What is the purpose of meshgrid in Python / NumPy?

Question

Can someone explain to me what is the purpose of meshgrid function in Numpy? I know it creates some kind of grid of coordinates for plotting, but I can\'t reall

Answer 1

Suppose you have a function:

def sinus2d(x, y):
    return np.sin(x) + np.sin(y)

and you want, for example, to see what it looks like in the range 0 to 2*pi. How would you do it? There np.meshgrid comes in:

xx, yy = np.meshgrid(np.linspace(0,2*np.pi,100), np.linspace(0,2*np.pi,100))
z = sinus2d(xx, yy) # Create the image on this grid

and such a plot would look like:

import matplotlib.pyplot as plt
plt.imshow(z, origin='lower', interpolation='none')
plt.show()

So np.meshgrid is just a convenience. In principle the same could be done by:

z2 = sinus2d(np.linspace(0,2*np.pi,100)[:,None], np.linspace(0,2*np.pi,100)[None,:])

but there you need to be aware of your dimensions (suppose you have more than two ...) and the right broadcasting. np.meshgrid does all of this for you.

Also meshgrid allows you to delete coordinates together with the data if you, for example, want to do an interpolation but exclude certain values:

condition = z>0.6
z_new = z[condition] # This will make your array 1D

so how would you do the interpolation now? You can give x and y to an interpolation function like scipy.interpolate.interp2d so you need a way to know which coordinates were deleted:

x_new = xx[condition]
y_new = yy[condition]

and then you can still interpolate with the "right" coordinates (try it without the meshgrid and you will have a lot of extra code):

from scipy.interpolate import interp2d
interpolated = interp2d(x_new, y_new, z_new)

and the original meshgrid allows you to get the interpolation on the original grid again:

interpolated_grid = interpolated(xx[0], yy[:, 0]).reshape(xx.shape)

These are just some examples where I used the meshgrid there might be a lot more.