blocks|key|147391|text|使用subplots并不太复杂，脊椎可能会很复杂。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|147392|愚蠢而简单的方式：|147393|%25matplotlib+inline
import+numpy+as+np
import+matplotlib.pyplot+as+plt

x+=+np.linspace(0.2,10,100)
fig,+ax+=+plt.subplots()
ax.plot(x,+1/x)
ax.plot(x,+np.log(x))
ax.set_aspect('equal')
ax.grid(True,+which='both')

ax.axhline(y=0,+color='k')
ax.axvline(x=0,+color='k')|code-block|syntax|javascript|147394|我得到了：|147395|​|147396|📷|atomic|147397|147398|(您看不到垂直轴，因为x下限为零。)|147399|使用简单脊椎的替代方案|147400|%25matplotlib+inline
import+numpy+as+np
import+matplotlib.pyplot+as+plt

x+=+np.linspace(0.2,10,100)
fig,+ax+=+plt.subplots()
ax.plot(x,+1/x)
ax.plot(x,+np.log(x))
ax.set_aspect('equal')
ax.grid(True,+which='both')

#+set+the+x-spine+(see+below+for+more+info+on+`set_position`)
ax.spines['left'].set_position('zero')

#+turn+off+the+right+spine/ticks
ax.spines['right'].set_color('none')
ax.yaxis.tick_left()

#+set+the+y-spine
ax.spines['bottom'].set_position('zero')

#+turn+off+the+top+spine/ticks
ax.spines['top'].set_color('none')
ax.xaxis.tick_bottom()|147401|147402|147403|147404|可选择使用seaborn+(我最喜欢)|147405|import+numpy+as+np
import+matplotlib.pyplot+as+plt
import+seaborn
seaborn.set(style='ticks')

x+=+np.linspace(0.2,10,100)
fig,+ax+=+plt.subplots()
ax.plot(x,+1/x)
ax.plot(x,+np.log(x))
ax.set_aspect('equal')
ax.grid(True,+which='both')
seaborn.despine(ax=ax,+offset=0)+#+the+important+part+here|147406|147407|147408|147409|使用脊椎的set_position方法|147410|以下是set_position+method+of+spines的文档|147411|147412|脊椎位置由(位置类型，数量)的2元组指定。职位类型包括：|blockquote|147413|147414|147415|147416|‘out’：将脊椎从数据区域向外放置指定数量的点。(负值指定将|unordered-list-item|147417|147418|spine+inward.)|147419|'axes‘：将脊椎放置在指定的轴坐标处(from+0.0-1.0).|147420|
|147421|'data’：将脊椎放置在指定的数据坐标。|147422|147423|此外，速记符号定义了一个特殊的位置：|147424|数据中心('axes',0.5)|147425|'zero‘->+(’|147426|147427|‘，0.0)+->+|147428|147429|所以你可以把左边的脊椎放在任何地方：|147430|ax.spines['left'].set_position((system,+poisition))|147431|其中，system是“向外”、“轴”或“数据”，并且position位于该坐标系中的位置。|147432|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/28c4ea7aae7ec38fec9e2597afda93ea.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/0a9d9d55b1564f62ded478f0b5ca58f4.png|2|https://ask.qcloudimg.com/http-save/yehe-900000/4afea239eb8add63193e95f71a874698.png|3|LINK|MUTABLE|url|http://matplotlib.org/api/spines_api.html#matplotlib.spines.Spine.set_position^0|2|8|0|0|0|0|0|0|1|0|0|0|0|0|0|0|0|1|1|0|0|5|7|0|0|0|0|1|2|0|0|5|C|0|3|C|3|T|3|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|1F|0|3|6|Q|8|0^^$0|@$1|2|3|4|5|6|7|2W|8|@$9|2X|A|2Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|2Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|30|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|31|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|32|8|@]|D|@]|E|$]]|$1|Q|3|R|5|S|7|33|8|@]|D|@$9|34|A|35|1|36]]|E|$]]|$1|T|3|P|5|6|7|37|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|38|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|39|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|J|7|3A|8|@]|D|@]|E|$K|L]]|$1|10|3|P|5|6|7|3B|8|@]|D|@]|E|$]]|$1|11|3|R|5|S|7|3C|8|@]|D|@$9|3D|A|3E|1|3F]]|E|$]]|$1|12|3|P|5|6|7|3G|8|@]|D|@]|E|$]]|$1|13|3|14|5|6|7|3H|8|@$9|3I|A|3J|B|C]]|D|@]|E|$]]|$1|15|3|16|5|J|7|3K|8|@]|D|@]|E|$K|L]]|$1|17|3|P|5|6|7|3L|8|@]|D|@]|E|$]]|$1|18|3|R|5|S|7|3M|8|@]|D|@$9|3N|A|3O|1|3P]]|E|$]]|$1|19|3|P|5|6|7|3Q|8|@]|D|@]|E|$]]|$1|1A|3|1B|5|6|7|3R|8|@$9|3S|A|3T|B|C]]|D|@]|E|$]]|$1|1C|3|1D|5|6|7|3U|8|@$9|3V|A|3W|B|C]]|D|@$9|3X|A|3Y|1|3Z]]|E|$]]|$1|1E|3|-4|5|6|7|40|8|@]|D|@]|E|$]]|$1|1F|3|1G|5|1H|7|41|8|@]|D|@]|E|$]]|$1|1I|3|-4|5|6|7|42|8|@]|D|@]|E|$]]|$1|1J|3|-4|5|6|7|43|8|@]|D|@]|E|$]]|$1|1K|3|-4|5|6|7|44|8|@]|D|@]|E|$]]|$1|1L|3|1M|5|1N|7|45|8|@]|D|@]|E|$]]|$1|1O|3|-4|5|6|7|46|8|@]|D|@]|E|$]]|$1|1P|3|1Q|5|6|7|47|8|@]|D|@]|E|$]]|$1|1R|3|1S|5|1N|7|48|8|@]|D|@]|E|$]]|$1|1T|3|1U|5|6|7|49|8|@]|D|@]|E|$]]|$1|1V|3|1W|5|1N|7|4A|8|@]|D|@]|E|$]]|$1|1X|3|-4|5|6|7|4B|8|@]|D|@]|E|$]]|$1|1Y|3|1Z|5|6|7|4C|8|@]|D|@]|E|$]]|$1|20|3|21|5|6|7|4D|8|@]|D|@]|E|$]]|$1|22|3|23|5|1N|7|4E|8|@]|D|@]|E|$]]|$1|24|3|1U|5|6|7|4F|8|@]|D|@]|E|$]]|$1|25|3|26|5|1N|7|4G|8|@]|D|@]|E|$]]|$1|27|3|-4|5|6|7|4H|8|@]|D|@]|E|$]]|$1|28|3|29|5|6|7|4I|8|@]|D|@]|E|$]]|$1|2A|3|2B|5|6|7|4J|8|@$9|4K|A|4L|B|C]]|D|@]|E|$]]|$1|2C|3|2D|5|6|7|4M|8|@$9|4N|A|4O|B|C]|$9|4P|A|4Q|B|C]]|D|@]|E|$]]|$1|2E|3|-4|5|6|7|4R|8|@]|D|@]|E|$]]]|2F|$2G|$5|2H|2I|2J|E|$2K|2L|2M|-4]]|2N|$5|2H|2I|2J|E|$2K|2O|2M|-4]]|2P|$5|2H|2I|2J|E|$2K|2Q|2M|-4]]|2R|$5|2S|2I|2T|E|$2U|2V]]]]

