Call Python skopt package from Julia

General Usage
1

When I try to call python skopt from Julia, I find a weird problem, here is a simple example that can reflect my question, if I define all variable range with Real, it will cause error in res = opt.tell(suggested, y) however, once I have one variable range defined with Integer, then it works well, where is the problem?

function blackbox(para)
    return (para[1]^2 - para[2])^2 + 2
end

using PyCall
skopt = pyimport("skopt")
py"""
from skopt.space import Real,Integer
space =[Real(-10, 10),
        Integer(-10, 10)]      # if use `Real(-10,10)`, it will cause error in  `res = opt.tell(suggested, y)`
                                        # but  if I use  Integer(-10, 10), it will run successfully, why?
"""

opt = skopt.Optimizer(py"space", n_initial_points=4, acq_func="EI", acq_optimizer="sampling")
suggested = opt.ask()
display(suggested)
y = blackbox(suggested)
display(y)
res = opt.tell(suggested, y)