How to create a multi-dimensional grid in python

Question

I have seen similar questions but none that need the format of the output array of shape (numpoints, dim)

Here is a simple example of what I have for dim=2

import numpy as np

bounds = [0.5, 0.5]
n = [10,10]
dim = 2
x = np.linspace(-bounds[0], bounds[0], n[0])
y = np.linspace(-bounds[1], bounds[1], n[1])
X, Y = np.meshgrid(x, y)

s = X.shape
data = np.zeros((n[0]*n[1],dim)) 

# convert mesh into point vector for which the model can be evaluated
c = 0
for i in range(s[0]):
    for j in range(s[1]):
        data[c,0] = X[i,j]
        data[c,1] = Y[i,j]
        c = c+1;
plt.scatter(data[:,0], data[:,1])

Is there a faster/better way of doing this so that the data are arranged in this way? I want a general method that could work for any dim.

Edit: Suggested answer does not work.

Does this answer your question? How to convert the output of meshgrid to the corresponding array of points? — Michael Szczesny
– Michael Szczesny, Commented Aug 11, 2022 at 15:00
@MichaelSzczesny when I use the suggested 'meshgrid2' function I get the error 'map' object not subscriptable. So yes, the question is similar — Kyriacos Xanthos
– Kyriacos Xanthos, Commented Aug 11, 2022 at 15:16

Peter · Accepted Answer · 2022-08-11 18:26:54Z

1

Yeah, that can be vectorized with

axis_coords = np.meshgrid(x, y, indexing='xy')
data = np.hstack([c.reshape(-1, 1) for c in axis_coords])

c.reshape(-1, 1) just reshapes c from (HxW to (H*W)x1) so that it can be stacked horizontally.

Note - if you're looking to generalize to more dims you probably want to switch to indexing='ij' so it's arranged by (row, column, dim2, dim3, ...) rather than (column, row, dim2, dim3, ...) since in numpy rows are considered the 0'th dimension and columns the 1st.

edited Aug 11, 2022 at 18:26

answered Aug 11, 2022 at 18:20

Peter

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2 Comments

Kyriacos Xanthos Over a year ago

Thanks for your answer, but I think this is not general enough for any dimension. I posted my answer, take a look at it and let me know.

Peter Over a year ago

Looks good to me - just watch out for the indexing='ij' vs indexing='xy' thing when generalising to higher dims.

Kyriacos Xanthos · Accepted Answer · 2022-08-12 07:10:18Z

1

I managed to solve my problem with this function that is general enough for any dim:

def get_grid_of_points(n, *args):
    ls = [np.linspace(-i,i,n) for i in args]
    mesh_ls = np.meshgrid(*ls)
    all_mesh = [np.reshape(x, [-1]) for x in mesh_ls]
    grid_points = np.stack(all_mesh, axis=1)
    return grid_points

get_grid_of_points(10, 0.5, 0.5)

answered Aug 12, 2022 at 7:10

Kyriacos Xanthos

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