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打印/获取字典中每个嵌套项目的完整路径 - Python3

热度:141   发布时间:2023-07-16 11:05:55.0

我有一个多层嵌套的 python 字典。 我正在尝试将此字典转换为看起来像人类编写的 Python 代码的字符串。 我已经在 stackoverflow 上搜索了一个解决方案,但找不到合适的解决方案 - 如果它回答了我的问题,请指出一个。

笔记:

1) 嵌套字典中某些键名在不同深度重复

2) 字典本身可以改变形状、内容,甚至键名,因此这个脚本应该适用于不同的情况

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',
                },
            },
        },
    },
}

成字符串:

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

我创建了这个接近的递归函数,但不恰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

您不应该在my_keys变量中累积所有键,而是仅根据当前字典中的当前键创建一个新字符串:

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

如果您希望函数返回包含字符串的列表,您可以使用:

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']
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