blocks|key|2908596|text|问题所在|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2908597|我担心您引用的matplotlib+git存储库讨论的错误修复只适用于plt.plot()，而不适用于plt.scatter()|offset|length|style|CODE|2908598|import+matplotlib.pyplot+as+plt

fig+=+plt.figure(figsize=(4,2))
ax+=+fig.add_subplot(121)
ax2+=+fig.add_subplot(122,+sharex=ax,+sharey=ax)
ax.plot([1,+2],[0.4,0.4],color='black',marker=',',lw=0,+linestyle="")
ax.set_title("ax.plot")
ax2.scatter([1,2],[0.4,0.4],color='black',marker=',',lw=0,+s=1)
ax2.set_title("ax.scatter")
ax.set_xlim(0,8)
ax.set_ylim(0,1)
fig.tight_layout()
print+fig.dpi+#prints+80+in+my+case
fig.savefig('plot.png',+dpi=fig.dpi)|code-block|syntax|javascript|2908599|​|2908600|📷|atomic|2908601|2908602|解决方案:设置markersize|2908603|解决方案是使用常用的"o"或"s"标记，但将标记大小设置为恰好一个像素。由于markersize是以点为单位给出的，因此需要使用数字dpi来计算一个像素的点的大小。这是72./fig.dpi。|2908604|2908605|For+aplot`，标记大小是直接|unordered-list-item|2908606|2908607|ax.plot(...，marker="o"，ms=72./fig.dpi)|2908608|For+a+ax.plot)标记大小是通过s参数给出的，该参数以平方点为单位，|2908609|2908610|标记(...，ax.scatter=‘o’，s=(72./fig.dpi)**2)|2908611|完整示例：|2908612|import+matplotlib.pyplot+as+plt

fig+=+plt.figure(figsize=(4,2))
ax+=+fig.add_subplot(121)
ax2+=+fig.add_subplot(122,+sharex=ax,+sharey=ax)
ax.plot([1,+2],[0.4,0.4],+marker='o',ms=72./fig.dpi,+mew=0,+
++++++++color='black',+linestyle="",+lw=0)
ax.set_title("ax.plot")
ax2.scatter([1,2],[0.4,0.4],color='black',+marker='o',+lw=0,+s=(72./fig.dpi)**2)
ax2.set_title("ax.scatter")
ax.set_xlim(0,8)
ax.set_ylim(0,1)
fig.tight_layout()
fig.savefig('plot.png',+dpi=fig.dpi)|2908613|2908614|2908615|2908616|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/ebac649fb9b3671e92e57f0df68e05c6.png|imageAlt|1|https://ask.qcloudimg.com/http-save/yehe-900000/331a90611ba017de81df4af5014838e4.png^0|0|Z|A|1F|D|0|0|0|0|1|0|0|0|0|A|3|E|3|2C|B|0|0|0|5|0|0|0|L|1|0|0|0|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1R|8|@$D|1S|E|1T|F|G]|$D|1U|E|1V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1X|8|@]|9|@]|A|$]]|$1|O|3|P|5|Q|7|1Y|8|@]|9|@$D|1Z|E|20|1|21]]|A|$]]|$1|R|3|N|5|6|7|22|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|23|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|24|8|@$D|25|E|26|F|G]|$D|27|E|28|F|G]|$D|29|E|2A|F|G]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|2B|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|2C|8|@$D|2D|E|2E|F|G]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|2F|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|2G|8|@]|9|@]|A|$]]|$1|13|3|14|5|Z|7|2H|8|@$D|2I|E|2J|F|G]]|9|@]|A|$]]|$1|15|3|-4|5|6|7|2K|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|2L|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|2M|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|J|7|2N|8|@]|9|@]|A|$K|L]]|$1|1C|3|N|5|6|7|2O|8|@]|9|@]|A|$]]|$1|1D|3|P|5|Q|7|2P|8|@]|9|@$D|2Q|E|2R|1|2S]]|A|$]]|$1|1E|3|N|5|6|7|2T|8|@]|9|@]|A|$]]|$1|1F|3|-4|5|6|7|2U|8|@]|9|@]|A|$]]]|1G|$1H|$5|1I|1J|1K|A|$1L|1M|1N|-4]]|1O|$5|1I|1J|1K|A|$1L|1P|1N|-4]]]]

<h3>The problem</h3>

I fear that the bugfix discussed at matplotlib git repository that you're citing is only valid for <code>plt.plot()</code> and not for <code>plt.scatter()</code>

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

fig = plt.figure(figsize=(4,2))
ax = fig.add_subplot(121)
ax2 = fig.add_subplot(122, sharex=ax, sharey=ax)
ax.plot([1, 2],[0.4,0.4],color='black',marker=',',lw=0, linestyle="")
ax.set_title("ax.plot")
ax2.scatter([1,2],[0.4,0.4],color='black',marker=',',lw=0, s=1)
ax2.set_title("ax.scatter")
ax.set_xlim(0,8)
ax.set_ylim(0,1)
fig.tight_layout()
print fig.dpi #prints 80 in my case
fig.savefig('plot.png', dpi=fig.dpi)
</code></pre>

<a href="https://i.stack.imgur.com/98OO7.png" rel="noreferrer"><img src="https://i.stack.imgur.com/98OO7.png" alt="enter image description here"></a>

<h3>The solution: Setting the markersize</h3>

The solution is to use a usual <code>"o"</code> or <code>"s"</code> marker, but set the markersize to be exactly one pixel. Since the markersize is given in points, one would need to use the figure dpi to calculate the size of one pixel in points. This is <code>72./fig.dpi</code>.

