matplotlib graph plotting

Question

I wish to plot a 2D scatter plot with values in the range [-0.5, 0.5] and [-0.5,0.5] for x and y coordinates respectively

fig1 = plt.figure(figsize=[7,7])
axes1 = fig1.add_axes([1,1,1,1])
axes1.set_xlabel("xlabel")
axes1.set_ylabel("ylabel")
axes1.set_title("Emotions Quadrants")
axes1.spines['left'].set_position(('axes', 0.5))
axes1.spines['bottom'].set_position(('axes', 0.5))
axes1.scatter(x_vals, y_vals, label="test")

above code scales and x and y coordinates according to to the data. I wish to show x and y coordinates from -0.5 to 0.5 equally spaced for both the axes and then plot the scatter graph

I wish to keep the axes intersecting at 0,0 and not scale according to the data points

Edit by taking x and y values as xvals = [0.237, 0.415, -0.264, 0.142, -0.037, -0.088, -0.143, -0.332, 0.166, -0.079] y vals= [0.265, 0.325, -0.069, 0.086, -0.136, 0.124, -0.051, -0.052, 0.280, 0.121]

I am getting following plot with @r-beginners answer [enter image description here][2]

I want the origin to be in the center and points not scaled and not shift downward [2]: https://i.sstatic.net/lxGIs.png

r-beginners · Accepted Answer · 2022-02-04 09:09:36Z

1

If you set the axis to the desired range and the frame to the center, you will get the graph you want.

import matplotlib.pyplot as plt
import numpy as np

#x_vals = np.linspace(-0.5,0.5,10)
#y_vals = np.linspace(-0.5,0.5,10)

x_vals = [0.237, 0.415, -0.264, 0.142, -0.037, -0.088, -0.143, -0.332, 0.166, -0.079]
y_vals= [0.265, 0.325, -0.069, 0.086, -0.136, 0.124, -0.051, -0.052, 0.280, 0.121]

x_vals_center = (np.max(x_vals) + np.min(x_vals)) / 2
y_vals_center = (np.max(y_vals) + np.min(y_vals)) / 2

fig1 = plt.figure(figsize=[7,7])
axes1 = fig1.add_axes([-0.5,-0.5,0.5,0.5])
axes1.set_xlabel("xlabel", labelpad=0)
axes1.set_ylabel("ylabel", labelpad=110)
axes1.set_title("Emotions Quadrants")
axes1.spines[['top', 'right']].set_visible(False)
#axes1.spines[['left','bottom']].set_position(('data', 0))

axes1.spines['left'].set_position(('data', x_vals_center))
axes1.spines['bottom'].set_position(('data', y_vals_center))

axes1.xaxis.set_ticks_position('bottom')
axes1.yaxis.set_ticks_position('left')

axes1.scatter(x_vals, y_vals, label="test")

plt.show()

edited Feb 4, 2022 at 9:09

answered Jan 31, 2022 at 7:05

r-beginners

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7 Comments

Sam Mend Over a year ago

Hi. Sorry but your method works only if the points are uniformly distrubuted along the y axis. there are more points in 3rd qudrant then it will not work

Sam Mend Over a year ago

try with xvals = [0.237, 0.415, -0.264, 0.142, -0.037, -0.088, -0.143, -0.332, 0.166, -0.079] y vals= [0.265, 0.325, -0.069, 0.086, -0.136, 0.124, -0.051, -0.052, 0.280, 0.121]

r-beginners Over a year ago

I understood what you pointed out. I changed it so that it always intersects at zero, without affecting the values I draw.

Sam Mend Over a year ago

Yes. It is almost correct. But I wanted the axes to be at the center of the image. Please see @Scott Boston answer. Thanks for your help though

r-beginners Over a year ago

The setting of set_position(('axes', 0.5)) only sets the axis to half of 1.0, so if the length of the axis changes it will not be the center you want.

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Scott Boston · Accepted Answer · 2022-02-04 05:10:05Z

IIUC,

x_vals = [0.237, 0.415, -0.264, 0.142, -0.037, -0.088, -0.143, -0.332, 0.166, -0.079] 
y_vals= [0.265, 0.325, -0.069, 0.086, -0.136, 0.124, -0.051, -0.052, 0.280, 0.121] 

fig1 = plt.figure(figsize=[7,7])
axes1 = fig1.add_axes([-0.5,-0.5,0.5,0.5])
axes1.set_xlabel("xlabel", labelpad=110)
axes1.set_ylabel("ylabel", labelpad=110)
axes1.set_title("Emotions Quadrants")
axes1.spines['right'].set_visible(False)
axes1.spines['top'].set_visible(False)

axes1.spines['left'].set_position(('axes', 0.5))
axes1.spines['bottom'].set_position(('axes', 0.5))

axes1.xaxis.set_ticks_position('bottom')
axes1.yaxis.set_ticks_position('left')

axes1.scatter(x_vals, y_vals, label="test")
axes1.set_ylabel('Arousal')
axes1.set_xlabel('Valence')

axes1.set_ylim(-.5,.5)
axes1.set_xlim(-.5,.5)

plt.show()

Output:

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