Odd indentation near origin of graph (sometimes)

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The discussion centers on an issue with an odd indentation in a 3D Gaussian plot generated using matplotlib when the random centroid is set near the origin. Users suggest that this artifact is likely due to the mesh configuration and recommend modifying the code to allow for either a polar or Cartesian mesh, along with options to specify the centroid and variance. A proposed solution involves adjusting the order of how the centroid and mesh are defined to achieve the desired centering effect. The modified code provides flexibility in plotting, enabling users to visualize the Gaussian distribution more accurately. Overall, the conversation highlights the importance of mesh choice and centroid placement in 3D visualizations.
MathewsMD
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Code:
from mpl_toolkits.mplot3d
import Axes3D
import matplotlib
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
import numpy as np
import matplotlib.pylab as plt
import math
import random from scipy
import integrate

step = 0.1
maxval = 1.0
fig = plt.figure()
ax = Axes3D(fig)
R1 = 4.
R2 = 7.
r = np.linspace(0,20,100)
p = np.linspace(0,2*np.pi,100)
R,P = np.meshgrid(r,p)
X,Y = R*np.cos(P),R*np.sin(P)
sigma0 = random.randint(4000., 7000.)/1000.
r0 = random.randint(0, R1*1000.)/1000. #random centroid
theta0 = random.uniform(0, np.pi*2)

Z = (np.e**((-(R**2 + r0**2- 2*R*r0*(np.cos(P)*np.cos(theta0) +  np.sin(P)*np.sin(theta0)))/(2*sigma0**2))))

ax.plot_surface(X, Y, Z, rstride=2, cstride=2, cmap=cm.jet)
ax.set_zlim3d(0,1)
ax.set_zlabel('Intensity')
plt.show()

When I run this code, it works. But due to the random number generator, sometimes the input values have the graph centred near the origin (e.g. set r0 = 5), and when this occurs, there seems to be an odd indentation in the graph itself. Maybe I'm missing something, but if I'm trying to display a 2D Gaussian, this shouldn't be there, right? If anyone has any thoughts on why this is occurring and any ideas to approach this problem, that would be greatly appreciated!
 
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I don't understand what you mean.

If you set r0 first and THEN determine r and p around it...would that center it as you desired? or would that defeat the purpose of what you are trying to do?
 
MathewsMD said:
Maybe I'm missing something, but if I'm trying to display a 2D Gaussian, this shouldn't be there, right?
It's an artifact of your mesh. I've modified your code a bit so I can choose between a polar mesh or a cartesian mesh, and so I can optionally specify the centroid and variance:
Code:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from scipy import random
import argparse

parser = argparse.ArgumentParser(description='Plot a 2D Gaussian.')
parser.add_argument(
    '-p', '--polar',
    help = 'plot with polar X,Y mesh (default: cartesian mesh)',
    action = 'store_true')
parser.add_argument(
    '-c', '--centroid',
    help = 'centroid coordinates (default: random)',
    type = float,
    nargs = 2)
parser.add_argument(
    '-s', '--sigma',
    help = 'variance (sigma) (default: random)',
    type = float,
    nargs = 1)
args = parser.parse_args()

if args.polar :
    r = np.linspace(0.0,20.0,100.0)
    p = np.linspace(0.0,2*np.pi,100.0)
    R,P = np.meshgrid(r,p)
    X,Y = R*np.cos(P),R*np.sin(P)
else :
    x = np.linspace(-20.0,20.0,101.0)
    y = np.linspace(-20.0,20.0,101.0)
    X,Y = np.meshgrid(x,y)

if args.centroid is not None :
    x0,y0 = args.centroid
else :
    x0,y0 = random.uniform(-4.0,4.0), random.uniform(-4.0,4.0)

if args.sigma is not None :
    sigma0 = args.sigma[0]
else :
    sigma0 = random.uniform(4.0, 7.0)

Z = np.e**(-((X-x0)**2 + (Y-y0)**2)/(2*sigma0**2))

fig = plt.figure()
ax = Axes3D(fig)
ax.plot_surface(X, Y, Z, rstride=2, cstride=2, cmap=cm.jet)
ax.set_zlim3d(0,1)
ax.set_zlabel('Intensity')
plt.show()

I get the following plot when I run this via python plot.py --centroid -5 5:
cartesian.jpg
Using a polar mesh (python plot.py --centroid -5 5 --polar) yields the following:
polar.jpg
 
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D H said:
It's an artifact of your mesh. I've modified your code a bit so I can choose between a polar mesh or a cartesian mesh, and so I can optionally specify the centroid and variance:
Code:
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
from matplotlib import cm
from matplotlib import pyplot as plt
from scipy import random
import argparse

parser = argparse.ArgumentParser(description='Plot a 2D Gaussian.')
parser.add_argument(
    '-p', '--polar',
    help = 'plot with polar X,Y mesh (default: cartesian mesh)',
    action = 'store_true')
parser.add_argument(
    '-c', '--centroid',
    help = 'centroid coordinates (default: random)',
    type = float,
    nargs = 2)
parser.add_argument(
    '-s', '--sigma',
    help = 'variance (sigma) (default: random)',
    type = float,
    nargs = 1)
args = parser.parse_args()

if args.polar :
    r = np.linspace(0.0,20.0,100.0)
    p = np.linspace(0.0,2*np.pi,100.0)
    R,P = np.meshgrid(r,p)
    X,Y = R*np.cos(P),R*np.sin(P)
else :
    x = np.linspace(-20.0,20.0,101.0)
    y = np.linspace(-20.0,20.0,101.0)
    X,Y = np.meshgrid(x,y)

if args.centroid is not None :
    x0,y0 = args.centroid
else :
    x0,y0 = random.uniform(-4.0,4.0), random.uniform(-4.0,4.0)

if args.sigma is not None :
    sigma0 = args.sigma[0]
else :
    sigma0 = random.uniform(4.0, 7.0)

Z = np.e**(-((X-x0)**2 + (Y-y0)**2)/(2*sigma0**2))

fig = plt.figure()
ax = Axes3D(fig)
ax.plot_surface(X, Y, Z, rstride=2, cstride=2, cmap=cm.jet)
ax.set_zlim3d(0,1)
ax.set_zlabel('Intensity')
plt.show()

I get the following plot when I run this via python plot.py --centroid -5 5:View attachment 83929Using a polar mesh (python plot.py --centroid -5 5 --polar) yields the following:
View attachment 83930


Thank you so much for the help!
 
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