Simple implementation of Latin Hypercube Sampling.
The example below shows how to sample from a random Latin Hypercube Design with five points for three inputs.
from simplelhs import LatinHypercubeSampling
lhs = LatinHypercubeSampling(3)
hc = lhs.random(5)
print(hc)
[[0.65830165 0.26660356 0.78491755]
[0.42168063 0.43244666 0.979281 ]
[0.39058169 0.76099351 0.34764726]
[0.07122137 0.15507069 0.58082752]
[0.87530571 0.94575193 0.03949576]]
The example below shows how to sample from a Maximin Latin Hypercube Design with five points for three inputs. Out of 1000 randomly sampled Latin Hypercube samples the sample with the maximal minimal distance between points is selected.
from simplelhs import LatinHypercubeSampling
lhs = LatinHypercubeSampling(3)
hc = lhs.maximin(5, 1000)
print(hc)
[[0.24607101 0.11399068 0.5456922 ]
[0.88731638 0.40600431 0.32305333]
[0.47416121 0.99487745 0.03087923]
[0.06288706 0.7227211 0.78248764]
[0.77081332 0.36862214 0.99449703]]
To scale the data to unit cube and back to its original range the functions normalise()
and unnormalise()
are provided. The example below scales the Maximin Latin Hypercube sample back to its original range.
from simplelhs import unnormalise
lower = np.array([0., -5., 10.,])
upper = np.array([1., 5., 40.])
hc_maximin_scaled = unnormalise(hc_maximin, lower, upper)
print(hc_maximin_scaled)
[[ 0.24607101 -3.86009322 26.37076609]
[ 0.88731638 -0.93995687 19.69159997]
[ 0.47416121 4.94877447 10.92637679]
[ 0.06288706 2.22721099 33.47462916]
[ 0.77081332 -1.31377864 39.83491091]]