Issues debugging infix grammar in pyparsing












1















I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:



import pyparsing as pyp

alphabet = "abcdefghijklmnopqrstuvwxyz"
alphabet = alphabet + alphabet.upper()

symbol = pyp.Word(alphabet)

predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"

parenthetic = pyp.Forward()

pyp_formula = pyp.infixNotation((predicate | parenthetic),
[
(pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
(pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
(pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
(pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
])

parenthetic << "(" + pyp_formula + ")"


When I run



parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')

parse_list = formula_parse.asList()

print(parse_list)


I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.



[[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
';', 'h', '(', 'D', ')']], '.']]


I've tried several alternatives but I don't seem to be able to get the right parse.



Any help is welcome!










share|improve this question



























    1















    I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:



    import pyparsing as pyp

    alphabet = "abcdefghijklmnopqrstuvwxyz"
    alphabet = alphabet + alphabet.upper()

    symbol = pyp.Word(alphabet)

    predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"

    parenthetic = pyp.Forward()

    pyp_formula = pyp.infixNotation((predicate | parenthetic),
    [
    (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
    (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
    (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
    (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
    (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
    (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
    (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
    (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
    ])

    parenthetic << "(" + pyp_formula + ")"


    When I run



    parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')

    parse_list = formula_parse.asList()

    print(parse_list)


    I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.



    [[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
    ':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
    ';', 'h', '(', 'D', ')']], '.']]


    I've tried several alternatives but I don't seem to be able to get the right parse.



    Any help is welcome!










    share|improve this question

























      1












      1








      1








      I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:



      import pyparsing as pyp

      alphabet = "abcdefghijklmnopqrstuvwxyz"
      alphabet = alphabet + alphabet.upper()

      symbol = pyp.Word(alphabet)

      predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"

      parenthetic = pyp.Forward()

      pyp_formula = pyp.infixNotation((predicate | parenthetic),
      [
      (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
      (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
      (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
      ])

      parenthetic << "(" + pyp_formula + ")"


      When I run



      parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')

      parse_list = formula_parse.asList()

      print(parse_list)


      I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.



      [[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
      ':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
      ';', 'h', '(', 'D', ')']], '.']]


      I've tried several alternatives but I don't seem to be able to get the right parse.



      Any help is welcome!










      share|improve this question














      I wrote this script in order to parse statements with a syntax similar to prolog, treating connectives as operators with precedence:



      import pyparsing as pyp

      alphabet = "abcdefghijklmnopqrstuvwxyz"
      alphabet = alphabet + alphabet.upper()

      symbol = pyp.Word(alphabet)

      predicate = symbol + "(" + pyp.ZeroOrMore(symbol + ",") + symbol + ")"

      parenthetic = pyp.Forward()

      pyp_formula = pyp.infixNotation((predicate | parenthetic),
      [
      (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
      (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
      (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
      (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
      ])

      parenthetic << "(" + pyp_formula + ")"


      When I run



      parse = pyp_formula.parseString('d(A, D), e(D) :- f(A), g(D); h(D).')

      parse_list = formula_parse.asList()

      print(parse_list)


      I get that "d(A, D), e(D)" has not even been split into two predicates and "f(A), g(D); h(D)" is treated as a single terminal within a non-terminal.



      [[[['d', '(', 'A', ',', 'D', ')', ',', 'e', '(', 'D', ')'],
      ':-', [['f', '(', 'A', ')', ',', 'g', '(', 'D', ')'],
      ';', 'h', '(', 'D', ')']], '.']]


      I've tried several alternatives but I don't seem to be able to get the right parse.



      Any help is welcome!







      python pyparsing






      share|improve this question













      share|improve this question











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      asked Nov 15 '18 at 21:49









      KipirpoKipirpo

      103




      103
























          1 Answer
          1






          active

          oldest

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          1














          There is no need to define parenthetic, infixNotation does that for you. And if you pyp.Group your predicate expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList helper as a short cut for expr, expr, expr: use pyp.delimitedList(expr). The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat() to turn on an internal parsing cache.



