Define and optimize the kriging mean term as a deterministic function dependent on space.

Python usage
1

Hello,

I am currently trying to build a kriging model, but I am having difficulty defining the mean term as a function dependent on space. Here is a brief description of what I can do and what I want to do.

Kriging is formulated: Y(\omega,x)=\mu(x)+W(\omega,x)
For ordinary kriging, the deterministic term is formulated: \mu(x)=\beta
In this case, the implementation in OpenTurns is as follows:
basis = ot.ConstantBasisFactory(n_DIM).build()
model = ot.KrigingAlgorithm(x_OBS, y_OBS, covarianceModel, basis)
[…]
model.setOptimizationBounds(scaleOptimizationBounds)
model.setOptimizationAlgorithm(local_solver)
model.run()

For my study, I define the deterministic term such as: \mu(x)=\beta.\phi(x)
Given that the formulation of \mu has changed, I need to adapt the implementation in OpenTurns by changing the definition of the term basis.

Could you help me define basis as the product of a scalar \beta and a function defined in Python \phi(x)? (Note that \beta need to be optimized during the building of the kriging model and \phi(x) is not analytical.)

Best regards,

Clément.

2

Hi Clément,

The first thing to do is to create an OpenTURNS function based on your Python function. Then, you encapsulate it into a Basis and you can build a KrigingAlgorithm instance based on it:
``python
import openturns as ot

def myFuncPy(x): # x has to be a list of scalars or an 1d numpy array or an OT Point
return [x[0]**3] # the return value has to be a list of scalars or an 1d numpy array or an OT Point

# Adapt the dimensions to your case
inDim = 1
outDim = 1
myFunc = ot.PythonFunction(inDim, outDim, myFuncPy)

basis = ot.Basis([myFunc])
inSample = ot.Uniform().getSample(5)
outSample = ot.Uniform().getSample(5)
cov = ot.ExponentialModel()
myKriging = ot.KrigingAlgorithm(inSample, outSample, cov, basis)
```

Note that the KrigingAlgorithm class will be deprecated in favor of the GaussianProcessRegression class.

Cheers

Régis

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