Orthogonal Latin Hypercube Sampling

olhs_sampling.py

"""Orthogonal Latin Hypercube Sampling.

---------------------------

MIT License

Copyright (c) 2021 Pamphile Tupui ROY

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"""
import functools

import numpy as np

from scipy import spatial
from scipy.stats import qmc
from scipy import optimize

import matplotlib.pyplot as plt
import seaborn as sns


def oa_lhs(p, d, seed):

    oa_sample = np.zeros(shape=(p**2, p+1))

    arrays = np.tile(np.arange(p), (2, 1))
    oa_sample[:, :2] = np.stack(np.meshgrid(*arrays), axis=-1).reshape(-1, 2)
    for p_ in range(1, p):
        oa_sample[:, 2+p_-1] = np.mod(oa_sample[:, 0] + p_*oa_sample[:, 1], p)

    # scramble the OA
    oa_sample_ = np.empty(shape=(p**2, p+1))
    for j in range(p+1):
        perms = np.random.permutation(p)
        for k in range(p):
            idx = np.where(oa_sample[:, j] == k)[0]
            oa_sample_[idx, j] = perms[k]
    
    oa_sample = oa_sample_

    # sample is a randomized OA from now
    # and the following is making it an OA-LHS

    oa_lhs_sample = np.zeros(shape=(p**2, p+1))
    for j in range(p+1):
        for k in range(p):
            idx = np.where(oa_sample[:, j] == k)[0]
            lhs = LatinHypercube(d=1, centered=True, seed=seed).random(p).flatten()
            oa_lhs_sample[:, j][idx] = lhs + oa_sample[:, j][idx]

    oa_lhs_sample /= p

    if d is not None:
        oa_lhs_sample = oa_lhs_sample[:, :d]
    return oa_lhs_sample


rng = np.random.default_rng()
p, d = 7, 3
oa_lhs_sample = oa_lhs(p, d, seed=rng)

sns.pairplot(pd.DataFrame(oa_lhs_sample), diag_kind="hist", corner=True, diag_kws={"bins": p})

Thanks for your inputs @kstoneriv3. I opened a PR if you did not see scipy/scipy#14546. Your test to check that we have an OA would be a nice addition. I will see how I can make this work. Or you're free to propose something over there.