将带有字典列表的 defaultdict(list) 字典转换为 csv 的最佳方法

Question

我的默认字典有一个地址键，并且有一个与该键匹配的字典列表。我想将此默认字典导出到 csv 文件。

见下文：

Right now my structure looks like this defaultdict(list)
#As you can see 1 key with multiple matching dictionaries. 
#And im just copying 1 address but I have ~10 w/ varying matches

defaultdic1 = 

defaultdict(list,
            {'Address_1': [{'Name': 'name',
               'Address_match': 'address_match_1',
               'ID': 'id',
               'Type': 'abc'},
              {'Name': 'name',
               'Address_match': 'address_match_2',
               'ID': 'id',
               'Type': 'abc'},
              {'Name': 'name',
               'Address_match': 'address_match_3',
               'ID': 'id',
               'Type': 'abc'}]})

我试过这样做：

json_data = json.dumps(data_json, indent=2)

jsondf = pd.read_json(json_data, typ = 'series')

and my result was this:

Address 1       [{'Name':'name', 'Address_match':'address_match_1' 'ID' : 'id', 'Type':'abc'} {'Name':'name', 'Address_match':'address_match_2' 'ID' : 'id', 'Type':'abc'}, {'Name':'name', 'Address_match':'address_match_3' 'ID' : 'id', 'Type':'abc'}]

结果/输出：

我想把它导出到一个 excel 文件中

更新我试过了。第一行正在打印密钥，但第二行仍在 {} 中，最好将它们从括号中取出并移到列中。有什么提示吗？

    for k, v in defaultdict.items():
        f.writerow([k])
        for values in v:
            f.writerow([values])

results in CSV are:

Address 1

{'Name':'name', 'Address_match':'address_match_1' 'ID' : 'id', 'Type':'abc'}
{'Name':'name', 'Address_match':'address_match_1' 'ID' : 'id', 'Type':'abc'}
{'Name':'name', 'Address_match':'address_match_2' 'ID' : 'id', 'Type':'abc'}

我希望我的结果是：

Address 1                Name, Address_match1, ID, Type
                         Name, Address_match2, ID, Type
                         Name, Address_match3, ID, Type

Address 2                Name1, Address_match1, ID, Type
                         Name1, Address_match1, ID, Type

Address 3                Name1, Address_match1, ID, Type
                         Name1, Address_match1, ID, Type

Answer 1

您的输入数据和输出数据不匹配，因此很难判断如何转换，但这里有一些东西可以使用您的 defaultdict 并将其转换为 CSV 文件：

import csv

dic1 = {'Address_2': 
    [
        {'Address 1': 
            [
                {'Name':'name', 'Address_match':'address_match_1', 'ID':'id', 'Type':'abc'}
            ]
        },
        {'Address 2': 
            [
                {'Name':'name', 'Address_match':'address_match_2', 'ID':'id', 'Type':'abc'}
            ]
        }, 
        {'Address 3': 
            [
                {'Name':'name', 'Address_match':'address_match_3', 'ID':'id', 'Type':'abc'}
            ]
        }
    ]
}

names = list(dic1['Address_2'][0]['Address 1'][0].keys())

myfile = csv.DictWriter( open('xxx.csv','w'), fieldnames = names  )
for row in dic1['Address_2']:
    myfile.writerow({'Name':list(row.keys())[0]})
    myfile.writerow(list(row.values())[0][0])

Answer 2

这就是最终解决它的原因！

names = list(dic1['Address_1'][0].keys())
f.close()
with open ("file.csv", "w", newline="") as f:
    writer = csv.writer(f)
    keys = names
    writer.writerow(["Address"] +(keys))
    for k, vl in defaultdict.items():
        for v in vl:
            writer.writerow([k] + [v[key] for key in keys])
f.close()