Issue
I would like a representation consisting of a scatter plot and 2 histograms on the right and below the scatter plot create. I have the following requirements: 1.) In the scatter plot, the apect ratio is equal so that the circle does not look like an ellipse. 2.) In the graphic, the subplots should be exactly as wide or high as the axes of the scatter plot.
This also works to a limited extent. However, I can't make the lower histogram as wide as the x axis of the scatter plot. How do I do that?
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
import random
#create some demo data
x = [random.uniform(-2.0, 2.0) for i in range(100)]
y = [random.uniform(-2.0, 2.0) for i in range(100)]
#create figure
fig = plt.figure()
gs = gridspec.GridSpec(2, 2, width_ratios = [3, 1], height_ratios = [3, 1])
ax = plt.subplot(gs[0])
# Axis labels
plt.xlabel('pos error X [mm]')
plt.ylabel('pos error Y [mm]')
ax.grid(True)
ax.axhline(color="#000000")
ax.axvline(color="#000000")
ax.set_aspect('equal')
radius = 1.0
xc = radius*np.cos(np.linspace(0,np.pi*2))
yc = radius*np.sin(np.linspace(0,np.pi*2))
plt.plot(xc, yc, "k")
ax.scatter(x,y)
hist_x = plt.subplot(gs[1],sharey=ax)
hist_y = plt.subplot(gs[2],sharex=ax)
plt.tight_layout() #needed. without no xlabel visible
plt.show()
what i want is:
Many thanks for your help!
Solution
The easiest (but not necessarily most elegant) solution is to manually position the lower histogram after applying the tight layout:
ax_pos = ax.get_position()
hist_y_pos = hist_y.get_position()
hist_y.set_position((ax_pos.x0, hist_y_pos.y0, ax_pos.width, hist_y_pos.height))
This output was produced by matplotlib version 3.4.3. For your example output, you're obviously using a different version, as I get a much wider lower histogram than you.
(I retained the histogram names as in your example although I guess the lower one should be hist_x
instead of hist_y
).
Answered By - Stef
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