OpenTURNS
Validate a polynomial chaosΒΆ
In this example, we show how to perform the draw validation of a polynomial chaos for the Ishigami function.
[1]:
import openturns as ot
from math import pi
Create the Ishigami test function.
[2]:
ot.RandomGenerator.SetSeed(0)
formula = ['sin(X1) + 7. * sin(X2)^2 + 0.1 * X3^4 * sin(X1)']
input_names = ['X1', 'X2', 'X3']
g = ot.SymbolicFunction(input_names, formula)
Create the probabilistic model
[3]:
distributionList = [ot.Uniform(-pi, pi)] * 3
distribution = ot.ComposedDistribution(distributionList)
Create a training sample
[4]:
N = 100
inputTrain = distribution.getSample(N)
outputTrain = g(inputTrain)
Create the chaos.
We could use only the input and output training samples: in this case, the distribution of the input sample is computed by selecting the best distribution that fits the data.
[5]:
chaosalgo = ot.FunctionalChaosAlgorithm(inputTrain, outputTrain)
Since the input distribution is known in our particular case, we instead create the multivariate basis from the distribution.
[6]:
multivariateBasis = ot.OrthogonalProductPolynomialFactory(distributionList)
totalDegree = 8
enumfunc = multivariateBasis.getEnumerateFunction()
P = enumfunc.getStrataCumulatedCardinal(totalDegree)
adaptiveStrategy = ot.FixedStrategy(multivariateBasis, P)
[7]:
selectionAlgorithm = ot.LeastSquaresMetaModelSelectionFactory()
projectionStrategy = ot.LeastSquaresStrategy(inputTrain, outputTrain, selectionAlgorithm)
[8]:
chaosalgo = ot.FunctionalChaosAlgorithm(inputTrain, outputTrain, distribution, adaptiveStrategy, projectionStrategy)
[9]:
chaosalgo.run()
result = chaosalgo.getResult()
metamodel = result.getMetaModel()
In order to validate the metamodel, we generate a test sample.
[10]:
n_valid = 1000
inputTest = distribution.getSample(n_valid)
outputTest = g(inputTest)
val = ot.MetaModelValidation(inputTest, outputTest, metamodel)
Q2 = val.computePredictivityFactor()
Q2
[10]:
0.9992361845215688
The Q2 is very close to 1: the metamodel is excellent.
[11]:
graph = val.drawValidation()
graph.setLegends([""])
graph.setTitle("Q2=%.2f%%" % (Q2*100))
graph
[11]:
The metamodel has a good predictivity, since the points are almost on the first diagonal.