如何将字典中的一些键分组到python中的嵌套键?

hanzgs • 3 年前 • 1135 次点击

我有一本字典清单

[{
'LoanID' : 001,
'Applicant_Income.1' : 5000,
'Applicant_Occupation.1' : "Manager",
'Applicant_Income.2' : 3000,
'Applicant_Occupation.2' : "Accountant",
'LoanAmount' : 10000
},
{
'LoanID' : 002,
'Applicant_Income.1' : 6000,
'Applicant_Occupation.1' : "Senior Manager",
'Applicant_Income.2' : 2000,
'Applicant_Occupation.2' : "Services",
'LoanAmount' : 8000
}]

我需要创建一个新的键“申请人”,并将“申请人收入”和“申请人职业”移到其中,结果列表中的每个dict如下所示

{
'LoanID' : 001,
'Applicant' : [
     { 
       'Applicant_Income' : 5000,
       'Applicant_Occupation' : "Manager"
     }'
     {
       'Applicant_Income' : 3000,
       'Applicant_Occupation' : "Accountant"
     }]
'LoanAmount' : 10000
},
{
'LoanID' : 002,
'Applicant' : [
     { 
       'Applicant_Income' : 6000,
       'Applicant_Occupation' : "Senior Manager"
     }'
     {
       'Applicant_Income' : 2000,
       'Applicant_Occupation' : "Services"
     }]
'LoanAmount' : 8000
}

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1135 次点击

文章 [ 2 ] | 最新文章 3 年前

• 1 楼

David 3 年前

另一个答案只有在收入和职业值的数量固定为2时才有效。

下面是一个更加动态的解决方案:
组织列表 是你在上面定义的列表吗
终结者 这是你想要的清单吗

def getApplicant(entry):
    tempAppList = []
    for key in [key for key, value in entry.items() if 'Applicant_Income' in key]:
        tempAppDict = {}
        tempAppDict['Applicant_Income'] = entry[key]
        tempAppDict['Applicant_Occupation'] = entry[key.replace('Income', 'Occupation')]
        tempAppList.append(tempAppDict)
    
    return tempAppList

finalList = []

for entry in orgList:
    tempDict = {}
    tempDict['LoanID'] = entry['LoanID']
    tempDict['Applicant'] = getApplicant(entry)
    tempDict['LoanAmount'] = entry['LoanAmount']
    
    finalList.append(tempDict)

输出:

[
    {
        "LoanID": "001",
        "Applicant": [
            {
                "Applicant_Income": 5000,
                "Applicant_Occupation": "Manager"
            },
            {
                "Applicant_Income": 3000,
                "Applicant_Occupation": "Accountant"
            }
        ],
        "LoanAmount": 10000
    },
    {
        "LoanID": "002",
        "Applicant": [
            {
                "Applicant_Income": 6000,
                "Applicant_Occupation": "Senior Manager"
            },
            {
                "Applicant_Income": 2000,
                "Applicant_Occupation": "Services"
            }
        ],
        "LoanAmount": 8000
    }
]

注意:整数上不允许使用前导零(就像LoadId上一样)。所以你需要用一个字符串

• 2 楼

catasaurus 3 年前

这应该管用( info 是您在上面定义的列表):

for d in range(len(info)):
    info[d]['Applicant'] = [{'Applicant_Income.1': info[d]['Applicant_Income.1'], 'Applicant_Occupation.1': info[d]['Applicant_Occupation.1']}, {'Applicant_Income.2': info[d]['Applicant_Income.2'], 'Applicant_Occupation.2': info[d]['Applicant_Occupation.2']}]
    del info[d]['Applicant_Income.1']
    del info[d]['Applicant_Occupation.1']
    del info[d]['Applicant_Income.2']
    del info[d]['Applicant_Occupation.2']

输出:

[{'LoanID': '002',
  'LoanAmount': 10000,
  'Applicant': [{'Applicant_Income.1': 5000,
    'Applicant_Occupation.1': 'Manager'},
   {'Applicant_Income.2': 3000, 'Applicant_Occupation.2': 'Accountant'}]},
 {'LoanID': '002',
  'LoanAmount': 8000,
  'Applicant': [{'Applicant_Income.1': 6000,
    'Applicant_Occupation.1': 'Senior Manager'},
   {'Applicant_Income.2': 2000, 'Applicant_Occupation.2': 'Services'}]}]

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