scatter plot with single pixel marker in matplotlib

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.

https://github.com/matplotlib/matplotlib/pull/695

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

My simplified dataset (data.csv)

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

My code is below.

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)

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

Any help is appreciated

The problem

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

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)

The solution: Setting the markersize

The solution is to use a usual "o" or "s" 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 72./fig.dpi.

For aplot`, the markersize is directly

ax.plot(..., marker="o", ms=72./fig.dpi)

For a scatter the markersize is given through the s argument, which is in square points,
```
ax.scatter(..., marker='o', s=(72./fig.dpi)**2)
```

Complete example:

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)

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.

https://matplotlib.org/devdocs/api/_as_gen/matplotlib.pyplot.scatter.html

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

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

Source: https://stackoverflow.com/a/45803960/4063622

来源：https://stackoverflow.com/questions/39753282/scatter-plot-with-single-pixel-marker-in-matplotlib

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python

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scatter-plot