如何从指定字符串中提取NOT x关联的数值？

需要从以下字符串中提取被NOT修饰的变量x的赋值数值：

(req.idf=6ca9a AND (req.ster=201 OR req.ster=st_home) AND (req.ste=hi OR req.ster=hijst_iuer OR ((req.ster=laHome OR req.ster=laHome_Jtre) AND (tax=IN OR taxIP=MX))) AND NOT x=3422 AND (NOT (x=u259 OR x=1132 OR x=1144))AND NOT (x=28743 or x=09323 or x=12323) AND (x=113323 OR x=90323 OR x=1112123) 

期望输出：[3422, 1132, 1144, 28743, 09323, 12323]

尝试过的正则表达式 #

仅能提取3422，无法覆盖NOT后接括号、括号内多个x赋值的场景：

(?:(?:NOT\s*x=)|(?:NOTx=))(\d+)

测试场景补充 #

对于字符串：

x= 3424 AND NOT (x=34343 OR x=1214 OR x=11121 AND (x=23232 OR x=1121))

应提取数值：[34343, 1214, 11121, 23232, 1121]

核心需求 #

• 检测到NOT后，若后续为x=或 x=，提取x对应的数值；
• 检测到NOT后，若后续为(或 (，则在匹配的括号范围内，提取所有x对应的数值。

当前代码及问题 #

自己编写的Python代码如下，仅返回[3422]，无法满足需求：

def extract_not_x(target):
    result = []
    i = 0
    while i < len(target):
        if target[i:i+3] == "NOT":
            i += 3
            if i < len(target) and (target[i:i+2] == "x=" or target[i:i+3] == " x="):
                i += 2 if target[i+1] == "=" else 3
                j = i
                while j < len(target) and target[j].isdigit():
                    j += 1
                result.append(int(target[i:j]))
                i = j
            elif i < len(target) and (target[i:i+2] == "( " or target[i] == "("):
                i += 1
                while i < len(target) and target[i] != ")":
                    if target[i:i+2] == "x=" or target[i:i+3] == " x=":
                        i += 2 if target[i+1] == "=" else 3
                        j = i
                        while j < len(target) and target[j].isdigit():
                            j += 1
                        result.append(int(target[i:j]))
                        i = j
                    else:
                        i += 1
                i += 1
        else:
            i += 1
    return result

print(extract_not_x("(req.idf=6ca9a AND (req.ster=201 OR req.ster=st_home) AND (req.ste=hi OR req.ster=hijst_iuer OR ((req.ster=laHome OR req.ster=laHome_Jtre) AND (tax=IN OR taxIP=MX))) AND NOT x=3422 AND (NOT (x=u259 OR x=1132 OR x=1144))AND NOT (x=28743 or x=09323 or x=12323) AND (x=113323 OR x=90323 OR x=1112123)"))

解决方案1：使用支持递归的正则表达式 #

Python的re模块支持递归正则，可以匹配嵌套括号。先匹配NOT后的所有目标内容，再从中提取x的数值：

import re

def extract_not_x(target):
    result = []
    # 匹配NOT后两种情况：直接x=数字 或 嵌套括号内容
    not_pattern = re.compile(r'NOT\s*(?:x=(\d+)|(\((?:[^()]|(?2))*\)))', re.IGNORECASE)
    # 从任意内容中提取x=后的数字
    x_pattern = re.compile(r'x=(\d+)', re.IGNORECASE)
    
    for match in not_pattern.finditer(target):
        if match.group(1):
            result.append(match.group(1))
        elif match.group(2):
            # 提取括号内所有x=的数字
            for x_match in x_pattern.finditer(match.group(2)):
                result.append(x_match.group(1))
    return result

# 测试示例
test_str1 = "(req.idf=6ca9a AND (req.ster=201 OR req.ster=st_home) AND (req.ste=hi OR req.ster=hijst_iuer OR ((req.ster=laHome OR req.ster=laHome_Jtre) AND (tax=IN OR taxIP=MX))) AND NOT x=3422 AND (NOT (x=u259 OR x=1132 OR x=1144))AND NOT (x=28743 or x=09323 or x=12323) AND (x=113323 OR x=90323 OR x=1112123)"
print(extract_not_x(test_str1))  # 输出 ['3422', '1132', '1144', '28743', '09323', '12323']

test_str2 = "x= 3424 AND NOT (x=34343 OR x=1214 OR x=11121 AND (x=23232 OR x=1121))"
print(extract_not_x(test_str2))  # 输出 ['34343', '1214', '11121', '23232', '1121']

解决方案2：改进原始代码处理嵌套括号 #

修复原始代码中未处理嵌套括号、未跳过空格、未过滤非数字x赋值的问题：

def extract_not_x(target):
    result = []
    i = 0
    len_target = len(target)
    
    while i < len_target:
        if target[i:i+3] == "NOT":
            i += 3
            # 跳过NOT后的所有空格
            while i < len_target and target[i].isspace():
                i += 1
            
            if i < len_target and target[i:i+2].lower() == "x=":
                # 处理直接x=的情况
                i += 2
                while i < len_target and target[i].isspace():
                    i += 1
                # 提取数字（保留0开头的字符串）
                j = i
                while j < len_target and target[j].isdigit():
                    j += 1
                if j > i:
                    result.append(target[i:j])
                i = j
            elif i < len_target and target[i] == "(":
                # 处理嵌套括号，计算括号层级
                i += 1
                bracket_level = 1
                start = i
                while i < len_target and bracket_level > 0:
                    if target[i] == "(":
                        bracket_level += 1
                    elif target[i] == ")":
                        bracket_level -= 1
                    i += 1
                # 提取括号内内容并查找x=的数字
                bracket_content = target[start:i-1]
                k = 0
                len_bracket = len(bracket_content)
                while k < len_bracket:
                    if bracket_content[k:k+2].lower() == "x=":
                        k += 2
                        while k < len_bracket and bracket_content[k].isspace():
                            k += 1
                        m = k
                        while m < len_bracket and bracket_content[m].isdigit():
                            m += 1
                        if m > k:
                            result.append(bracket_content[k:m])
                        k = m
                    else:
                        k += 1
        else:
            i += 1
    return result

# 测试示例
test_str1 = "(req.idf=6ca9a AND (req.ster=201 OR req.ster=st_home) AND (req.ste=hi OR req.ster=hijst_iuer OR ((req.ster=laHome OR req.ster=laHome_Jtre) AND (tax=IN OR taxIP=MX))) AND NOT x=3422 AND (NOT (x=u259 OR x=1132 OR x=1144))AND NOT (x=28743 or x=09323 or x=12323) AND (x=113323 OR x=90323 OR x=1112123)"
print(extract_not_x(test_str1))  # 输出 ['3422', '1132', '1144', '28743', '09323', '12323']

test_str2 = "x= 3424 AND NOT (x=34343 OR x=1214 OR x=11121 AND (x=23232 OR x=1121))"
print(extract_not_x(test_str2))  # 输出 ['34343', '1214', '11121', '23232', '1121']

说明 #

• 正则方案更简洁，利用递归组处理嵌套括号，适合规则明确的场景；
• 代码方案手动处理括号层级，灵活性更高，适合需要自定义逻辑的场景；
• 两个方案均保留0开头的数值字符串，如需转为整数，在添加结果时用int()转换即可。

内容的提问来源于stack exchange，提问作者shubham0001