python之根据另一个列表中的值操作字典列表

所以基本上我有三个这样的列表:

list1 = [{ 
    'IP' : "1.1.1.1",  
    'ID' : 1, 
    }, 
    { 
    'IP' : "2.2.2.2",  
    'ID' : 2 
    }] 
 
list2 = [{ 
    'vulnerability_id' : 4567,  
    'ID' : 1, 
    }, 
    { 
    'vulnerability_id' : 6578,  
    'ID' : 2 
    }] 
 
list3 = [{ 
    'vulnerability_id' : 4567,  
    'description' : 'blah', 
    }, 
    { 
    'vulnerability_id' : 6578,  
    'description' : 'blah blah' 
    }] 

我需要做的是获取基于 IP 的漏洞描述，但这可能需要某种列表理解，我不确定如何去做。 (并将其放入新词典中)

我需要检查我拥有的 IP 值，比如 1.1.1.1，然后将它的 ID 与 list2 进行比较，然后比较 list3 中关联的“vulnerability_id”。

任何帮助将不胜感激，如果这没有意义，请说出来，我会尝试和扩展。

请您参考如下方法:

这在计算方面确实很繁重，但您只有列表作为数据源，所以，这是我的“解决方案”:

results = [] 
for ip_info in list1: 
    result = {} 
    result['ip'] = ip_info['IP'] 
    result['vulnerability'] = next(( 
            vuln_info['vulnerability_id'] 
            for vuln_info in list2 
            if vuln_info['ID'] == ip_info['ID'] 
        ),None) 
 
    result['description'] = next(( 
            desc_info['description'] 
            for desc_info in list3 
            if desc_info['vulnerability_id'] == result['vulnerability'] 
        ),None) 
 
    results.append(result) 

结果:

[{'description': 'blah', 'ip': '1.1.1.1', 'vulnerability': 4567}, 
 {'description': 'blah blah', 'ip': '2.2.2.2', 'vulnerability': 6578}] 

编辑:根据我的回答和@Alex Hall 的回答进行改进:

def find(l, match, v, k): 
    return next((x[k] for x in l if x[match] == v), None) 
 
results = [] 
for ip_info in list1: 
    ip = ip_info['IP'] 
    _id = ip_info['ID'] 
    vul = find(list2, 'ID', _id, 'vulnerability_id') 
    desc = find(list3, 'vulnerability_id', vulnerability, 'description')     
    results.append(dict(ip=ip, vulnerability=vul, description=desc))