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?

@MichaelSzczesny it actually does what I wanted to do. (But it turns out it's too memory heavy lol.) — JustAnEuropean, Mar 28 '22 at 19:07
Change the `dtype` to `np.uint8`? `np.indices((100,100,100), dtype=np.uint8).reshape(3,-1).T` — Michael Szczesny, Mar 28 '22 at 19:09
@MichaelSzczesny I've tried that but it's still not enough for 10 dimensions and a side_len of 10. I think I misunderstand the problem and have asked the wrong question, but I believe your answer to it is correct. I still haven't understood Tim's answer. — JustAnEuropean, Mar 28 '22 at 19:12

Tim Roberts · Accepted Answer · 2022-03-28T20:16:33.873

2

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.

edited Mar 28 '22 at 20:16

answered Mar 28 '22 at 18:42

Tim Roberts

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Hey, thanks, but I have run this code and it doesn't look like it outputs the right thing. After I changed `a = np.arange(0,100,101)` to `a = np.arange(0,100,1)` (because otherwise it was just [0]) I got a single column of 3 times repeated values of 0..99. – JustAnEuropean Mar 28 '22 at 19:36
Oops, I actually used `linspace`, not `arange`, and I used too many parens in the `meshgrid` call. Fixed now. Your `arange` should also work. It will produce integers instead of floats. – Tim Roberts Mar 28 '22 at 20:15

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

1 Answers1