Halton Sequence in python

halton.py

"""Halton low discrepancy sequence.

This snippet implements the Halton sequence following the generalization of
a sequence of *Van der Corput* in n-dimensions.

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

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Copyright (c) 2017 Pamphile Tupui ROY

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"""
import numpy as np


def primes_from_2_to(n):
    """Prime number from 2 to n.

    From `StackOverflow <https://stackoverflow.com/questions/2068372>`_.

    :param int n: sup bound with ``n >= 6``.
    :return: primes in 2 <= p < n.
    :rtype: list
    """
    sieve = np.ones(n // 3 + (n % 6 == 2), dtype=np.bool)
    for i in range(1, int(n ** 0.5) // 3 + 1):
        if sieve[i]:
            k = 3 * i + 1 | 1
            sieve[k * k // 3::2 * k] = False
            sieve[k * (k - 2 * (i & 1) + 4) // 3::2 * k] = False
    return np.r_[2, 3, ((3 * np.nonzero(sieve)[0][1:] + 1) | 1)]


def van_der_corput(n_sample, base=2):
    """Van der Corput sequence.

    :param int n_sample: number of element of the sequence.
    :param int base: base of the sequence.
    :return: sequence of Van der Corput.
    :rtype: list (n_samples,)
    """
    sequence = []
    for i in range(n_sample):
        n_th_number, denom = 0., 1.
        while i > 0:
            i, remainder = divmod(i, base)
            denom *= base
            n_th_number += remainder / denom
        sequence.append(n_th_number)

    return sequence


def halton(dim, n_sample):
    """Halton sequence.

    :param int dim: dimension
    :param int n_sample: number of samples.
    :return: sequence of Halton.
    :rtype: array_like (n_samples, n_features)
    """
    big_number = 10
    while 'Not enought primes':
        base = primes_from_2_to(big_number)[:dim]
        if len(base) == dim:
            break
        big_number += 1000

    # Generate a sample using a Van der Corput sequence per dimension.
    sample = [van_der_corput(n_sample + 1, dim) for dim in base]
    sample = np.stack(sample, axis=-1)[1:]

    return sample


print(van_der_corput(10))
# [0.0, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875, 0.0625, 0.5625]
print(halton(2, 5))
# [[ 0.5         0.33333333]
#  [ 0.25        0.66666667]
#  [ 0.75        0.11111111]
#  [ 0.125       0.44444444]
#  [ 0.625       0.77777778]]

@molinav Thanks for finding this out! I am currently working on a PR to have this in SciPy. So this is useful.

hello @tupui! How can I use this to generate Halton sequence in a given rectangle? :)

Hi @jdavidd, once you generate a sample, you just have to scale the values from [0, 1) to [a, b), b>a the bounds you want.
For instance if you have two parameters the first range is [-2, 6] and the second [0, 5]:

bounds = [[-2, 0], [6, 5]]
bounds = np.array(bounds)
min_ = np.min(bounds, axis=0)
max_ = np.max(bounds, axis=0)
sample = sample * (max_ - min_) + min_

But have a look at the PR I have in scipy for more stuff like discrepancy: scipy/scipy#10844