Python 3-基于字符串的字典搜索列表

Question

说我有绳子

"((attr1=25 and attr2=8) or attr3=15)"

或

"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"

或

"(attrXYZ=10)"

甚至是

"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"

以及包含字典的列表，其中每个字典都可能或不具有字符串中指定的属性。Python中是否有一种简单的方法来过滤与这种类型的字符串查询相匹配的字典？

Answer 1

免责声明:这是一个非常懒惰和不安全的解决方案，它使用了Python、评估和exec这两个最不光彩的函数，但如果输出与您提供的输出形状完全相同，则可以工作。

我们的策略是编辑输入，使其看起来类似Python自然理解的语法，而不是创建自己的解析器。为此，我们将使用dis模块(用于Python字节码的反汇编程序)来获取字符串中的所有名称。

import dis 

class Number:
    def __init__(self, n, exists=True):
        self.n = n
        self.exists = exists

    def __lt__(self, other):
        return self.n < other if self.exists else False

    def __le__(self, other):
        return self.n <= other if self.exists else False

    def __eq__(self, other):
        return self.n == other if self.exists else False

    def __ne__(self, other):
        return self.n != other if self.exists else False

    def __gt__(self, other):
        return self.n > other if self.exists else False

    def __ge__(self, other):
        return self.n >= other if self.exists else False


def clear_entries(entry):
    entry_output = entry.replace('!=', '<>').replace('=','==').replace('<>','!=')
    return entry_output

def check_condition(dict_, str_):
    str_ = clear_entries(str_)

    for k, v in dict_.items():
        exec("{0} = {1}".format(k, v))

    all_names = dis.Bytecode(str_).codeobj.co_names
    l_ = locals()
    non_defined_names = [v for v in all_names if v not in l_]

    for name in non_defined_names:
        exec("{0} = Number(0, exists=False)".format(name))  # the number value does not matter here (because of the 'exists' flag)

    if eval(str_):
        return True

    return False

测试

if __name__ == '__main__':
    entries = [
        "((attr1=25 and attr2=8) or attr3=15)",
        "((attr1>25 and attr2<50) or (attr3=10 and attr4=20))",
        "(2<attrXYZ<10)",
        "(attr1=20 and attr2=20 and attr3=20 and attr4=20)",
        "(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"
    ]

    dicts = [
        {'attr1': 25, 'attr2': 8, 'attr3': 123},
        {'attr1': 1, 'attr2': 8, 'attr3': 123},
        {'attr1': 26, 'attr2': 8, 'attr3': 123, 'attr4': 1},
        {'attr1': 1, 'attr2': 50, 'attr3': 1, 'attr4': 20},
        {'attr1': -1, 'attr2': 50, 'attr3': 1, 'attr4': 20},
        {'attrXYZ': 3},
        {'attrXYZ': 10},
        {'attr1': 20}

    ]

    for entry in entries:
        for d in dicts:
            print(check_condition(d, entry), '"{0}"'.format(entry), d)

结果

(True, '"((attr1=25 and attr2=8) or attr3=15)"', {'attr1': 25, 'attr2': 8, 'attr3': 123})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attr1': 1, 'attr2': 8, 'attr3': 123})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attr1': 26, 'attr2': 8, 'attr3': 123, 'attr4': 1})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attr1': 1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attr1': -1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attrXYZ': 3})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attrXYZ': 10})
(False, '"((attr1=25 and attr2=8) or attr3=15)"', {'attr1': 20})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attr1': 25, 'attr2': 8, 'attr3': 123})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attr1': 1, 'attr2': 8, 'attr3': 123})
(True, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attr1': 26, 'attr2': 8, 'attr3': 123, 'attr4': 1})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attr1': 1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attr1': -1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attrXYZ': 3})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attrXYZ': 10})
(False, '"((attr1>25 and attr2<50) or (attr3=10 and attr4=20))"', {'attr1': 20})
(False, '"(2<attrXYZ<10)"', {'attr1': 25, 'attr2': 8, 'attr3': 123})
(False, '"(2<attrXYZ<10)"', {'attr1': 1, 'attr2': 8, 'attr3': 123})
(False, '"(2<attrXYZ<10)"', {'attr1': 26, 'attr2': 8, 'attr3': 123, 'attr4': 1})
(False, '"(2<attrXYZ<10)"', {'attr1': 1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"(2<attrXYZ<10)"', {'attr1': -1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(True, '"(2<attrXYZ<10)"', {'attrXYZ': 3})
(False, '"(2<attrXYZ<10)"', {'attrXYZ': 10})
(False, '"(2<attrXYZ<10)"', {'attr1': 20})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attr1': 25, 'attr2': 8, 'attr3': 123})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attr1': 1, 'attr2': 8, 'attr3': 123})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attr1': 26, 'attr2': 8, 'attr3': 123, 'attr4': 1})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attr1': 1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attr1': -1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attrXYZ': 3})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attrXYZ': 10})
(False, '"(attr1=20 and attr2=20 and attr3=20 and attr4=20)"', {'attr1': 20})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attr1': 25, 'attr2': 8, 'attr3': 123})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attr1': 1, 'attr2': 8, 'attr3': 123})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attr1': 26, 'attr2': 8, 'attr3': 123, 'attr4': 1})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attr1': 1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attr1': -1, 'attr2': 50, 'attr3': 1, 'attr4': 20})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attrXYZ': 3})
(False, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attrXYZ': 10})
(True, '"(attr1=20 or (attr2=20 and attr3=20 and attr4=20 and attr1231231=1))"', {'attr1': 20})

Answer 2

只有在确保查询字符串是安全的情况下，才能这样做。

(编辑:你真的应该使用像拟解析这样的东西，而不是做一些快速和肮脏的事情。)

如果源来自不受信任的输入，请执行而不是，在查询字符串上使用exec。

import re

QUERY_EXEC_RE = re.compile('(\w+)=')

def _matches(query_exec, d):
    a = []
    exec('a.append({0})'.format(query_exec), globals(), locals())
    return a[0]

def query_dicts(query, dicts):
    query_exec = QUERY_EXEC_RE.sub(r'd.get("\1") == ', query)
    return [d for d in dicts if _matches(query_exec, d)]

示例：

query = "((attr1=25 and attr2=8) or attr3=15)"
dicts = [
    dict(attr1=1, attr2=2, attr3=3),
    dict(attr1=25, attr2=7, attr3=12),
    dict(attr1=24, attr2=8, attr3=13),
    dict(attr1=25, attr2=8, attr3=14),
    dict(attr1=5, attr2=1, attr3=15),
    dict(attr3=15),
    dict(attr1=25, attr2=8),
]
answer = query_dicts(query, dicts)
print(answer)

[{'attr1': 25, 'attr2': 8, 'attr3': 14},
 {'attr1': 5, 'attr2': 1, 'attr3': 15},
 {'attr3': 15},
 {'attr1': 25, 'attr2': 8}]