Question

我需要获得两个嵌套字典的delta：

我正在使用这样的函数来获取嵌套字典

def _get_data(self):
    duplicates = defaultdict(list) # to append tuples into a dictionary
    counter_dict = AutoVivification()
    vpls_dict = AutoVivification()

    fpcs = self._get_slot_fpcs_online()
    pattern = "GOT:\s+(\d+).*([0-9A-F]{2,2}\:[0-9A-F]{1,2}\:[0-9A-F]{1,2}\:[0-9A-F]{1,2}\:[0-9A-F]{1,2}\:[0-9A-F]{1,2})\s+\d{4}\s+(\d\s+\d).*(\d\s+\d+/\d+)"
    regex = re.compile(pattern,re.IGNORECASE)

    for i in fpcs:
        if i == '11':
            for pfe in range(2):
                cmd = self._conn.rpc.request_pfe_execute(target='fpc' + str(i),command='show l2metro '+str(pfe)+' mac hw')                  
                cmd_str = etree.tostring(cmd)
                for x in regex.findall(cmd_str):
                    if x[2] =='0   0' and x[3] != '7 255/255':  
                        duplicates[i].append(x)
        else:
            cmd = self._conn.rpc.request_pfe_execute(target='fpc' + str(i),command='show l2metro 0 mac hw') 
            cmd_str = etree.tostring(cmd)
            for x in regex.findall(cmd_str):
                if x[2] =='0   0' and x[3] != '7 255/255':  
                    duplicates[i].append(x)

    for k,v in duplicates.iteritems():
        for j in v:
            cmd_vpls = self._conn.rpc.get_l2_learning_routing_instances()
            vpls_instance = ''.join(cmd_vpls.xpath("//l2ald-rtb-entry[l2rtb-id=" + '"' + str(j[0]) + '"'"]/l2rtb-name//text()")[0])
            vpls_dict[k][j[1]][j[3]][j[0]][vpls_instance] = self._conn.cli('show configuration routing-instances '+ vpls_instance + ' forwarding-options family vpls filter',warning=False).split('\n')[1].replace('input','').replace(';','')
            counter_cmd = self._conn.rpc.get_firewall_filter_information(filtername=str(vpls_dict[k][j[1]][j[3]][j[0]][vpls_instance]).strip())
            counter_dict[k][j[1]][j[3]][vpls_instance][vpls_dict[k][j[1]][j[3]][j[0]][vpls_instance].strip()] = ''.join(counter_cmd.xpath('./filter-information/policer/packet-count//text()')).replace('\n','')
    return counter_dict

counter_dict结果看起来像：

{＆＃39; 10＆＃39;：{＆＃39; 07：72：9d：dc：4c＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39 ; 128379＆＃39;：{＆＃39; CDALJ1 / 17223002010＆＃39;：＆＃39; 91304＆＃39;}}}，＆＃39; 00：0f：bb：fa：25：fd＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 232367＆＃39;：{＆＃39; CDALJ1 / 14100001093228＆＃39;：＆＃39; 1585097＆＃39;}}}，＆＃39; 00：1b：c0：f2：f4：fa＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 156420＆＃39;：{＆＃39; CDALJ1 / 08903762011＆＃39;：＆＃39; 0＆＃39;}，＆＃39; 166980＆＃39;：{＆＃39; CDALJ1 / 19369922011＆＃39;：＆＃39; 0＆＃39;}}}，＆＃39; 88：e0：f3：61：d8：01＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 182099＆＃39;：{＆＃39; CDALJ1 / 11274452012＆＃39;：＆＃39; 0＆＃39;}}}，＆＃39; ec：13：db：0a：95：01＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 182099＆＃39;：{＆＃39; CDALJ1 / 11274452012＆＃39;：＆＃39; 0＆＃39;}}}}，

