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fix has_choice_labelling(), which did not return anything

refactoring
Sebastian Junges 5 years ago
parent
commit
8ed76f5feb
  1. 2
      src/storage/model.cpp

2
src/storage/model.cpp

@ -169,7 +169,7 @@ void define_sparse_model(py::module& m) {
// Models with double numbers
py::class_<SparseModel<double>, std::shared_ptr<SparseModel<double>>, ModelBase> model(m, "_SparseModel", "A probabilistic model where transitions are represented by doubles and saved in a sparse matrix");
model.def_property_readonly("labeling", &getLabeling<double>, "Labels")
.def("has_choice_labeling", [](SparseModel<double> const& model) {model.hasChoiceLabeling();}, "Does the model have an associated choice labelling?")
.def("has_choice_labeling", [](SparseModel<double> const& model) {return model.hasChoiceLabeling();}, "Does the model have an associated choice labelling?")
.def_property_readonly("choice_labeling", [](SparseModel<double> const& model) {return model.getChoiceLabeling();}, "get choice labelling")
.def("has_choice_origins", [](SparseModel<double> const& model) {return model.hasChoiceOrigins();}, "has choice origins?")
.def_property_readonly("choice_origins", [](SparseModel<double> const& model) {return model.getChoiceOrigins();})
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