n dimensional grid in Python / numpy

Question

I have an unknown number n of variables that can range from 0 to 1 with some known step s, with the condition that they sum up to 1. I want to create a

Answer 1

EDIT

Here is a better solution. It basically partitions the number of steps into the amount of variables to generate all the valid combinations:

def partitions(n, k):
    if n < 0:
        return -partitions(-n, k)
    if k <= 0:
        raise ValueError('Number of partitions must be positive')
    if k == 1:
        return np.array([[n]])
    ranges = np.array([np.arange(i + 1) for i in range(n + 1)])
    parts = ranges[-1].reshape((-1, 1))
    s = ranges[-1]
    for _ in range(1, k - 1):
        d = n - s
        new_col = np.concatenate(ranges[d])
        parts = np.repeat(parts, d + 1, axis=0)
        s = np.repeat(s, d + 1) + new_col
        parts = np.append(parts, new_col.reshape((-1, 1)), axis=1)
    return np.append(parts, (n - s).reshape((-1, 1)), axis=1)

def make_grid_part(n, step):
    num_steps = round(1.0 / step)
    return partitions(num_steps, n) / float(num_steps)

print(make_grid_part(3, 0.33333))

Output:

array([[ 0.        ,  0.        ,  1.        ],
       [ 0.        ,  0.33333333,  0.66666667],
       [ 0.        ,  0.66666667,  0.33333333],
       [ 0.        ,  1.        ,  0.        ],
       [ 0.33333333,  0.        ,  0.66666667],
       [ 0.33333333,  0.33333333,  0.33333333],
       [ 0.33333333,  0.66666667,  0.        ],
       [ 0.66666667,  0.        ,  0.33333333],
       [ 0.66666667,  0.33333333,  0.        ],
       [ 1.        ,  0.        ,  0.        ]])

For comparison:

%timeit make_grid_part(5, .1)
>>> 338 µs ± 2.25 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit make_grid_simple(5, .1)
>>> 26.4 ms ± 806 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

make_grid_simple actually runs out of memory if you push it just a bit further.

---

Here is one simple way:

def make_grid_simple(n, step):
    num_steps = round(1.0 / step)
    vs = np.meshgrid(*([np.linspace(0, 1, num_steps + 1)] * n))
    all_combs = np.stack([v.flatten() for v in vs], axis=1)
    return all_combs[np.isclose(all_combs.sum(axis=1), 1)]

print(make_grid_simple(3, 0.33333))

Output:

[[ 0.          0.          1.        ]
 [ 0.33333333  0.          0.66666667]
 [ 0.66666667  0.          0.33333333]
 [ 1.          0.          0.        ]
 [ 0.          0.33333333  0.66666667]
 [ 0.33333333  0.33333333  0.33333333]
 [ 0.66666667  0.33333333  0.        ]
 [ 0.          0.66666667  0.33333333]
 [ 0.33333333  0.66666667  0.        ]
 [ 0.          1.          0.        ]]

However, this is not the most efficient way to do it, since it is simply making all the possible combinations and then just picking the ones that add up to 1, instead of generating only the right ones in the first place. For small step sizes, it may incur in too high memory cost.