<ul>
<li><code>For a</code>plot`, the markersize is directly

<pre><code>ax.plot(..., marker="o", ms=72./fig.dpi)
</code></pre></li>
<li>For a <code>scatter</code> the markersize is given through the <code>s</code> argument, which is in square points, 

<pre><code>ax.scatter(..., marker='o', s=(72./fig.dpi)**2)
</code></pre></li>
</ul>

Complete example:

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

fig = plt.figure(figsize=(4,2))
ax = fig.add_subplot(121)
ax2 = fig.add_subplot(122, sharex=ax, sharey=ax)
ax.plot([1, 2],[0.4,0.4], marker='o',ms=72./fig.dpi, mew=0, 
 color='black', linestyle="", lw=0)
ax.set_title("ax.plot")
ax2.scatter([1,2],[0.4,0.4],color='black', marker='o', lw=0, s=(72./fig.dpi)**2)
ax2.set_title("ax.scatter")
ax.set_xlim(0,8)
ax.set_ylim(0,1)
fig.tight_layout()
fig.savefig('plot.png', dpi=fig.dpi)
</code></pre>

<a href="https://i.stack.imgur.com/3dOH6.png" rel="noreferrer"><img src="https://i.stack.imgur.com/3dOH6.png" alt="enter image description here"></a>

blocks|key|291119|text|对于任何还在尝试解决这个问题的人来说，我找到的解决方案是在plt.scatter中指定s参数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|291120|S参数表示要打印的点的面积。|291121|它似乎不是很完美，因为s=1似乎覆盖了我屏幕的4个像素，但这肯定比我能找到的任何其他东西都要小。|291122|https://matplotlib.org/devdocs/api/_as_gen/matplotlib.pyplot.scatter.html|offset|length|291123|形状s:标量或array_like，|291124|+(n，)，可选|blockquote|291125|291126|以点%5E2为单位的大小。默认值为rcParams'lines.markersize‘**+2。|291127|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|21|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|11|8|@]|9|@$H|12|I|13|1|14]]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|16|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|18|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|G]]]]

For anyone still trying to figure this out, the solution I found was to specify the s argument in plt.scatter.

The s argument refers to the area of the point you are plotting.

It doesn't seem to be quite perfect, since s=1 seems to cover about 4 pixels of my screen, but this definitely makes them smaller than anything else I've been able to find.

<a href="https://matplotlib.org/devdocs/api/_as_gen/matplotlib.pyplot.scatter.html" rel="nofollow noreferrer">https://matplotlib.org/devdocs/api/_as_gen/matplotlib.pyplot.scatter.html</a>

<blockquote>
 s : scalar or array_like, shape (n, ), optional 
 size in points^2. Default is rcParams['lines.markersize'] ** 2.
</blockquote>

blocks|key|2908645|text|将plt.scatter()参数设置为linewidths=0，并计算出参数s的正确值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2908646|来源：https://stackoverflow.com/a/45803960/4063622|2908647|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/45803960/4063622^0|1|D|J|C|12|1|0|3|18|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|H|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Set the <code>plt.scatter()</code> parameter to <code>linewidths=0</code> and figure out the right value for the parameter <code>s</code>.

Source: <a href="https://stackoverflow.com/a/45803960/4063622">https://stackoverflow.com/a/45803960/4063622</a>

I am trying to plot a large dataset with a scatter plot. 
I want to use matplotlib to plot it with single pixel marker. 
It seems to have been solved.

<a href="https://github.com/matplotlib/matplotlib/pull/695" rel="noreferrer">https://github.com/matplotlib/matplotlib/pull/695</a>

But I cannot find a mention of how to get a single pixel marker. 

My simplified dataset (data.csv)

<pre><code>Length,Time
78154393,139.324091
84016477,229.159305
84626159,219.727537
102021548,225.222662
106399706,221.022827
107945741,206.760239
109741689,200.153263
126270147,220.102802
207813132,181.67058
610704756,50.59529
623110004,50.533158
653383018,52.993885
659376270,53.536834
680682368,55.97628
717978082,59.043843
</code></pre>

My code is below.

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

inputfile='data.csv'
iplevel = pd.read_csv(inputfile)
base = os.path.splitext(inputfile)[0]

fig = plt.figure()
plt.yscale('log')
#plt.xscale('log')
plt.title(' My plot: '+base)
plt.xlabel('x')
plt.ylabel('y')
plt.scatter(iplevel['Time'], iplevel['Length'],color='black',marker=',',lw=0,s=1)
fig.tight_layout()
fig.savefig(base+'_plot.png', dpi=fig.dpi)
</code></pre>

You can see below that the points are not single pixel. 

<a href="https://i.stack.imgur.com/Z6sB6.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Z6sB6.png" alt="data_plot.png"></a>

Any help is appreciated

scatter plot with single pixel marker in matplotlib

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试用散点图绘制一个大型数据集。我想使用matplotlib用单像素标记来绘制它。这个问题似乎已经解决了。但我找不到如何获得单像素标记的提法。我的简化数据集(data.csv)Length,Time78154393,139.32409184016477,229.15930584626159,219.7275371...

问matplotlib中使用单像素标记的散点图
EN

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