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











      share|improve this question




      share|improve this question










      asked Nov 15 '18 at 21:49









      KipirpoKipirpo

      103




      103
























          1 Answer
          1






          active

          oldest

          votes


















          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























            Your Answer






            StackExchange.ifUsing("editor", function () {
            StackExchange.using("externalEditor", function () {
            StackExchange.using("snippets", function () {
            StackExchange.snippets.init();
            });
            });
            }, "code-snippets");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "1"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53328371%2fissues-debugging-infix-grammar-in-pyparsing%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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
































                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Stack Overflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53328371%2fissues-debugging-infix-grammar-in-pyparsing%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Florida Star v. B. J. F.

                    Danny Elfman

                    Lugert, Oklahoma