Using <code>subplots</code> is not too complicated, the spines might be.

<h2>Dumb, simple way:</h2>

<pre><code>%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(0.2,10,100)
fig, ax = plt.subplots()
ax.plot(x, 1/x)
ax.plot(x, np.log(x))
ax.set_aspect('equal')
ax.grid(True, which='both')

ax.axhline(y=0, color='k')
ax.axvline(x=0, color='k')
</code></pre>

And I get:

<img src="https://i.stack.imgur.com/k3Uv7.png" alt="with axlines">

(you can't see the vertical axis since the lower x-limit is zero.)

<h2>Alternative using simple spines</h2>

<pre><code>%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(0.2,10,100)
fig, ax = plt.subplots()
ax.plot(x, 1/x)
ax.plot(x, np.log(x))
ax.set_aspect('equal')
ax.grid(True, which='both')

# set the x-spine (see below for more info on `set_position`)
ax.spines['left'].set_position('zero')

# turn off the right spine/ticks
ax.spines['right'].set_color('none')
ax.yaxis.tick_left()

# set the y-spine
ax.spines['bottom'].set_position('zero')

# turn off the top spine/ticks
ax.spines['top'].set_color('none')
ax.xaxis.tick_bottom()
</code></pre>

<img src="https://i.stack.imgur.com/KYDMJ.png" alt="with_spines">

<h2>Alternative using <code>seaborn</code> (my favorite)</h2>

<pre><code>import numpy as np
import matplotlib.pyplot as plt
import seaborn
seaborn.set(style='ticks')

x = np.linspace(0.2,10,100)
fig, ax = plt.subplots()
ax.plot(x, 1/x)
ax.plot(x, np.log(x))
ax.set_aspect('equal')
ax.grid(True, which='both')
seaborn.despine(ax=ax, offset=0) # the important part here
</code></pre>

<img src="https://i.stack.imgur.com/FHALV.png" alt="with_seaborn">

<h2>Using the <code>set_position</code> method of a spine</h2>

Here are the docs for a the <a href="http://matplotlib.org/api/spines_api.html#matplotlib.spines.Spine.set_position" rel="noreferrer"><code>set_position</code> method of spines</a>:

