend+displayed
ax2+=+sns.lineplot(x='cars',+y='miles',+hue='brand',+data=df,+legend=None)|code-block|syntax|javascript|2972414|entityMap^0|M|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

If you are using seaborn you can use the parameter <code>legend</code>. Even if you are ploting more than once in the same figure. Example with some df
<pre><code>import seaborn as sns

# Will display legend
ax1 = sns.lineplot(x='cars', y='miles', hue='brand', data=df)

# No legend displayed
ax2 = sns.lineplot(x='cars', y='miles', hue='brand', data=df, legend=None)
</code></pre>

To add a legend to a matplotlib plot, one simply runs <code>legend()</code>.

How to remove a legend from a plot?

(The closest I came to this is to run <code>legend([])</code> in order to empty the legend from data. But that leaves an ugly white rectangle in the upper right corner.)

Remove the legend on a matplotlib figure

要向matplotlib图中添加图例，只需运行legend()。如何从绘图中删除图例? How to remove a plot？(我最接近的做法是运行legend([])，以便从数据中清空图例。但这会在右上角留下一个丑陋的白色矩形。)

问移除matplotlib图形上的图例
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回答 9

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