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})

@kstoneriv3 thanks for the feedbacks!

How about using below in the scrambling instead of using for loop? This fancy indexing runs OK in my local python environment.

    # scramble the OA
    perms = np.random.permutation(p)
    oa_sample_ = perms[oa_sample]

I was not sure about this because it would mean that the permutations are not independent. So we change all the 0 for a 1 in all columns. Whereas the current version changes only one column at a time. If we go with your version, I don't see how this is doing any scrambling. In the end we just renamed the symbols if we do this.

• Are we assuming that d<=p+1?

Yes.

• In line 71,

    if d is not None:
        idx = seed.permutation(p+1)[:d]
        oa_lhs_sample = oa_lhs_sample[:, idx]

feels more natural to me as all the columns have equal probability to be selected.

Totally, it's a mistake of mine. Thanks.

• The code for generating an orthogonal array might be broken, as there does not seem to be exactly one point in each cell of 7x7 grid. There should be test for making sure that the generated samples are actually OAs of strength 2.

I could not run your snippet.

I was not sure about this because it would mean that the permutations are not independent. So we change all the 0 for a 1 in all columns. Whereas the current version changes only one column at a time. If we go with your version, I don't see how this is doing any scrambling. In the end, we just renamed the symbols if we do this.

Ah, you are right! I now have an intuition that permutations should be independently sampled for each column. Then the code near line 71 does not need any fix.

I could not run your snippet.

If runs once you define oa_lhs_sample and p.

import numpy as np
# define OA
p = 2
oa_lhs_sample = np.array([[0, 0, 1, 1], [0, 1, 0, 1], [0, 1, 1, 0]]).T / 2 + 0.25

for i in range(p):
    for j in range(i+1, p):
        samples_2d = oa_lhs_sample[:, [i, j]]
        cart_prod = ((samples_2d * p) // 1).astype(int)  # should be an cartesian product of [0, 1, ..., p-1] x [0, 1, ..., p-1] 
        count_cells = np.zeros([p, p])
        count_cells[cart_prod] += 1  # all pairs of 0 to p-1 must be counted once
        assert np.all(count_cells == 1)

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.