Question

我希望将未知结构的JSON层次结构扁平化为字典，以捕获字典结果中的完整键层次结构以唯一地标识它。

到目前为止，我能够递归地打印所有父/子节点的key：value对，但是遇到了麻烦：

（1）弄清楚如何传递父级层次结构键来执行递归（子级）执行，然后在退出子级键时将其重置。

（2）写一个字典结果-当我在递归函数中定义字典时，最终会创建多个字典...为了避免这种情况，是否需要将该函数包装在主函数中？

谢谢！

# flatten/enumerate example I'm using

with open('_json\\file.json') as f:
    data = json.load(f)

def parse_json_response(content):

    if len (content.keys()) > 1 :
        for key, value in content.items():
            if type(value) is dict:
                parse_json_response(value)
            else:
                print('{}:{}'.format(key,value))
    else:
        print(value)

if __name__ == '__main__':
    parse_json_response(data)

# current result as print
id = 12345
firstName = John
lastName = Smith
DOB = 1980-01-01
phone = 123
line1 = Unit 4
line2 = 3 Main st

# desired result to dictionary {}
id = 12345
fields.firstName = John
fields.lastName = Smith
fields.DOB = 1980-01-01
fields.phone = 123
fields.address.residential.line1 = Unit 4
fields.address.residential.line2 = 3 Main st

Answer 1

尝试以下操作：

test = {
    "ID": "12345",
    "fields": {
        "firstName": "John",
        "lastName": "Smith",
        "DOB": "1980-01-01",
        "phoneLand": "610292659333",
        "address": {
            "residential": {
                "line1": "Unit 4",
                "line2": "3 Main st"
            }
        }
    }
}

def func(d, parent=""):
    for key, value in d.items():
        if isinstance(value, dict):
            func(value, parent=parent+key+".")
        else:
            print(f"{parent+key} = {value}")


func(test)

结果：

ID = 12345
fields.firstName = John
fields.lastName = Smith
fields.DOB = 1980-01-01
fields.phoneLand = 610292659333
fields.address.residential.line1 = Unit 4
fields.address.residential.line2 = 3 Main st

Answer 2

通过跟踪父级并在正确的位置递归，可以创建展平的字典（而不只是打印值）。可能看起来像这样：

d = {
    "ID": "12345",
    "fields": {
        "firstName": "John",
        "lastName": "Smith",
        "DOB": "1980-01-01",
        "phoneLand": "610292659333",
        "address": {
            "residential": {
                "line1": "Unit 4",
                "line2": "3 Main st"
            }
        }
    }
}

def flattenDict(d, parent=None):
    ret = {}
    for k, v in d.items():
        if parent:
            k = f'{parent}.{k}'
        if isinstance(v, dict):
            ret.update(flattenDict(v, k))
        else:
            ret[k] = v
    return ret

flat = flattenDict(d)

flat将是：

{'ID': '12345',
 'fields.firstName': 'John',
 'fields.lastName': 'Smith',
 'fields.DOB': '1980-01-01',
 'fields.phoneLand': '610292659333',
 'fields.address.residential.line1': 'Unit 4',
 'fields.address.residential.line2': '3 Main st'}

您还可以将输出安排为生成元组的生成器。然后，您可以将其传递给dict()以获得相同的结果：

def flattenDict(d):
    for k, v in d.items():
        if isinstance(v, dict):
            yield from ((f'{k}.{kk}', v) for kk, v in flattenDict(v))
        else:
            yield (k, v)

dict(flattenDict(d))

递归枚举JSON层次结构的父级/子级到字典

2 个答案: