如何将字符串分解为嵌套令牌？

Question

我有由布尔项和方程组成的字符串，如下所示x=1 AND (x=2 OR x=3) AND NOT (x=4 AND x=5) AND (x=5) AND y=1我想将 它们分成由 分隔的组，同时将括号作为分组运算符。例如，上面字符串的结果将是xAND[['x=1'], ['x=2', 'x=3'], ['x=4'], ['x=5'], ['x=5']]x=2并且位于同一组中，因为它们由 分组，而不是由 分隔。最后一个等式被忽略，因为它不以 开头。x=3()ANDx更新另一个例子是x=1 AND (x=2 OR (x=3 AND x=4))其中每个等式应该是分开的[['x=1'], ['x=2', [['x=3'], ['x=4']]]我发现最接近的是这篇文章，但我不知道如何根据我的需求修改它。

Answer 1

正如您可能在另一个问题中看到的那样，解析诸如此类的中缀表示法最好使用帮助程序（以前称为 ）进行解析。以下是用于您的问题的基础知识：infixNotationoperatorPrecedenceinfixNotation

import pyparsing as pp

# define expressions for boolean operator keywords, and for an ident

# (which we take care not to parse an operator as an identifier)

AND, OR, NOT = map(pp.Keyword, "AND OR NOT".split())

any_keyword = AND | OR | NOT

ident = pp.ungroup(~any_keyword + pp.Char(pp.alphas))

ident.setName("ident")

# use pyparsing_common.number pre-defined expression for any numeric value

numeric_value = pp.pyparsing_common.number

# define an expression for 'x=1', 'y!=200', etc.

comparison_op = pp.oneOf("= != < > <= >=")

comparison = pp.Group(ident + comparison_op + numeric_value)

comparison.setName("comparison")

# define classes for the parsed results, where we can do further processing by

# node type later

class Node:

    oper = None

    def __init__(self, tokens):

        self.tokens = tokens[0]

    def __repr__(self):

        return "{}:{!r}".format(self.oper, self.tokens.asList())

class UnaryNode(Node):

    def __init__(self, tokens):

        super().__init__(tokens)

        del self.tokens[0]

class BinaryNode(Node):

    def __init__(self, tokens):

        super().__init__(tokens)

        del self.tokens[1::2]

class NotNode(UnaryNode):

    oper = "NOT"

class AndNode(BinaryNode):

    oper = "AND"

class OrNode(BinaryNode):

    oper = "OR"

# use infixNotation helper to define recursive expression parser,

# including handling of nesting in parentheses

expr = pp.infixNotation(comparison,

        [

            (NOT, 1, pp.opAssoc.RIGHT, NotNode),

            (AND, 2, pp.opAssoc.LEFT, AndNode),

            (OR, 2, pp.opAssoc.LEFT, OrNode),

        ])

现在尝试在测试字符串上使用此解析器。expr

test = "x=1 AND (x=2 OR x=3 OR y=12) AND NOT (x=4 AND x=5) AND (x=6) AND y=7"

try:

    result = expr.parseString(test, parseAll=True)

    print(test)

    print(result)

except pp.ParseException as pe:

    print(pp.ParseException.explain(pe))

给：

x=1 AND (x=2 OR x=3 OR y=12) AND NOT (x=4 AND x=5) AND (x=6) AND y=7

[AND:[['x', '=', 1], OR:[['x', '=', 2], ['x', '=', 3], ['y', '=', 12]], NOT:[AND:[['x', '=', 4], ['x', '=', 5]]], ['x', '=', 6], ['y', '=', 7]]]

从这一点开始，折叠嵌套的 AND 节点并删除非比较可以使用常规 Python 完成。x

Answer 2

我想你可以做这样的事情：

operators = ["AND NOT", "AND"]

sepChar = ":"

yourInputString = yourInputString.replace("(","").replace(")","") # remove the parenthesis

# Replace your operators with the separator character

for op in operators:

    yourInputString = yourInputString.replace(op,sepChar)

# output of your string so far

# yourInputString

# 'x=1 : x=2 OR x=3 : x=4 : x=5 : x=5 : y=1'

# Create a list with the separator character

operationsList = youtInputString.split(sepChar) 

# operationsList

# ['x=1', 'x=2 OR x=3', 'x=4', 'x=5', 'x=5', 'y=1']

# For the second result, let's do another operation list:

operators2 = ["OR"]

output = []

# Loop to find the other operators

for op in operationsList:

    for operator in operators2:

        if operator in op:

            op = op.split(operator)

    output.append(op)

# output:

# [['x=1'], ['x=2', 'x=3'], ['x=4'], ['x=5'], ['x=5'],['y=1']]

        

在这种情况下，我使用“：”作为分隔字符，但您可以根据需要进行更改。如果有帮助，请让我知道！

编辑

对于括号嵌套方法，我带来了一些精彩的东西：

import re

operators = ["AND NOT","AND","OR"]

# Substitute parenthesis

yourInputString = yourInputString.replace("(","[").replace(")","]")

# yourInputString

# "[x=1 AND [x=2 OR x=3] AND NOT [x=4 AND x=5] AND [x=5] AND y=1]"

# Replace your operators

for op in operators:

    yourInputString = yourInputString(op,",")

# yourInputString

# "[x=1 , [x=2 , x=3] , [x=4 , x=5] , [x=5] , y=1]"

# Find matches like x = 5 and substitue with 'x = 5'

compiler = re.compile(r"[xyz]{1}=\d")

matches = compiler.findall(yourInputString)

# matches

# ['x=1', 'x=2', 'x=3', 'x=4', 'x=5', 'x=5', 'y=1']

# Convert the list into unique outputs

matches = list(set(matches))

# matches

# ['x=1', 'x=2', 'x=3', 'x=4', 'x=5', 'y=1']

# Replace your matches to add quotes to each element

for match in matches:

    yourInputString = yourInputString.replace(match,f"'{match}'")

# yourInputString

# "['x=1' , ['x=2' , 'x=3'] , ['x=4' , 'x=5'] , ['x=5'] , 'y=1']"

# Here is the special move, convert your text into list

myList = eval(yourInputString)

# myList

# ['x=1', ['x=2', 'x=3'], ['x=4', 'x=5'], ['x=5'], 'y=1']