递归地访问嵌套字典的路径和值

In Python 2.7, how does one dynamically access and print out the keys and values of a nested dictionary? Here's a nonsensical example: https://jsoneditoronline.org/?id=da7a486dc2e24bf8b94add9f04c71b4d

在Python 2.7中，如何动态地访问和打印嵌套字典的键和值?这里有一个荒谬的例子:https://jsoneditoronline.org/?

Normally, I would do something like:

通常，我会这样做:

import json

json_sample = 'sample_dict.json'
json_file = open(json_sample, 'r')
json_data = json.load(json_file)

items = json_data['sample_dict']

for item in items:
    dict_id = item['dict_id']
    person = item['person']['person_id']
    family = item['family']['members']

    print dict_id
    print person
    print family

I can hard code it like this and it'll give me desirable results, but how would I access each of the keys and values dynamically so that:

我可以像这样硬编码，它会给我理想的结果，但是我如何动态地访问每个键和值，以便:

The first row just prints the keys (dict_id, person['person_id'], person['name'], family['members']['father'])
第一行只打印键(dict_id、person['person_id]]、person['name']、family['members']['father'])
The second row prints the values respectively (5, 15, "Martin", "Jose")
第二行分别打印值(5、15、“Martin”、“Jose”)

The end result should be in a CSV file.

最终结果应该在CSV文件中。

1 个解决方案

#1

You can use a recursive visitor/generator which returns all the path/value pairs of the leaves:

您可以使用递归的访问者/生成器返回所有路径/值对的叶子:

def visit_dict(d, path=[]):
    for k, v in d.items():
        if not isinstance(v, dict):
            yield path + [k], v
        else:
            yield from visit_dict(v, path + [k])

(replace the yield from ... with the appropriate equivalent if using Python < 3.4)

(替换掉……如果使用Python < 3.4，则使用适当的等效值)

Getting the keys:

把钥匙:

>>> ','.join('/'.join(k) for k, v in visit_dict(json_data['sample_dict'][0]))
'dict_id,person/person_id,person/name,person/age,family/person_id,family/members/father,family/members/mother,family/members/son,family/family_id,items_id,furniture/type,furniture/color,furniture/size,furniture/purchases'

and the values:

和值:

>>> ','.join(str(v) for k, v in visit_dict(json_data['sample_dict'][0]))
'5,15,Martin,18,20,Jose,Maddie,Jerry,2,None,Chair,Brown,Large,[]'

#1