<blockquote>
 Spine position is specified by a 2 tuple of (position type, amount).
 The position types are:
 
 <ul>
 <li>'outward' : place the spine out from the data area by the specified number of points. (Negative values specify placing the 
 spine inward.)</li>
 <li>'axes' : place the spine at the specified Axes coordinate (from
 0.0-1.0).</li>
 <li>'data' : place the spine at the specified data coordinate.</li>
 </ul>
 
 Additionally, shorthand notations define a special positions:
 
 <ul>
 <li>'center' -> ('axes',0.5)</li>
 <li>'zero' -> ('data', 0.0)</li>
 </ul>
</blockquote>

So you can place, say the left spine anywhere with:

<code>ax.spines['left'].set_position((system, poisition))</code>

where <code>system</code> is 'outward', 'axes', or 'data' and <code>position</code> in the place in that coordinate system.

blocks|key|150193|text|对于那些想要像我一样搜索它的人，让我来回答这个(相当古老的)问题。尽管它建议所有工作解决方案，但我认为(唯一)提供的答案way太复杂了，当涉及到问题中描述的如此简单的情况时(注意:此方法需要您指定所有轴端点)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|150194|我在matplotlib's+pyplot上的one+of+the+first+tutorials中找到了一个简单有效的解决方案。在创建plot之后，添加以下行就足够了|150195|plt.axis([xmin,+xmax,+ymin,+ymax])|CODE|150196|如下例所示：|150197|from+matplotlib+import+pyplot+as+plt

xs+=+[1,2,3,4,5]
ys+=+[3,5,1,2,4]

plt.scatter(xs,+ys)
plt.axis([0,6,0,6])++#this+line+does+the+job
plt.show()|code-block|syntax|javascript|150198|这将产生以下结果：|150199|matplotlib+plot+with+axes+endpoints+specified|150200|entityMap|0|LINK|mutability|MUTABLE|url|https://matplotlib.org/api/pyplot_api.html|1|https://matplotlib.org/3.3.3/tutorials/introductory/pyplot.html#working-with-text|2|https://i.stack.imgur.com/yI2bM.png^0|1O|17|0|2|J|0|N|Q|1|0|0|Y|0|0|0|0|0|19|2|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]|$9|1E|A|1F|1|1G]]|E|$]]|$1|H|3|I|5|6|7|1H|8|@$9|1I|A|1J|B|J]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1K|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|1L|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|6|7|1M|8|@]|D|@]|E|$]]|$1|T|3|U|5|6|7|1N|8|@]|D|@$9|1O|A|1P|1|1Q]]|E|$]]|$1|V|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]|13|$5|Y|Z|10|E|$11|14]]|15|$5|Y|Z|10|E|$11|16]]]]

Let me answer to this (rather old) question for those who will search for it as I just did. Although it suggested working solutions, I consider the (only) provided answer as way too complex, when it comes to such a simple situation like that described in the question (note: this method requires you to specify all axes endpoints).
I found a simple working solution in <a href="https://matplotlib.org/3.3.3/tutorials/introductory/pyplot.html#working-with-text" rel="nofollow noreferrer">one of the first tutorials</a> on <a href="https://matplotlib.org/api/pyplot_api.html" rel="nofollow noreferrer">matplotlib's pyplot</a>. It is sufficient to add the following line after the creation of the plot
<code>plt.axis([xmin, xmax, ymin, ymax])</code>
as in the following example:
<pre><code>from matplotlib import pyplot as plt

xs = [1,2,3,4,5]
ys = [3,5,1,2,4]

plt.scatter(xs, ys)
plt.axis([0,6,0,6]) #this line does the job
plt.show()
</code></pre>
which produces the following result:
<a href="https://i.stack.imgur.com/yI2bM.png" rel="nofollow noreferrer">matplotlib plot with axes endpoints specified</a>

I have following simple plot, and I would like to display the origin axis (x, y). I already have grid, but I need the x, y axis to be emphasized.

<img src="https://i.stack.imgur.com/a9hAQ.png" alt="enter image description here">

this is my code:

<pre><code>x = linspace(0.2,10,100)
plot(x, 1/x)
plot(x, log(x))
axis('equal')
grid()
</code></pre>

I have seen <a href="https://stackoverflow.com/questions/4694478/center-origin-in-matplotlib">this</a> question. The accepted answer suggests to use "Axis spine" and just links to some example. The example is however too complicated, using subplots. I am unable to figure out, how to use "Axis spine" in my simple example.

show origin axis (x,y) in matplotlib plot

数据中心

我有以下简单的图，我想显示原点轴(x，y)。我已经有了网格，但我需要强调x，y轴。​​这是我的代码：x = linspace(0.2,10,100)plot(x, 1/x)plot(x, log(x))axis('equal')grid()我已经看到了的问题。公认的答案是建议使用"Axis spine“，并且只需链接到...

问在matplotlib图中显示原点轴(x，y)
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