"""
Kriging: choose a polynomial trend on the beam model
====================================================
"""
# %%
# The goal of this example is to show how to configure the trend in a Kriging metamodel.
# This example focuses on three polynomial trends:
#
# * :class:`~openturns.ConstantBasisFactory`,
# * :class:`~openturns.LinearBasisFactory`,
# * :class:`~openturns.QuadraticBasisFactory`.
#
# In the :doc:`/auto_meta_modeling/kriging_metamodel/plot_kriging_chose_trend` example,
# we give another example of this procedure.
#
# For this purpose, we use the :ref:`cantilever beam ` example.
# %%
# Definition of the model
# -----------------------
# %%
from openturns.usecases import cantilever_beam
import openturns as ot
from openturns.viewer import View
ot.RandomGenerator.SetSeed(0)
ot.Log.Show(ot.Log.NONE)
# %%
# We load the use case :
cb = cantilever_beam.CantileverBeam()
# %%
# We define the function which evaluates the output depending on the inputs.
model = cb.model
# %%
# Then we define the distribution of the input random vector.
dimension = cb.dim # number of inputs
myDistribution = cb.distribution
# %%
# Create the design of experiments
# --------------------------------
# %%
# We consider a simple Monte-Carlo sampling as a design of experiments.
# This is why we generate an input sample using the :meth:`~openturns.Distribution.getSample` method of the distribution.
# Then we evaluate the output using the `model` function.
# %%
sampleSize_train = 10
X_train = myDistribution.getSample(sampleSize_train)
Y_train = model(X_train)
# %%
# Create the metamodel
# --------------------
# %%
# In order to create the Kriging metamodel, we first select a constant trend with the `ConstantBasisFactory` class.
# Then we use a squared exponential covariance kernel.
# The `SquaredExponential` kernel has one amplitude coefficient and 4 scale coefficients.
# This is because this covariance kernel is anisotropic : each of the 4 input variables is associated with its own scale coefficient.
# %%
basis = ot.ConstantBasisFactory(dimension).build()
covarianceModel = ot.SquaredExponential(dimension)
# %%
# Typically, the optimization algorithm is quite good at setting sensible optimization bounds.
# In this case, however, the range of the input domain is extreme.
# %%
print("Lower and upper bounds of X_train:")
print(X_train.getMin(), X_train.getMax())
# %%
# We need to manually define sensible optimization bounds.
# Note that since the amplitude parameter is computed analytically (this is possible when the output dimension is 1), we only need to set bounds on the scale parameter.
# %%
scaleOptimizationBounds = ot.Interval(
[1.0, 1.0, 1.0, 1.0e-10], [1.0e11, 1.0e3, 1.0e1, 1.0e-5]
)
# %%
# Finally, we use the `KrigingAlgorithm` class to create the Kriging metamodel.
# It requires a training sample, a covariance kernel and a trend basis as input arguments.
# We need to set the initial scale parameter for the optimization. The upper bound of the input domain is a sensible choice here.
# We must not forget to actually set the optimization bounds defined above.
# %%
covarianceModel.setScale(X_train.getMax())
algo = ot.KrigingAlgorithm(X_train, Y_train, covarianceModel, basis)
algo.setOptimizationBounds(scaleOptimizationBounds)
# %%
# Run the algorithm and get the result.
# %%
algo.run()
result = algo.getResult()
krigingWithConstantTrend = result.getMetaModel()
# %%
# The `getTrendCoefficients` method returns the coefficients of the trend.
# %%
print(result.getTrendCoefficients())
# %%
# The constant trend always has only one coefficient (if there is one single output).
# %%
print(result.getCovarianceModel())
# %%
# Setting the trend
# -----------------
# %%
covarianceModel.setScale(X_train.getMax())
basis = ot.LinearBasisFactory(dimension).build()
algo = ot.KrigingAlgorithm(X_train, Y_train, covarianceModel, basis)
algo.setOptimizationBounds(scaleOptimizationBounds)
algo.run()
result = algo.getResult()
krigingWithLinearTrend = result.getMetaModel()
result.getTrendCoefficients()
# %%
# The number of coefficients in the linear and quadratic trends depends on the number of inputs, which is
# equal to
#
# .. math::
# dim = 4
#
#
# in the cantilever beam case.
#
# We observe that the number of coefficients in the trend is 5, which corresponds to:
#
# * 1 coefficient for the constant part,
# * dim=4 coefficients for the linear part.
# %%
covarianceModel.setScale(X_train.getMax())
basis = ot.QuadraticBasisFactory(dimension).build()
algo = ot.KrigingAlgorithm(X_train, Y_train, covarianceModel, basis)
algo.setOptimizationBounds(scaleOptimizationBounds)
algo.run()
result = algo.getResult()
krigingWithQuadraticTrend = result.getMetaModel()
result.getTrendCoefficients()
print(algo.getOptimizationBounds())
print(result.getCovarianceModel())
# %%
# This time, the trend has 15 coefficients:
#
# * 1 coefficient for the constant part,
# * 4 coefficients for the linear part,
# * 10 coefficients for the quadratic part.
#
# This is because the number of coefficients in the quadratic part has
#
# .. math::
# \frac{dim (dim+1)}{2}=\frac{4\times 5}{2}=10
#
#
# coefficients, associated with the symmetric matrix of the quadratic function.
# %%
# Validate the metamodel
# ----------------------
# %%
# We finally want to validate the Kriging metamodel. This is why we generate a validation sample with size 100 and we evaluate the output of the model on this sample.
# %%
sampleSize_test = 100
X_test = myDistribution.getSample(sampleSize_test)
Y_test = model(X_test)
# %%
def drawMetaModelValidation(X_test, Y_test, krigingMetamodel, title):
val = ot.MetaModelValidation(Y_test, krigingMetamodel(X_test))
R2 = val.computeR2Score()[0]
graph = val.drawValidation().getGraph(0, 0)
graph.setLegends([""])
graph.setLegends(["%s, R2 = %.2f%%" % (title, 100 * R2), ""])
graph.setLegendPosition("upper left")
return graph
# %%
grid = ot.GridLayout(1, 3)
grid.setTitle("Different trends")
graphConstant = drawMetaModelValidation(
X_test, Y_test, krigingWithConstantTrend, "Constant"
)
graphLinear = drawMetaModelValidation(X_test, Y_test, krigingWithLinearTrend, "Linear")
graphQuadratic = drawMetaModelValidation(
X_test, Y_test, krigingWithQuadraticTrend, "Quadratic"
)
grid.setGraph(0, 0, graphConstant)
grid.setGraph(0, 1, graphLinear)
grid.setGraph(0, 2, graphQuadratic)
_ = View(grid, figure_kw={"figsize": (13, 4)})
# %%
# We observe that the three trends perform very well in this case.
# With more coefficients, the Kriging metamodel is more flexibile and can adjust better to the training sample.
# This does not mean, however, that the trend coefficients will provide a good fit for the validation sample.
#
# The number of parameters in each Kriging metamodel is the following:
#
# * the constant trend Kriging has 6 coefficients to estimate: 5 coefficients for the covariance matrix and 1 coefficient for the trend,
# * the linear trend Kriging has 10 coefficients to estimate: 5 coefficients for the covariance matrix and 5 coefficients for the trend,
# * the quadratic trend Kriging has 20 coefficients to estimate: 5 coefficients for the covariance matrix and 15 coefficients for the trend.
#
# In the cantilever beam example, fitting the metamodel to a training sample with only 10 points is made much easier because the function to approximate is almost linear.
# In this case, a quadratic trend is not advisable because it can interpolate all points in the training sample.