从嵌套字典中提取匹配值

Question

如果值与列表中的值匹配，我将尝试从嵌套字典中提取值

data = [
            {
                "id": 12345678,
                "list_id": 12345,
                "creator_id": 1234567,
                "entity_id": 1234567,
                "created_at": "2020-01-30T00:43:55.256-08:00",
                "entity": {
                    "id": 123456,
                    "type": 0,
                    "first_name": "John",
                    "last_name": "Doe",
                    "primary_email": "john@fakemail.com",
                    "emails": [
                        "john@fakemail.com"
                    ]
                }
            },
            {
                "id": 12345678,
                "list_id": 12345,
                "creator_id": 1234567,
                "entity_id": 1234567,
                "created_at": "2020-01-30T00:41:54.375-08:00",
                "entity": {
                    "id": 123456,
                    "type": 0,
                    "first_name": "Jane",
                    "last_name": "Doe",
                    "primary_email": "jane@fakemail.com",
                    "emails": [
                        "jane@fakemail.com"
                    ]
                }
            }
        ]

代码如下

match_list = ['jane@fakemail.com',[]]
first_names = []
email = []
for i in match_list:
    for record in data:
        if 'primary_email' == i:
            email.append(record.get('entity',{}).get('primary_email', None))
            first_names.append(record.get('entity',{}).get('first_name', None))       
print(first_names)
print(email)

它不返回匹配值，只返回空列表。这里的任何帮助都将不胜感激

预期产量为

first_names = ['Jane'] and email = ['jane@fakemail.com']

Answer 1

在您的代码中，当您比较'primary_email'==i时，总是会得到一个空列表，它总是False

将其更改为record['entity']['primary_email']==i

这里不需要使用get。因为如果mail与任何primary_email不匹配，那么什么也不会发生primary_email将仅在满足条件{}时添加

试试这个，我重构了你的代码

In [25]: for mail in match_list:
    ...:     for d in data:
    ...:         if d['entity']['primary_email']==mail:
    ...:             first_name.append(d['entity']['first_name'])
    ...:             emails.append(d['entity']['primary_email'])

输出

In [26]: emails
Out[26]: ['jane@fakemail.com']

In [27]: first_name
Out[27]: ['Jane']

Answer 2

将临时值存储在变量中，以使代码更易于处理：

emails = []
names = []
match_list = ['jane@fakemail.com',[]]


for item in data:
    entry = item.get('entity', {})

    fName = entry.get('first_name', '')
    pMail = entry.get('primary_email', '')

    if pMail in match_list:
        print (fName)
        print (pMail)

        emails.append(pMail)
        names.append(fName)

输出：

Jane
jane@fakemail.com

Answer 3

在代码的第6行

    if 'primary_email' == i:

您正在将match_list（即“i”）中的元素与名为“primary_email”的字符串（而不是实际的email）进行比较。自从jane@fakemail.com'不等于'primary_email'（字面上是字符串）

改用

if record['entity']['primary_email'] == i:

您的代码应该按预期工作