Making a scatter marker as large as the Axes

Question

Im trying to scatter a single (square) marker such that it fills the whole figure (no more, no less).

As for simplification, I'm creating a figure such that x- and y- axes both go from -0.5 to 0.5. That is, the plotting area is the unit square, centred at the origin.

The marker now shall be scattered at the origin. What size should it be so that it occupies exactly the unit square?

I looked at this Finding the right marker size for a scatter plot and this pyplot scatter plot marker size but couldn't get it right so far.

This is what I tried:

fig, ax = plt.subplots(figsize=(4,4));

ax.set_aspect('equal');
ax.set_xlim(-0.5, 0.5);
ax.set_ylim(-0.5, 0.5);

figsize = fig.get_size_inches()[0]

dpi = fig.dpi

print(f'figsize = {int(figsize)}')

print(f'dpi = {int(dpi)}')

print(f'figure is {int(figsize*dpi)} x {int(figsize*dpi)} pixels\n')

print(f'setting the marker size to be {int(figsize*dpi)}**2 = {int((figsize*dpi)**2)}')

ax.scatter(0, 0, s=(figsize*dpi)**2, marker='s');

It turns out that the marker (blue area) does fill the unit square but it is actually filling way more than that. After manually trying different sizes, the right value seems to be around 46000 (opposed to the 82944 suggested at the second post).

You are using the figure size, but expect the marker to be as large as the axes box? — ImportanceOfBeingErnest, Aug 28 '19 at 16:29
I believe yes, since the axes match the figure boundaries (?) — L. B., Aug 28 '19 at 16:31
No they don't (as seen by the numbers 0.5 outside the axes, but inside the figure). — ImportanceOfBeingErnest, Aug 28 '19 at 16:34
oh, thanks, I see. I think I'll have to find where the axes are located in the figure. — L. B., Aug 28 '19 at 16:41

score 1 · Accepted Answer · answered Aug 28 '19 at 16:58

You will need to apply the aspect, then get the axes width and transform it to display space (or transform the axes position first, then get its width). This can be used to calculate the width of the axes in units of points.
The square of that number is the markersize of the scatter if it shall be as large as the axes.

import matplotlib.pyplot as plt

fig, ax = plt.subplots(figsize=(4,4))

ax.set_xlim(-0.5, 0.5)
ax.set_ylim(-0.5, 0.5)

ax.set_aspect('equal')
ax.apply_aspect()
s = ax.get_position().transformed(fig.transFigure).width*72/fig.dpi
ax.scatter(0, 0, s=s**2, marker='s'); 
plt.show()

Making a scatter marker as large as the Axes

1 Answers1