python - 遍历多层字典

问题描述

这是我的数据

data = [
    {
        "title": "Main Topic 1",
        "num": "Type 1",
        "text": "Some Text",
        "sub": [
            {
                "title": "Sub Topic 1",
                "num": "1",
                "text": "Some more Text",
                "sub": [
                    {
                        "num": "(a)",
                        "text": "This is the actual text for the topic 1(a)",
                    },
                    {
                        "num": "(b)",
                        "text": "This is the actual text for the topic 1(b)",
                    },
                ],
            },
            {
                "title": "Sub Topic 2",
                "num": "2",
                "text": "This is the actual text for the topic 2",
            },
        ],
    },
    {
        "title": "Main Topic 2",
        "num": "Type 2",
        "text": "Some Text",
        "sub": [
            {
                "title": "Sub Topic 3",
                "num": "3",
                "text": "Some more Text",
                "sub": [
                    {
                        "num": "(a)",
                        "text": "This is the actual text for the topic 3(a)",
                    },
                    {
                        "num": "(b)",
                        "text": "This is the actual text for the topic 3(b)",
                    },
                ],
            },
            {
                "title": "Sub Topic 4",
                "num": "4",
                "text": "This is the actual text for the topic 4",
            },
        ],
    },
]

现在，我想要这样的输出：

{'title': 'Main Topic 1~Sub Topic 1~NA', 'num': 'Type 1~1~(a)', 'text': 'This is the actual text for the topic 1(a)'}
{'title': 'Main Topic 1~Sub Topic 1~NA', 'num': 'Type 1~1~(b)', 'text': 'This is the actual text for the topic 1(b)'}
{'title': 'Main Topic 1~Sub Topic 2', 'num': 'Type 1~2', 'text': 'This is the actual text for the topic 2'}
{'title': 'Main Topic 2~Sub Topic 3~NA', 'num': 'Type 2~3~(a)', 'text': 'This is the actual text for the topic 3(a)'}
{'title': 'Main Topic 2~Sub Topic 3~NA', 'num': 'Type 2~3~(b)', 'text': 'This is the actual text for the topic 3(b)'}
{'title': 'Main Topic 2~Sub Topic 4', 'num': 'Type 2~4', 'text': 'This is the actual text for the topic 4'}

这是我实现这一目标的代码：

def get_each_provision(title, num, text):
    provision = {}
    provision['title'] = title
    provision['num'] = num
    provision['text'] = text
    return provision


def get_consolidated_provisions(data):
    provisions = []
    for level1 in data:

        title_level1 = level1['title']
        num_level1 = level1['num']
        text_level1 = level1['text']

        if 'sub' in level1:

            level2_subs = level1['sub']
            for level2 in level2_subs:

                title_level2 = '%s~%s'%(title_level1, level2['title'])
                num_level2 = '%s~%s'%(num_level1, level2['num'])
                text_level2 = level2['text']

                if 'sub' in level2:

                    level3_subs = level2['sub']
                    for level3 in level3_subs:

                        title = '%s~%s'%(title_level2, level3.get('title', 'NA'))
                        num = '%s~%s'%(num_level2, level3['num'])
                        text = level3['text']
                        provisions.append(get_each_provision(title, num, text))

                else:
                    provisions.append(get_each_provision(title_level2, num_level2, text_level2))

        else:
            provisions.append(get_each_provision(title_level1, num_level1, text_level1))

    return provisions

print('----------------------------------------------')
provisions = get_consolidated_provisions(data)
for each_provision in provisions:
    print(each_provision)

它按预期工作正常。我想要实现的是 - 基本上从每个字典和子字典中获取最低级别的“文本”（在键“sub”下）我的问题有两个方面：（1）有没有更好的方法来实现这一点？(2) 如果有另一级字典列表，我的代码将中断。我可以申请另一个级别，但希望不要。

如果您想知道，上面的变量“data”是通过提取pdf文件实现的json格式。数据提取成功，如变量“数据”所示。这个想法是识别每个小节及其“num”和“title”的前导序列。

需要注意的一点：（1）最低级别没有键'title'和（2）最低级别的字典没有键'sub'。两者都如变量数据所示。

标签： pythondictionary

def flatten(items): new_list = [] for i in items: if "sub" in i: new_dict = {} for k, v in i.items(): if not k == "sub": new_dict[k] = v new_list.append(new_dict) new_list += flatten(i["sub"]) else: new_list.append(i) return new_list

python - 遍历多层字典

问题描述

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