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.

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

MIT License

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]]

I am experiencing a performance leak when computing the Halton sequences with Python 3.6.7 and NumPy 1.13.3. The issue is triggered because the function halton is passing np.int64 instances to the base argument from function van_der_corput, and the builtin divmod inside van_der_corput seems to be very inefficient if not used with native Python integers.

An example of what IPython3 returns in my laptop for divmod:

In [2]: native_two, numpy_two = 2, np.int64(2)

In [3]: %timeit divmod(10, native_two)
The slowest run took 25.93 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 89.8 ns per loop

In [4]: %timeit divmod(10, numpy_two)
The slowest run took 68.61 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 222 ns per loop

Because of this performance leak, I see these timings for the van_der_corput function when trying to generate 100000 samples:

In [5]: %timeit van_der_corput(100000, native_two)
1 loop, best of 3: 325 ms per loop

In [6]: %timeit van_der_corput(100000, numpy_two)
1 loop, best of 3: 3.58 s per loop

And finally, the halton function is suffering the consequences to generate e.g. 100000 samples from a 5D-space:

In [7]: %timeit halton(5, 100000)
1 loop, best of 3: 10.6 s per loop

I could workaround this issue with two different possibilities:

• change line 81 with base = primes_from_2_to(big_number)[:dim].tolist(), to convert the array into a list of native Python integers, or
• force the argument types before line 59 with n_sample, base = int(n_sample), int(base) (I find this better because the performance leak is inside van_der_corput, not in halton).

I decided to apply the second solution, and after the change the performance leak is not there anymore:

In [9]: %timeit halton(5, 100000)
1 loop, best of 3: 959 ms per loop

I do not know if this issue can be reproduced with other versions, but at least I found it interesting enough to be commented. I hope this can be useful to others.

@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