Generating Latin hypercube samples with numpy

Question

I wrote some code to generate Latin hypercube samples for interpolation over high-dimensional parameter spaces. Latin hypercubes are essentially collections of points on a hypercube that are placed on a cubic/rectangular grid, which possess the property that no two points share any individual coordinate, and every row/column/higher-dimensional-axis is sampled once. e.g., for 2 dimensions and 4 total samples, this is a Latin hypercube:

| | | |x|
| |x| | |
|x| | | |
| | |x| |

One algorithm for generating this is to generate D random permutations of the integers 0 through N-1, where D is the number of dimensions and N is the desired number of samples. Concatenating these together into a DxN matrix gives a list of coordinates which will form a Latin hypercube.

I am also interested in generating a special type of Latin hypercube called a symmetric Latin hypercube. This property essentially means that the hypercube is invariant under spatial inversion. One way of expressing this condition, is that (if the samples are zero-indexed) if the hypercube has the sample (i, j, ..., k), for some integer indices i,j,k, then it also has the sample (n-1-i, n-1-j, ..., n-1-k), where n is the number of samples. For 2D and n=4, this is an example of a symmetric Latin hypercube:

| |x| | |
| | | |x|
|x| | | |
| | |x| |

The condensed version of why I want to do this is that I want to evaluate a function which is expensive to compute, and depends on many parameters. I pre-generate a list of values for this function at a set of points and use interpolation over the parameters to extend this list to any desired point within a set range. Latin hypercube samples are used for the pre-computation instead of a grid, with the idea that they will more efficiently sample the behavior of the function and result in lower interpolation errors.

I have tested the code below both using automated tests and by visual inspection of the results for the trivial 2D case. Statistical tests of higher dimensional cases have convinced me it works successfully there as well. The code can produce either random Latin hypercubes or a rather nongeneral set of symmetric ones (I am convinced my second algorithm cannot in principle generate any possible symmetric Latin hypercube, just a specific type). Also, the symmetric algorithm doesn't really work properly for odd numbers of samples, so I have outright blocked such samples from being generated.

My code can be seen along with specification and performance tests here, although not the visual tests at the moment.

I am aware that my project structure is probably overkill for the amount of code here. I am mostly looking for any insights regarding my algorithms, as while they are very fast even for large dimensions and sample sizes, I don't doubt improvements exist. However any comments are welcome.

import numpy as np


class LatinSampler:
    def _check_num_is_even(self, num):
        if num % 2 != 0:
            raise ValueError("Number of samples must be even")

    def get_lh_sample(self, param_mins, param_maxes, num_samples):
        dim = param_mins.size

        latin_points = np.array([np.random.permutation(num_samples) for i in range(dim)]).T

        lengths = (param_maxes - param_mins)[None, :]
        return lengths*(latin_points + 0.5)/num_samples + param_mins[None, :]

    def get_sym_sample(self, param_mins, param_maxes, num_samples):
        self._check_num_is_even(num_samples)

        dim = param_mins.size

        even_nums = np.arange(0, num_samples, 2)
        permutations = np.array([np.random.permutation(even_nums) for i in range(dim)])
        inverses = (num_samples - 1) - permutations

        latin_points = np.concatenate((permutations,inverses), axis=1).T

        lengths = (param_maxes - param_mins)[None, :]
        return  lengths*(latin_points + 0.5)/num_samples + param_mins[None, :]

Edit: I glossed over this, but this code doesn't return Latin hypercubes per se, but rather Latin hypercubes stretched/squeezed and shifted so that the points can be over arbitrary ranges of values. However the core code will generate a Latin hypercube if the last manipulation (multiply by lengths and add param_mins) is omitted.

Answer 1

Once I understood what you wanted to accomplish, your code became a lot more clear to me and I can provide some thoughts.

Regarding the symmetrical version, you are indeed only sampling from a subset of the available symmetrical latin hypercubes. Half of the parameter space will never be chosen as samples. Consider the 2D example with n=4:

even_nums = [0, 2] and, hence, A,B,C,D are the 4 possible samples that can be generated with this method. In fact there's two possible Latin Hypercubes: [A,D] and [B,C]. A',B',C',D' denote their respective inverse positions that are then added. As you can see, half of the fields are empty and will never be chosen as a sample.

Instead of permuting indexes, I would sample pairs of symmetrical points directly, as it allows you to control for the constraint while still achieving uniform sampling:

import random
import numpy as np

num_samples = 5
dim = 2

available_indices = [set(range(num_samples)) for _ in range(dim)]
samples = []

# if num_samples is odd, we have to chose the midpoint as a sample
if num_samples % 2 != 0:
    k = num_samples//2
    samples.append([k] * dim)
    for idx in available_indices:
        idx.remove(k)

# sample symmetrical pairs
for _ in range(num_samples//2):
    sample1 = list()
    sample2 = list()

    for idx in available_indices:
        k = random.sample(idx, 1)[0]
        sample1.append(k)
        sample2.append(num_samples-1-k)
        idx.remove(k)
        idx.remove(num_samples-1-k)

    samples.append(sample1)
    samples.append(sample2)

samples = np.array(samples)

Answer 2

You can use the LatinHypercube class which I added to SciPy in version 1.7. This can replace most of the code here.

To generate a LHS, one simply can generate a uniform sample U(0, 1) and shift it with random permutations of integers, like this:

import numpy as np

n, d = ...
rng = np.random.default_rng()
samples = rng.uniform(size=(n, d))
perms = np.tile(np.arange(1, n + 1), (d, 1))
for i in range(d):
    rng.shuffle(perms[i, :])
perms = perms.T

samples = (perms - samples) / n