With python's rdflib I can parse a SPARQL query string and translate its algebraic form with following syntax:
statement = "Select ?s ?p ?o where {?s ?p ?o.}"
query_tree = parser.parseQuery(statement) # query parse-tree
q_algebra = algebra.translateQuery(query_tree) # query algebra
algebra.pprintAlgebra(q_algebra)
This is how the algebra tree of above query looks like:
SelectQuery(
p = Project(
p = BGP(
triples = [(rdflib.term.Variable('s'), rdflib.term.Variable('p'), rdflib.term.Variable('o'))]
_vars = {rdflib.term.Variable('s'), rdflib.term.Variable('p'), rdflib.term.Variable('o')}
)
PV = [rdflib.term.Variable('s'), rdflib.term.Variable('p'), rdflib.term.Variable('o')]
_vars = {rdflib.term.Variable('s'), rdflib.term.Variable('p'), rdflib.term.Variable('o')}
)
datasetClause = None
PV = [rdflib.term.Variable('s'), rdflib.term.Variable('p'), rdflib.term.Variable('o')]
_vars = {rdflib.term.Variable('s'), rdflib.term.Variable('p'), rdflib.term.Variable('o')}
)
After updating the query algebra by e.g. using algebra.traverse(q_algebra, update_function) I want to get a query string again. I searched quiet a lot through the rdflib module but could not find a way how to back translate the algebra. Is there a function that takes the SPARQL query algebra as an input and returns a SPARQL query string?
