pythonstatisticsskopt

Can one constrain the outcome of skopt.Lhs.generate?


Let's say I have an array like this:

from skopt.space import Space
from skopt.sampler import Lhs
import numpy as np

np.random.seed(42)

rows = 5
cols = 3

dummy = np.zeros((rows, cols))

array([[0., 0., 0.],
       [0., 0., 0.],
       [0., 0., 0.],
       [0., 0., 0.],
       [0., 0., 0.]])

and I now would like to use skopt.Lhs.generate to fill certain positions of this array with a 1 whereby I would like to exclude certain positions stored in ignore:

ignore = np.array([
    [3, 1],
    [4, 1]
])

How would I do this best?

I can do

space = Space([(0, rows - 1), (0, cols - 1)])

lhs = Lhs(criterion="maximin", iterations=1000)
lh = np.array(lhs.generate(space.dimensions, 3))

dummy[lh[:, 0], lh[:, 1]] = 1

which gives

array([[0., 0., 1.],
       [1., 0., 0.],
       [0., 0., 0.],
       [0., 0., 0.],
       [0., 1., 0.]])

but as one can see the position 4, 1 is occupied but it shouldn't.

One way could be to put the lhs.generate call inside a while loop and then always checks whether any element is in ignore but I am wondering whether there is a more straightforward solution to this.


Solution

  • Just for completeness, a solution using a while loop could look like this:

    from skopt.space import Space
    from skopt.sampler import Lhs
    import numpy as np
    
    def contains_element(a, b):
        return any(li in a for li in b)
    
    np.random.seed(42)
    
    rows = 5
    cols = 3
    
    dummy = np.zeros((rows, cols))
    
    ignore = [
        [3, 1],
        [4, 1]
    ]
    
    space = Space([(0, rows - 1), (0, cols - 1)])
    
    contains_row = True
    while contains_row:
    
        lhs = Lhs(criterion="maximin", iterations=1000)
        lh = lhs.generate(space.dimensions, 3)
        contains_row = contains_element(lh, ignore)
        print(lh)
        print(contains_row)
        print('-----------------------\n')
    
    lh = np.array(lh)
    dummy[lh[:, 0], lh[:, 1]] = 1
    
    print(dummy)
    

    which prints

    [[2, 0], [0, 2], [4, 1]]
    True
    -----------------------
    
    [[2, 0], [1, 2], [4, 1]]
    True
    -----------------------
    
    [[4, 2], [0, 1], [2, 0]]
    False
    -----------------------
    
    [[0. 1. 0.]
     [0. 0. 0.]
     [1. 0. 0.]
     [0. 0. 0.]
     [0. 0. 1.]]
    

    So, here it takes 3 iterations until positions are found which are not in ignore. If anyone knows a more straightforward solution, please let me know!