打印/获取字典中每个嵌套项的完整路径-Python3

我有一个多层嵌套的python字典。 我试图将这个字典转换成一个字符串，看起来像人类编写的python代码。 我已经搜索了stackoverflow的解决方案，但找不到一个合适的-请指出一个，如果它回答了我的问题

注:

1） 某些键名在嵌套字典中的不同深度重复

2） 字典本身可以更改形状、内容，甚至可能更改键名，因此此脚本应适用于不同的情况

3） 为了简化这个问题，我将所有嵌套列表转换为字典-这些字典的键是转换为字符串类型的数字-但是使用嵌套列表的解决方案是可以的

目标是转换：

sample_dictionary = {
    'id' : '123',
    'type' : 'testing',
    'records' : {
        '0' : {
            'record_id' : '324',
            'record_type' : 'test1',
        },
        '1' : {
            'record_id' : '121',
            'record_type' : 'test2',
            'sub_records' : {
                '0' : {
                    'sub_record_id' : 'sub_578',
                    'sub_record_type' : 'sub_test',
                },
            },
        },
    },
}

放入字符串：

sample_dictionary[id] = 123
sample_dictionary[type] = testing
sample_dictionary[records][0][record_id] = 324
sample_dictionary[records][0][record_type] = test1
sample_dictionary[records][1][record_id] = 121
sample_dictionary[records][1][record_id] = test2
sample_dictionary[records][1][sub_records][0][sub_record_id] = sub_578
sample_dictionary[records][1][sub_records][0][sub_record_type] = sub_test

我创建了这个递归函数，该函数非常接近，但不适当地包含了额外的键：

def FindValues(dictionary, my_keys=''):

    for key, value in dictionary.items():

        my_keys += '[' + key + ']'

        if type(value) is dict:

            FindValues(value, my_keys)

        else: 

            result = 'sample_dictionary' + my_keys + ' = ' + value

            print(result)


FindValues(sample_dictionary)

导致：

sample_dictionary[id] = 123
sample_dictionary[id][type] = testing
sample_dictionary[id][type][records][0][record_id] = 324
sample_dictionary[id][type][records][0][record_id][record_type] = test1
sample_dictionary[id][type][records][0][1][record_id] = 121
sample_dictionary[id][type][records][0][1][record_id][record_type] = test2
sample_dictionary[id][type][records][0][1][record_id][record_type][sub_records][0][sub_record_id] = sub_578
sample_dictionary[id][type][records][0][1][record_id][record_type][sub_records][0][sub_record_id][sub_record_type] = sub_test

---

您不应在

my_keys

变量中累加所有键，而是仅基于当前dict中的当前键创建新字符串：

def FindValues(dictionary, my_keys=''):

    for key, value in dictionary.items():

        current_key = my_keys + '[' + key + ']'

        if type(value) is dict:

            FindValues(value, current_key)

        else: 

            result = 'sample_dictionary' + current_key  + ' = ' + value

            print(result)

FindValues(sample_dictionary)

输出：

sample_dictionary[id] = 123
sample_dictionary[type] = testing
sample_dictionary[records][0][record_id] = 324
sample_dictionary[records][0][record_type] = test1
sample_dictionary[records][1][record_id] = 121
sample_dictionary[records][1][record_type] = test2
sample_dictionary[records][1][sub_records][0][sub_record_id] = sub_578
sample_dictionary[records][1][sub_records][0][sub_record_type] = sub_test

['sample_dictionary[id] = 123',
 'sample_dictionary[type] = testing',
 'sample_dictionary[records][0][record_id] = 324',
 'sample_dictionary[records][0][record_type] = test1',
 'sample_dictionary[records][1][record_id] = 121',
 'sample_dictionary[records][1][record_type] = test2',
 'sample_dictionary[records][1][sub_records][0][sub_record_id] = sub_578',
 'sample_dictionary[records][1][sub_records][0][sub_record_type] = sub_test']

---

如果希望函数返回包含字符串的列表，可以使用：

def FindValues(dictionary, my_keys=''):

    output = []

    for key, value in dictionary.items():

        current_key = my_keys + '[' + key + ']'

        if type(value) is dict:

            output += FindValues(value, current_key)

        else: 

            output.append('sample_dictionary' + current_key  + ' = ' + value)

    return output

FindValues(sample_dictionary)

输出：

sample_dictionary[id] = 123
sample_dictionary[type] = testing
sample_dictionary[records][0][record_id] = 324
sample_dictionary[records][0][record_type] = test1
sample_dictionary[records][1][record_id] = 121
sample_dictionary[records][1][record_type] = test2
sample_dictionary[records][1][sub_records][0][sub_record_id] = sub_578
sample_dictionary[records][1][sub_records][0][sub_record_type] = sub_test

['sample_dictionary[id] = 123',
 'sample_dictionary[type] = testing',
 'sample_dictionary[records][0][record_id] = 324',
 'sample_dictionary[records][0][record_type] = test1',
 'sample_dictionary[records][1][record_id] = 121',
 'sample_dictionary[records][1][record_type] = test2',
 'sample_dictionary[records][1][sub_records][0][sub_record_id] = sub_578',
 'sample_dictionary[records][1][sub_records][0][sub_record_type] = sub_test']

---

谢谢-我一直在努力解决这个问题-我需要的最后一部分是返回函数的输出（作为字符串列表，或以新行分隔的单个字符串）-有没有可能帮上忙？