          Here is how your code looks with these changes:



          import pyparsing as pyp
          pyp.ParserElement.enablePackrat()

          alphabet = "abcdefghijklmnopqrstuvwxyz"
          alphabet = alphabet + alphabet.upper()

          symbol = pyp.Word(alphabet)

          predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")

          pyp_formula = pyp.infixNotation(predicate,
          [
          (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
          (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
          (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
          (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
          (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
          (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
          (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
          (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
          ])

          pyp_formula.runTests(
          '''
          d(A, D) -> h(D).
          d(A, D), e(D) :- f(A), g(D); h(D).
          '''
          )


          And the results:



          d(A, D) -> h(D).
          [[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
          [0]:
          [[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
          [0]:
          [['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
          [0]:
          ['d', '(', 'A', 'D', ')']
          [1]:
          ->
          [2]:
          ['h', '(', 'D', ')']
          [1]:
          .


          d(A, D), e(D) :- f(A), g(D); h(D).
          [[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
          [0]:
          [[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
          [0]:
          [[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
          [0]:
          [['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
          [0]:
          ['d', '(', 'A', 'D', ')']
          [1]:
          ,
          [2]:
          ['e', '(', 'D', ')']
          [1]:
          :-
          [2]:
          [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
          [0]:
          [['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
          [0]:
          ['f', '(', 'A', ')']
          [1]:
          ,
          [2]:
          ['g', '(', 'D', ')']
          [1]:
          ;
          [2]:
          ['h', '(', 'D', ')']
          [1]:
          .





          share|improve this answer























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            1 Answer
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            1 Answer
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            active

            oldest

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            active

            oldest

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            1














            There is no need to define parenthetic, infixNotation does that for you. And if you pyp.Group your predicate expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList helper as a short cut for expr, expr, expr: use pyp.delimitedList(expr). The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat() to turn on an internal parsing cache.



            Here is how your code looks with these changes:



            import pyparsing as pyp
            pyp.ParserElement.enablePackrat()

            alphabet = "abcdefghijklmnopqrstuvwxyz"
            alphabet = alphabet + alphabet.upper()

            symbol = pyp.Word(alphabet)

            predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")

            pyp_formula = pyp.infixNotation(predicate,
            [
            (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
            (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
            (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
            (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
            (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
            (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
            (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
            (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
            ])

            pyp_formula.runTests(
            '''
            d(A, D) -> h(D).
            d(A, D), e(D) :- f(A), g(D); h(D).
            '''
            )


            And the results:



            d(A, D) -> h(D).
            [[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
            [0]:
            [[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
            [0]:
            [['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
            [0]:
            ['d', '(', 'A', 'D', ')']
            [1]:
            ->
            [2]:
            ['h', '(', 'D', ')']
            [1]:
            .


            d(A, D), e(D) :- f(A), g(D); h(D).
            [[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
            [0]:
            [[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
            [0]:
            [[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
            [0]:
            [['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
            [0]:
            ['d', '(', 'A', 'D', ')']
            [1]:
            ,
            [2]:
            ['e', '(', 'D', ')']
            [1]:
            :-
            [2]:
            [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
            [0]:
            [['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
            [0]:
            ['f', '(', 'A', ')']
            [1]:
            ,
            [2]:
            ['g', '(', 'D', ')']
            [1]:
            ;
            [2]:
            ['h', '(', 'D', ')']
            [1]:
            .





            share|improve this answer




























              1














              There is no need to define parenthetic, infixNotation does that for you. And if you pyp.Group your predicate expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList helper as a short cut for expr, expr, expr: use pyp.delimitedList(expr). The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat() to turn on an internal parsing cache.



              Here is how your code looks with these changes:



              import pyparsing as pyp
              pyp.ParserElement.enablePackrat()

              alphabet = "abcdefghijklmnopqrstuvwxyz"
              alphabet = alphabet + alphabet.upper()

              symbol = pyp.Word(alphabet)

              predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")

              pyp_formula = pyp.infixNotation(predicate,
              [
              (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
              (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
              (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
              (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
              (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
              (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
              (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
              (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
              ])

              pyp_formula.runTests(
              '''
              d(A, D) -> h(D).
              d(A, D), e(D) :- f(A), g(D); h(D).
              '''
              )


              And the results:



              d(A, D) -> h(D).
              [[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
              [0]:
              [[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
              [0]:
              [['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
              [0]:
              ['d', '(', 'A', 'D', ')']
              [1]:
              ->
              [2]:
              ['h', '(', 'D', ')']
              [1]:
              .


              d(A, D), e(D) :- f(A), g(D); h(D).
              [[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
              [0]:
              [[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
              [0]:
              [[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
              [0]:
              [['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
              [0]:
              ['d', '(', 'A', 'D', ')']
              [1]:
              ,
              [2]:
              ['e', '(', 'D', ')']
              [1]:
              :-
              [2]:
              [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
              [0]:
              [['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
              [0]:
              ['f', '(', 'A', ')']
              [1]:
              ,
              [2]:
              ['g', '(', 'D', ')']
              [1]:
              ;
              [2]:
              ['h', '(', 'D', ')']
              [1]:
              .





              share|improve this answer


























                1












                1








                1







                There is no need to define parenthetic, infixNotation does that for you. And if you pyp.Group your predicate expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList helper as a short cut for expr, expr, expr: use pyp.delimitedList(expr). The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat() to turn on an internal parsing cache.



