## How to plot a 2d gaussian with different sigma?

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I am trying to make and plot a 2d gaussian with two different standard deviations. They give the equation on mathworld: http://mathworld.wolfram.com/GaussianFunction.html but I can't seem to get a proper 2D array which centers it around zero.

I got this, but it does not quite work.

x = np.array([np.arange(size)])
y = np.transpose(np.array([np.arange(size)]))

psf  = 1/(2*np.pi*sigma_x*sigma_y) * np.exp(-(x**2/(2*sigma_x**2) + y**2/(2*sigma_y**2)))


Your function is centred on zero but your coordinate vectors are not. Try:

size = 100
sigma_x = 6.
sigma_y = 2.

x = np.linspace(-10, 10, size)
y = np.linspace(-10, 10, size)

x, y = np.meshgrid(x, y)
z = (1/(2*np.pi*sigma_x*sigma_y) * np.exp(-(x**2/(2*sigma_x**2)
+ y**2/(2*sigma_y**2))))

plt.contourf(x, y, z, cmap='Blues')
plt.colorbar()
plt.show()


Visualizing the bivariate Gaussian distribution, The multivariate Gaussian distribution of an $n$-dimensional vector in Matplotlib v.1.4 – you can replace it with any other sane colormap, such as mu, Sigma) # Create a surface plot and projected filled contour plot under� sigma = abs (peaks (100))+0.1; % Preallocate: If2 = NaN (size (I)); % Loop through each row and column: for row = 1:size (I,1) for col = 1:size (I,2) % Perform Gaussian filter on the entire grid using sigma (row,col): tmp = imgaussfilt (I,sigma (row,col)); % Only save the value corresponding to this row and col:

Your Gaussian is centered on (0,0) so set up the axes around this origin. For example,

In : size = 200

In : sigma_x,sigma_y = 50, 20

In : x = np.array([np.arange(size)]) - size/2
In : y = np.transpose(np.array([np.arange(size)])) - size /2
In : psf  = 1/(2*np.pi*sigma_x*sigma_y) *
np.exp(-(x**2/(2*sigma_x**2) + y**2/(2*sigma_y**2)))

In : pylab.imshow(psf)
Out: <matplotlib.image.AxesImage at 0x10bc07f10>

In : pylab.show() 2d gaussian function - MATLAB Answers, I need to plot a 2d gaussian function, where x and y corresponds to the image pixels, my code uses a nested for loop which makes my program function mat = gauss2d(mat, sigma, center) The above is very much different than Alan's "1. How can I plot the 1$\sigma$ contour of the 2D Gaussian? It encompasses $\approx 0.39$ of the volume under the surface (being analogous to the 0.68 fraction marking the 1$\sigma$ region in the 1D Gaussian case). So, the precise questions are (a) at what height of the PDF should I place the 1$\sigma$ contour?

Probably this answer is too late for @Coolcrab , but I would like to leave it here for future reference.

You can use a multivariate Gaussian formula as follows changing the mean elements changes the origin, while changing the covariance elements changes the shape (from circle to ellipse). Here is the code:

import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D

# Our 2-dimensional distribution will be over variables X and Y
N = 40
X = np.linspace(-2, 2, N)
Y = np.linspace(-2, 2, N)
X, Y = np.meshgrid(X, Y)

# Mean vector and covariance matrix
mu = np.array([0., 0.])
Sigma = np.array([[ 1. , -0.5], [-0.5,  1.]])

# Pack X and Y into a single 3-dimensional array
pos = np.empty(X.shape + (2,))
pos[:, :, 0] = X
pos[:, :, 1] = Y

def multivariate_gaussian(pos, mu, Sigma):
"""Return the multivariate Gaussian distribution on array pos."""

n = mu.shape
Sigma_det = np.linalg.det(Sigma)
Sigma_inv = np.linalg.inv(Sigma)
N = np.sqrt((2*np.pi)**n * Sigma_det)
# This einsum call calculates (x-mu)T.Sigma-1.(x-mu) in a vectorized
# way across all the input variables.
fac = np.einsum('...k,kl,...l->...', pos-mu, Sigma_inv, pos-mu)

return np.exp(-fac / 2) / N

# The distribution on the variables X, Y packed into pos.
Z = multivariate_gaussian(pos, mu, Sigma)

# plot using subplots
fig = plt.figure()
ax1 = fig.add_subplot(2,1,1,projection='3d')

ax1.plot_surface(X, Y, Z, rstride=3, cstride=3, linewidth=1, antialiased=True,
cmap=cm.viridis)
ax1.view_init(55,-70)
ax1.set_xticks([])
ax1.set_yticks([])
ax1.set_zticks([])
ax1.set_xlabel(r'$x_1$')
ax1.set_ylabel(r'$x_2$')

ax2 = fig.add_subplot(2,1,2,projection='3d')
ax2.contourf(X, Y, Z, zdir='z', offset=0, cmap=cm.viridis)
ax2.view_init(90, 270)

ax2.grid(False)
ax2.set_xticks([])
ax2.set_yticks([])
ax2.set_zticks([])
ax2.set_xlabel(r'$x_1$')
ax2.set_ylabel(r'$x_2$')

plt.show()


Matplotlib — 0.1.0 documentation, The example below illustrates plotting several lines with different format styles in one command plt.title('Gaussian distribution') plt.text(-2, 0.45, r'$\mu=0,\ \sigma =1$') plt.xlim(-4, Make a contour plot of a single 2d gaussian centred on 0,0. The graph of the 2D Gaussian function is obtained by rotating the 1D function graphs around the vertical $$z$$-axis. Isotropic Gaussian is separable into y and x components. Thus, applying 2d Gaussian blur with sigma = to 'ideal' slanted edge is similar to applying 1d Gaussian blur with sigma = to every scan line, where k is the edge slope.

NumPy: Generate a generic 2D Gaussian-like array, Write a NumPy program to generate a generic 2D Gaussian-like array. g = np. exp(-( (d-mu)**2 / ( 2.0 * sigma**2 ) ) ) print("2D Gaussian-like� This shows an example of a gaussian distribution with various parameters. Make a contour plot of a single 2d gaussian centred on 0,0. A scatter plot is a type of plot that shows the data as a collection of points. contour2D extends R's contour function.

Bivariate Gaussian — astroML 0.4 documentation, An example of data generated from a bivariate Gaussian distribution. Draw 10^ 5 points from a multivariate normal distribution # # we use the bins[-1], bins[1 ], bins[-1]]) # draw 1, 2, 3-sigma ellipses over the distribution for N in (1, 2,� Gaussian processes can also be used in the context of mixture of experts models, for example. Implementations of the robust point set registration algorithm described in "Robust Point Set Registration Using Gaussian Mixture Models", Bing Jian and Baba C. Obtain 3D Point Cloud 2. The 2D Gaussian code can optionally fit a tilted Gaussian.

Plotting a Gaussian normal curve with Python and Matplotlib , A mean resistance of 979.8 kΩ and a standard deviation of 73.10 kΩ represents the sample of resistors. The desired resistance tolerance for the 1� 2d gaussian function. % The above is very much different than Alan's "1./2*pi*sigma^2" The function has to give a final plot of the gaussian bump using the

##### Comments
• Your x and y axes start at 0, so you only get a quarter of your 2D Gaussian (that falling in the positive xy quadrant. Try subtracting size/2 from each array.