＆＃39; 11＆＃39;：{＆＃39; 00：0c：07：ac：75＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 232173＆＃39;：{＆＃39; CDALJ1 / 14100001093242＆＃39;：＆＃39; 0＆＃39;}}}，＆＃39; 00：00：0c：07：ac：f5＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 293667＆＃39;：{＆＃39; CDALJ1 / 14100001095054＆＃39;：＆＃39; 2723092＆＃39;}}}，＆＃39; 00：00：0c：07：ac：f6＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 298967＆＃39;：{＆＃39; CDALJ1 / 14100001095106＆＃39;：＆＃39; 0＆＃39;}}}，＆＃39; 00：00：0c：07：ac：f7＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 298969＆＃39;：{＆＃39; CDALJ1 / 14100001095107＆＃39;：＆＃39; 0＆＃39;}}}，＆＃39; 07：72：9d：dc：4c＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 128379＆＃39;：{＆＃39; CDALJ1 / 17223002010＆＃39;：＆＃39; 91304＆＃39;}}}

[......]

我试图获得内部价值保持字典结构的增量：

mac_dict1 = _get_data（）

{＆＃39; 10＆＃39;：{＆＃39; 07：72：9d：dc：4c＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39 ; 128379＆＃39;：{＆＃39; CDALJ1 / 17223002010＆＃39;：＆＃39; 91304＆＃39;}}}，＆＃39; 00：0f：bb：fa：25：fd＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 232367＆＃39;：{＆＃39; CDALJ1 / 14100001093228＆＃39;：＆＃39; 1585097＆＃39;}}}

sleep5

mac_dict2 = _get_data（）

{＆＃39; 10＆＃39;：{＆＃39; 07：72：9d：dc：4c＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39 ; 128379＆＃39;：{＆＃39; CDALJ1 / 17223002010＆＃39;：＆＃39; 91310＆＃39;}}}，＆＃39; 00：0f：bb：fa：25：fd＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 232367＆＃39;：{＆＃39; CDALJ1 / 14100001093228＆＃39;：＆＃39; 1585100＆＃39;}}}

result = get_diff（mac_dict1，mac_dict2）

结果应该提供如下结果：

{＆＃39; 10＆＃39;：{＆＃39; 07：72：9d：dc：4c＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39 ; 128379＆＃39;：{＆＃39; CDALJ1 / 17223002010＆＃39;：＆＃39; 6＆＃39;}}}，＆＃39; 00：0f：bb：fa：25：fd＆＃39;：{＆＃39; 0 255/255＆＃39;：{＆＃39; 232367＆＃39;：{＆＃39; CDALJ1 / 14100001093228＆＃39;：＆＃39; 3＆＃39;}}}

你能否提供关于如何做到这一点（不是代码）的任何提示或提示？

由于

Answer 1

您可以使用简单的递归函数执行此操作，将str值转换为int，并在dict的情况下减去或递归：

def subtract(x, y):
    if isinstance(x, dict) and isinstance(y, dict):
        return {key: subtract(x[key], y[key]) for key in x if key in y}
    else:
        return str(int(x) - int(y))

使用给定输入运行它：

d1 = {
    '10': {
        '00:07:72:9d:dc:4c': {'0 255/255': {'128379': {'CDALJ1/17223002010': '91304'}}},
        '00:0f:bb:fa:25:fd': {'0 255/255': {'232367': {'CDALJ1/14100001093228': '1585097'}}}
    }
}
d2 = {
    '10': {
        '00:07:72:9d:dc:4c': {'0 255/255': {'128379': {'CDALJ1/17223002010': '91310'}}},
        '00:0f:bb:fa:25:fd': {'0 255/255': {'232367': {'CDALJ1/14100001093228': '1585100'}}}
    }
}

print subtract(d2, d1)

格式化输出：

{
    '10': {
        '00:07:72:9d:dc:4c': {'0 255/255': {'128379': {'CDALJ1/17223002010': '6'}}}, 
        '00:0f:bb:fa:25:fd': {'0 255/255': {'232367': {'CDALJ1/14100001093228': '3'}}}
    }
}

请注意，如果您的输入包含应该被解释为dict的{{1}}和str之外的任何其他类型的值，示例代码将失败。

减去两个嵌套字典python的值

1 个答案: