Question

我有一个嵌套字典，如下所示：

d= {"key1":"A", "key2":"B", "score1":0.1, "score2":0.4, "depth":0, 
"chain":[
 {"key1":"A1", "key2":"B1", "score1":0.2, "score2":0.5, "depth":1, 
  "chain":[{"key1":"A11", "key2":"B11","score1":0.3, "score2":0.6, "depth":2}, 
         {"key1":"A12", "key2":"B12","score1":0.5, "score2":0.7, "depth":2}]
 },
 {"key1":"A2", "key2":"B2","score1":0.1, "score2":0.2,"depth":1,
  "chain":[{None, None, None, None, None}, 
         {"key1":"A22", "key2":"B22","score1":0.1, "score2":0.5, "depth":2}]
 }
    ]
}

我想创建一个函数，当我调用fun(key1, d)时，它可以返回一个保留原始层次结构的字典，但在每个级别内，它将返回key1的值，并总结得分1的值和得分2，如下所示：

{"A":0.5, "depth":0,
"chain":[
 {"A1":0.7, "depth":1,
  "chain":[{"A11":0.9,"depth":2}, 
         {"A12":1.3, "depth":2}]
 },
 {"A2":0.3,"depth":1,
  "chain":[None, 
         {"A22":0.6, "depth":2}]
 }
    ]
}

我该怎么做？我试过了

def gen_dict_extract(key, input_dic):
    return {input_dic[key]:input_dic["score1"]+input_dic["score2"],
        "depth":input_dic["depth"], 
        "chain": gen_dict_extract(key,input_dic["chain"])}

Answer 1

您尝试过的解决方案有两个问题：

chain不保证存在并且
chain是词典列表，您将其视为单个词典

希望以下内容可以满足您的要求：

def gen_dict_extract(key, input_dic):
    rv = {
        input_dic[key]: input_dic["score1"] + input_dic["score2"],
        "depth": input_dic["depth"],
    }
    if "chain" in input_dic:
        rv["chain"] = [gen_dict_extract(key, x) for x in input_dic["chain"]]
    return rv

Answer 2

由于“链”列表中没有“ None”，因此以下功能最终起作用，它根据提供的@dvk解决方案进行了一些细微更新：

def gen_dict_extract(key, input_dic):
rv = {
    input_dic[key]: input_dic["score1"] + input_dic["score2"],
    "depth": input_dic["depth"],
}
if "chain" in input_dic:
    rv["chain"]=[]
    for x in input_dic["chain"]:
        if x is not None:
            rv["chain"].insert(input_dic["chain"].index(x),gen_dict_extract(key, x))
return rv

返回嵌套字典中某些键的值，并形成层次结构字典

2 个答案: