使用二元和一元运算符、保留字且不带括号解析表达式

问题描述

我正在尝试解析由二元运算符 +、一元运算符 not 和可以是不是 的任何字母字符串的标识符组成的表达式不是

I'm trying to parse expressions made of the binary operator +, the unary operator not and identifiers that can be any alphabetical string that isn't not

from pyparsing import (
    CaselessKeyword,
    Combine,
    Word,
    alphas,
    opAssoc,
    infixNotation,
)

identifier = Combine(~CaselessKeyword('not') + Word(alphas))
expression = infixNotation(identifier, [
  ('+', 2, opAssoc.LEFT),
  (CaselessKeyword('not'), 1, opAssoc.RIGHT),
]

运行

expression.parseString('a + (not b)')

满足我的期望

[['a', '+', ['not', 'b']]]

但是，没有括号

expression.parseString('a + not b')

我只得到第一个令牌:

['a']

如何在没有括号的情况下定义要工作的语言?

How can I define the language to work as I would like without the parentheses?

(在实际情况下有更多的运算符和保留字:这是解析 S3 Select 语言的一步)

(In the real case there are more operators and reserved words: this is a step towards parsing the S3 Select language)

推荐答案

在S3中NOT比AND更高:

运算符优先级下表按降序显示了运算符的优先级.

(来自 S3亚马逊网站).

在那个表中，NOT 在上面 AND.

所以你的代码应该是:

identifier = Combine(~CaselessKeyword('not') + Word(alphas))
expression = infixNotation(identifier, [
    (CaselessKeyword('not'), 1, opAssoc.RIGHT),
    ('+', 2, opAssoc.LEFT),
])

顺便说一句 - 如果NOT 被列为低于二进制 +"，则 a + not b 是无效的表达.+ 需要两个运算符:一个是 a，但 not b 不是有效的操作数.

BTW - If "NOT is listed as a lower than the binary +", than a + not b is not valid expression. + needs two operators: one is a, but not b is not valid operand.

顺便说一句(来自评论):请不要在同一个表达式中混用算术运算符 + 和逻辑运算符 NOT.1 + not 2 不是一个有效的表达式.每种语言都决定如何解析这种奇怪的表达式.

BTW2 (from comments):Please don't mix + which is an arithmetic operator and NOT which is a logic operator in the same expression. 1 + not 2 is not a valid expression.Every language decide how to parse that's kinds of strange expressions.

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