How to create a uniform grid of n-dimensional vectors using numpy (in other words, uniformly fill a n-dimensional hypercube)?

Question

In one dimension a "grid" would be an array of numbers between let's say 0 and 100. In two dimensions the grid would be an array of points like [0, 0] [0, 1] [0, 2]... [1,0], [1, 1] ... [99, 99]. In three dimensions and more dimensions it would look similar.

My current output is like that:

It doesn't create every combination of values for each value in the nth - 1 column.

The code I use is:

import numpy as np

class Cube:
    side_len = 100
    def __init__(self, n):
        current_point = np.zeros(n)
        self.arr = []
        for i in range(n):
            for j in range(Cube.side_len):
                self.arr.append(current_point.copy())
                current_point[i] += 1.0
        self.arr.append([Cube.side_len for _ in range(n)])
        self.arr = np.array(self.arr)
        np.random.shuffle(self.arr)

if __name__ == '__main__':
    cube(10)

I tried also with meshgrid but I could not understand the documentation. I wanted it to be a shallow list of points but I get X, Y and I don't get what I am supposed to do with that?

Answer 1

Here's the way you do it. Meshgrid with 3 dimensions returns a list of three things, which are the values for the 3 axes to get a uniform spread of points. You can then use vstack to stack those together, and transpose to get a list of 3D coordinates:

>>> import numpy as np
>>> a = np.linspace(0,100,101)
>>> x = np.meshgrid( a, a, a )
>>> y = np.vstack(list(map(np.ravel,x))).T
>>> y
array([[  0.,   0.,   0.],
       [  0.,   0.,   1.],
       [  0.,   0.,   2.],
       ...,
       [100., 100.,  98.],
       [100., 100.,  99.],
       [100., 100., 100.]])
>>>

Credit to this post: How to convert the output of meshgrid to the corresponding array of points?

Remember that the grid is the size of one axis cubed, so these very quickly get large.