如何将此python循环转换为列表理解？

如何解决如何将此python循环转换为列表理解？

没关系。正如 juanpa.arrivillaga 所指出的，列表推导不返回字典 >

背景

我有以下 python 代码（请参阅下一段代码）。我想尝试优化它以比较常规循环和列表理解之间的执行时间。

常规循环版本

def flatten_json( nested_dict,flattened_dict={},superior_level_key: str = ""):

    for key,value in nested_dict.items():
        if type(nested_dict[key]) is dict:
            flattened_dict = flatten_json(
                nested_dict[key],flattened_dict,"{}_".format(key))
        else:
            flattened_dict['{}{}'.format(superior_level_key,key)] = value

    return flattened_dict


import json

with open('json.json') as j:
    d = json.load(j)
    print(flatten_json(d,{},""))

当前的、失败的、列表理解版本

def flatten_json(nested_dict,superior_level_key: str = ""):

    return [flatten_json(nested_dict[key],"{}_".format(key))
            if type(nested_dict) is dict
            else value for key,value in nested_dict.items()]


import json

with open('json.json') as j:
    d = json.load(j)
    print(flatten_json(d,""))

错误

列表理解版本抛出以下错误：

Traceback (most recent call last):
  File "p1.py",line 13,in <module>
    print(flatten_json(d,""))
  File "p1.py",line 3,in flatten_json
    return [flatten_json(nested_dict[key],"{}_".format(key))
  File "p1.py",in <listcomp>
    return [flatten_json(nested_dict[key],line 5,in flatten_json
    else value for key,value in nested_dict.items()]
AttributeError: 'float' object has no attribute 'items'

问题

为什么会抛出该错误以及如何修复它？

输入

{
    "_score": 1.0,"_index": "sirene_prod","_id": "AXSp612eur2DngRir4BH","_type": "sirene_prod","_source": {
        "enseigne": "","codpos": {
            "cp": "17300","bur_distrib": "300","depet": "17"
        },"id": "ddf9e5b2aa0099ff6934a3d83b1678f64e27859e377362ef8682a9b1","l3_normalisee": "","apet700": "10.71C","sigle": "","siren": "793120569","libapen": "Boulangerie et boulangerie-patisserie","apen700": "10.71C","cedex": "","typvoie": "AV","numvoie": 33,"nom": "","depet_limit": [
            "16","24","33","79","85"
        ],"libcom": "ROCHEFORT","l2_normalisee": {
            "text": "","initial": ""
        },"libvoie": "GAMBETTA","nic": "00017","prenom": "","nomen_long": {
            "text": "LA PASSION DU PAIN","nom": "LA PASSION DU PAIN","initial": "LPDP"
        },"indrep": ""
    }
}

输出

{'_score': 1.0,'_index': 'sirene_prod','_id': 'AXSp612eur2DngRir4BH','_type': 'sirene_prod','_surce_enseigne': '','codpos_cp': '17300','codpos_bur_distrib': '300','codpos_depet': '17','_soure_id': 'ddf9e5b2aa0099ff6934a3d83b1678f64e27859e377362ef8682a9b1','_source_l3_normalisee': '','_surce_apet700': '10.71C','_source_sigle': '','_source_siren': '793120569','_source_libapen': 'Bouangerie et boulangerie-patisserie','_source_apen700': '10.71C','_source_cedex': '','_source_typvie': 'AV','_source_numvoie': 33,'_source_nom': '','_source_depet_limit': ['16','24','33','79' '85'],'_source_libcom': 'ROCHEFORT','l2_normalisee_text': '','l2_normalisee_nom': '','l2_normamalisee_initial': '','_source_libvoie': 'GAMBETTA','_source_nic': '00017','_source_prenom': '',renom': '','nomen_long_text': 'LA PASSION DU PAIN','nomen_long_nom': 'LA PASSION DU P_long_initiaAIN','nomen_long_initial': 'LPDP','_source_indrep': ''}

解决方法

这是一个迭代解决方案，它应该比您当前的递归解决方案更快：

def flatten_json_iterative(nested):
    flattened = {}
    stack = [(nested,"")]
    push_to_stack = stack.append
    pop_from_stack = stack.pop
    while stack:
        nested_dict,superior_key = pop_from_stack()
        for key,value in nested_dict.items():
            if isinstance(value,dict):
                push_to_stack((value,f"{key}_"))
            else:
                flattened[f"{superior_key}{key}"] = value
    return flattened

在回复中：

In [8]: flatten_json_iterative(data)
Out[8]:
{'_score': 1.0,'_index': 'sirene_prod','_id': 'AXSp612eur2DngRir4BH','_type': 'sirene_prod','_source_enseigne': '','_source_id': 'ddf9e5b2aa0099ff6934a3d83b1678f64e27859e377362ef8682a9b1','_source_l3_normalisee': '','_source_apet700': '10.71C','_source_sigle': '','_source_siren': '793120569','_source_libapen': 'Boulangerie et boulangerie-patisserie','_source_apen700': '10.71C','_source_cedex': '','_source_typvoie': 'AV','_source_numvoie': 33,'_source_nom': '','_source_depet_limit': ['16','24','33','79','85'],'_source_libcom': 'ROCHEFORT','_source_libvoie': 'GAMBETTA','_source_nic': '00017','_source_prenom': '','_source_indrep': '','nomen_long_text': 'LA PASSION DU PAIN','nomen_long_nom': 'LA PASSION DU PAIN','nomen_long_initial': 'LPDP','l2_normalisee_text': '','l2_normalisee_nom': '','l2_normalisee_initial': '','codpos_cp': '17300','codpos_bur_distrib': '300','codpos_depet': '17'}

顺便说一句：

您的方法不起作用，因为您正在检查 nested_dict 是否为 dict 然后对其进行递归，if type(nested_dict) is dict 相反，您想检查 value 是一个字典，然后递归它。但这不会对你有多大帮助。我不够聪明，无法找到一种使用递归和字典理解的方法来展平像这样的嵌套字典。