                Here is how your code looks with these changes:



                import pyparsing as pyp
                pyp.ParserElement.enablePackrat()

                alphabet = "abcdefghijklmnopqrstuvwxyz"
                alphabet = alphabet + alphabet.upper()

                symbol = pyp.Word(alphabet)

                predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")

                pyp_formula = pyp.infixNotation(predicate,
                [
                (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
                (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
                (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
                ])

                pyp_formula.runTests(
                '''
                d(A, D) -> h(D).
                d(A, D), e(D) :- f(A), g(D); h(D).
                '''
                )


                And the results:



                d(A, D) -> h(D).
                [[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
                [0]:
                [[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
                [0]:
                [['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
                [0]:
                ['d', '(', 'A', 'D', ')']
                [1]:
                ->
                [2]:
                ['h', '(', 'D', ')']
                [1]:
                .


                d(A, D), e(D) :- f(A), g(D); h(D).
                [[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
                [0]:
                [[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
                [0]:
                [[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
                [0]:
                [['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
                [0]:
                ['d', '(', 'A', 'D', ')']
                [1]:
                ,
                [2]:
                ['e', '(', 'D', ')']
                [1]:
                :-
                [2]:
                [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
                [0]:
                [['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
                [0]:
                ['f', '(', 'A', ')']
                [1]:
                ,
                [2]:
                ['g', '(', 'D', ')']
                [1]:
                ;
                [2]:
                ['h', '(', 'D', ')']
                [1]:
                .





                share|improve this answer













                There is no need to define parenthetic, infixNotation does that for you. And if you pyp.Group your predicate expression, then the symbol and symbol args will get grouped for you. pyparsing also has the delimitedList helper as a short cut for expr, expr, expr: use pyp.delimitedList(expr). The default delimiter is ',', but you can define other delimiters. Lastly, with a list of operators this long, it is likely that you will have performance issues with any non-trivial parse - add a call to ParserElement.enablePackrat() to turn on an internal parsing cache.



                Here is how your code looks with these changes:



                import pyparsing as pyp
                pyp.ParserElement.enablePackrat()

                alphabet = "abcdefghijklmnopqrstuvwxyz"
                alphabet = alphabet + alphabet.upper()

                symbol = pyp.Word(alphabet)

                predicate = pyp.Group(symbol + "(" + pyp.delimitedList(symbol) + ")")

                pyp_formula = pyp.infixNotation(predicate,
                [
                (pyp.oneOf('~'), 1, pyp.opAssoc.RIGHT),
                (pyp.oneOf(','), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf(';'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf('::'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf('->'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf(':-'), 2, pyp.opAssoc.LEFT),
                (pyp.oneOf('--'), 1, pyp.opAssoc.LEFT),
                (pyp.oneOf('.'), 1, pyp.opAssoc.LEFT),
                ])

                pyp_formula.runTests(
                '''
                d(A, D) -> h(D).
                d(A, D), e(D) :- f(A), g(D); h(D).
                '''
                )


                And the results:



                d(A, D) -> h(D).
                [[[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']]
                [0]:
                [[['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']], '.']
                [0]:
                [['d', '(', 'A', 'D', ')'], '->', ['h', '(', 'D', ')']]
                [0]:
                ['d', '(', 'A', 'D', ')']
                [1]:
                ->
                [2]:
                ['h', '(', 'D', ')']
                [1]:
                .


                d(A, D), e(D) :- f(A), g(D); h(D).
                [[[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']]
                [0]:
                [[[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]], '.']
                [0]:
                [[['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']], ':-', [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]]
                [0]:
                [['d', '(', 'A', 'D', ')'], ',', ['e', '(', 'D', ')']]
                [0]:
                ['d', '(', 'A', 'D', ')']
                [1]:
                ,
                [2]:
                ['e', '(', 'D', ')']
                [1]:
                :-
                [2]:
                [[['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']], ';', ['h', '(', 'D', ')']]
                [0]:
                [['f', '(', 'A', ')'], ',', ['g', '(', 'D', ')']]
                [0]:
                ['f', '(', 'A', ')']
                [1]:
                ,
                [2]:
                ['g', '(', 'D', ')']
                [1]:
                ;
                [2]:
                ['h', '(', 'D', ')']
                [1]:
                .






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 16 '18 at 4:01









                PaulMcGPaulMcG

                46.7k969111




                46.7k969111
































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