Hi Sofiane, Thanks for the trick : I did not know this constructor.
However, my particular need was really to get over the linear model. My particular applications are the calibration of a model based on the BLUE or ridge regression. In these cases, we really need to solve the basic linear least squares problem, without any link to a functional basis. How would you do that ? Best regards, Michaël Michaël BAUDIN Ingénieur - Chercheur EDF – R&D Département PRISME 6, quai Watier 78401 CHATOU michael.bau...@edf.fr<mailto:michael.bau...@edf.fr> Tél. : 01 30 87 81 82 Fax : 01 30 87 82 13 De : sofiane_had...@yahoo.fr [mailto:sofiane_had...@yahoo.fr] Envoyé : mardi 2 avril 2019 09:00 À : users@openturns.org; BAUDIN Michael <michael.bau...@edf.fr> Objet : Re: [ot-users] Linear least squares always insert a constant Hi Michael, The LinearModelAlgorithm indeed relies on a linear basis (if no one is provided). However you can define your own basis and provide it to the algorithm class : """ import openturns as ot ot.RandomGenerator.SetSeed(0) g = ot.SymbolicFunction(['x', 'y'], ['0.5+sin(x)-2*y']) npoints = 50 x = ot.Uniform(-2,2).getSample(npoints) x.stack(ot.Normal(0, 3).getSample(npoints)) x.setDescription(["x", "y"]) y = g(x) # Create basis B = ( [x,y]-->x, [x,y]-->y ) input_description = x.getDescription() basis = ot.Basis([ot.SymbolicFunction(input_description, [description_i]) for description_i in input_description]) algo = ot.LinearModelAlgorithm(x, basis, y) algo.run() ... """ The trend coefficients's size is 2 corresponding to the basis size Sofiane Le jeudi 28 mars 2019 à 20:09:52 UTC+1, BAUDIN Michael <michael.bau...@edf.fr<mailto:michael.bau...@edf.fr>> a écrit : Dear all ! The LinearLeastSquares class in OT 1.12 always inserts a constant in the model, be it wanted by the user or not: https://github.com/openturns/openturns/blob/d0802a1b17b60bd86afa234662a047bc4f04492f/lib/src/Base/MetaModel/LinearLeastSquares.cxx#L105 In the API, this term corresponds to the getConstant() method. The same is true for LinearModelFactory. In the demo script in PS, I use linear least squares to approximate the sine function with the polynomial basis 1, x, x^2, x^3. The linear model involves 4 coefficients. An intercept is always added leading to 5 estimated coefficients, that I do not want. g = ot.SymbolicFunction(['x'], ['0.5+sin(x)']) npoints = 50 x=ot.Uniform(-2,2).getSample(npoints).sort() y = g(x) # Create input basis = ot.SymbolicFunction(['x'], ['1','x','x^2','x^3']) inputData = basis(x) With LinearLeastSquares, I get a constant: myLeastSquares = ot.LinearLeastSquares(inputData, y) myLeastSquares.run() beta0 = myLeastSquares.getConstant()[0] With LinearModelFactory, I get 5 coefficicents instead of 4: LMF = ot.LinearModelFactory() linearModel = LMF.build(inputData, y) beta = linearModel.getRegression() As far as I can see, the LinearModelAlgorithm in OT 1.13 has the same behaviour: https://github.com/openturns/openturns/blob/ce1bc890a907faeecde495f5528ed42e401153c7/lib/src/Uncertainty/Algorithm/MetaModel/LinearModel/LinearModelAlgorithm.cxx#L65 I assume that the constant is always there, so that the method prevents from having a bias in the estimate. But in cases where you want really to perform linear least squares, then there is an issue. As far as I can see, the LeastSquaresMethod is the right tool. Unfortunately, this cannot be used from the Python API. Am I correct ? Best regards, Michaël PS import openturns as ot from openturns.viewer import View g = ot.SymbolicFunction(['x'], ['0.5+sin(x)']) npoints = 50 x=ot.Uniform(-2,2).getSample(npoints).sort() y = g(x) # Create input basis = ot.SymbolicFunction(['x'], ['1','x','x^2','x^3']) inputData = basis(x) # Solve myLeastSquares = ot.LinearLeastSquares(inputData, y) myLeastSquares.run() beta0 = myLeastSquares.getConstant()[0] print("beta0=%s" % (beta0)) beta = myLeastSquares.getLinear() print("beta=%s" % (beta)) # Check responseSurface = myLeastSquares.getResponseSurface() ypredicted = responseSurface(inputData) # graph = ot.Graph("Linear Model","x","y",True,"topleft") curve = ot.Curve(x,ypredicted) curve.setLegend("Linear Model") graph.add(curve) cloud = ot.Cloud(x,y) cloud.setColor("red") cloud.setLegend("Data") graph.add(cloud) View(graph) # ot.ResourceMap.SetAsString('R-executable-command','bla\\bla\\R.exe') LMF = ot.LinearModelFactory() linearModel = LMF.build(inputData, y) beta = linearModel.getRegression() print(beta) Ce message et toutes les pièces jointes (ci-après le 'Message') sont établis à l'intention exclusive des destinataires et les informations qui y figurent sont strictement confidentielles. 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