从python中的字符串解析嵌套逻辑

Question

我需要把表格的字符串

'a==1&&(b==2||(c==3||d==4&&e==5)&&f==6)||(g==7&&h==8)'

进入要过滤的sqlalchemy逻辑（通过or_和and_）。第一步实际上是将字符串解析为有用的内容。我想把它做成表格：

^{pr2}$

然后递归地进行解析。不幸的是，我在这个解析步骤上遇到了麻烦，以前从未使用过pyparsing。在

编辑（解决方案）

我在so或interblag上找不到一个简单的解决方案，但是在对文档进行了大量挖掘之后，我设法将以下简单表达式组合在一起：

from pyparsing import Word, alphanums, nestedExpr

content  = Word( alphanums + '=' ) | '||' | '&&'
unnester = nestedExpr( content = content )

a = 'a==3||b==1&&(c==4||(d==1||e==5)&&f==9)'
unnester.parseString( '(' + a + ')' ).asList()

只要在迭代过程中执行一个扁平化步骤，这看起来就可以很好地工作。在

Answer 1

我也没有真正使用过pyparsing，但下面是一个直接的python实现，它可以满足您的需要：

import re
from collections import namedtuple

TOKENIZER = re.compile(r"\w+==\w+|&&|\|\||[()]").findall

Node = namedtuple("Node", ["parent", "children"])

def syntax_tree(text, tokenizer, brackets):
    root = cur_node = Node(None, [])
    stack = []
    for token in tokenizer(text):
        if token == brackets["("]:
            stack.append(token)
            new_node = Node(cur_node, [])
            cur_node.children.append(new_node)
            cur_node = new_node
        elif token == brackets[")"]:
            if stack and stack.pop() == brackets[")"]:
                cur_node = cur_node.parent
            else:
                raise Exception("Parse error: unmatched parentheses")
        else:
            cur_node.children.append(token)

        if stack:
            raise Exception("Parse error: unmatched parentheses")

    return root

def listify(root):
    if isinstance(root, Node):
        return [listify(item) for item in root.children]
    else:
        return root


if __name__ == "__main__":
    expr = "a==1&&(b==2||(c==3||d==4&&e==5)&&f==6)||(g==7&&h==8)"

    tree = syntax_tree(expr, TOKENIZER, {"(": "(", ")": ")"})
    obj = listify(tree)

Answer 2

Pyparsing有一个用于解析附加分组操作的内置函数，包括操作符优先级的识别，称为infixNotation（以前称为operatorPrecedence）。请参见此示例代码中如何使用它来解析示例表达式：

from pyparsing import Word, alphas, nums, oneOf, Group, infixNotation, opAssoc

sample = 'a==1&&(b==2||(c==3||d==4&&e==5)&&f==6)||(g==7&&h==8)'

# define some basic elements
varname = Word(alphas)
integer = Word(nums).setParseAction(lambda t:int(t[0]))

# Use varname and integer to define a comparison expression
comparisonOp = oneOf("< == > <= >= !=")
term = varname | integer
comparisonExpr = Group(term + comparisonOp + term)

# Use pyparsing builtin 'infixNotation' to implement parser for
# parenthetically grouped expression of various operators (formerly
# named 'operatorPrecedence')
logicalExpr = infixNotation(comparisonExpr,
    [
    ('&&', 2, opAssoc.LEFT),
    ('||', 2, opAssoc.LEFT),
    ])

# parse out your sample expression, use pprint to do some pretty-printing
import pprint
pprint.pprint(logicalExpr.parseString(sample).asList())

印刷品：

^{pr2}$