python - 以大多数pythonic方式从字典列表生成配置样式文件

Question

我有一个字典列表，

[{'section_id': 1, 
'parent_sec_id': 0, 
'sec_name': 'apple', 
'key1': 'val1'},
{'section_id': 2, 
'parent_sec_id': 0, 
'sec_name': 'banana', 
'key2': 'val2'},
{'section_id': 3, 
'parent_sec_id': 1, 
'sec_name': 'orange', 
'key3': 'val3'},
{'section_id': 4, 
'parent_sec_id': 2, 
'sec_name': 'guava', 
'key4': 'val4'},
{'section_id': 5, 
'parent_sec_id': 3, 
'sec_name': 'grape', 
'key5': 'val5'}]

每个字典都有一个字典标识符作为“section_id”，还有一个键作为“parent_section_id”，它告诉它是否是任何其他字典的子字典。所以基本上，如果 parent_section_id 设置为 0（零），那么它是一个父字典，否则它是该部分 id 提到的字典的子字典。
现在从上面的字典列表中，我被要求实现以下格式（是的，我被问到，面试的一部分）：

apple
{
    'key1': 'val1'

    orange
    {
        'key3': 'val3'

        grape
        {
            'key5': 'val5'
        }
    }

}

banana
{
    'key2': 'val2'

    guava
    {
        'key4': 'val4'
    }
}

有人告诉我这是用于为任何程序编写配置文件的格式。我只是好奇从这个字典列表中生成文件的最佳方法是什么。

标签： pythonpython-3.xpython-2.7listdictionary

Answer 1

不是很优雅，但是您可以在 a 中收集您的项目collections.defaultdict()，然后将您的字典路径输出到一个新文件。

基本思想是首先收集值为 0 的根父 ID，然后将后续子字典添加到这些根。您可以将每个列表中的最后一个值用作最近添加项目的父 ID。

演示：

from collections import defaultdict

def group_sections(data, parent_id, section_id, root_id = 0):
    """Groups sections into dictionary of lists, connecting on parent keys"""

    groups = defaultdict(list)

    # Separate root and rest of children
    roots = [dic for dic in data if dic[parent_id] == root_id]
    children = [dic for dic in data if dic[parent_id] != root_id]

    # Add roots first
    for root in roots:
        groups[root[section_id]].append(root)

    # Append children next
    for child in children:
        for key, collection in list(groups.items()):

            # Get most recently added child
            recent = collection[-1]

            # Only add child if equal to parent
            if child[parent_id] == recent[section_id]:
                groups[key].append(child)

    # Filter out result dictionary to not include parent and section ids
    return {
        k1: [
            {k2: v2 for k2, v2 in d.items() if k2 != parent_id and k2 != section_id}
            for d in v2
        ]
        for k1, v2 in groups.items()
    }

def write_config_file(filename, data, name_key):
    """Write config file, using dictionary of lists"""

    # Writes n tabs to string
    tab_str = lambda n: "\t" * n

    with open(filename, mode="w") as config_file:
        for group in data.values():
            tabs = 0
            for dic in group:
                for key in dic:

                    # Write name key
                    if key == name_key:
                        config_file.write(
                            "%s%s\n%s{\n" % (tab_str(tabs), dic[key], tab_str(tabs))
                        )
                        tabs += 1

                    # Otherwise write key-value pairs
                    else:
                        config_file.write(
                            "%s'%s': '%s'\n" % (tab_str(tabs), key, dic[key])
                        )

            # Write ending curly braces
            for i in range(tabs - 1, -1, -1):
                config_file.write("%s}\n" % (tab_str(i)))

if __name__ == "__main__":
    list_dicts = [
        {"section_id": 1, "parent_sec_id": 0, "sec_name": "apple", "key1": "val1"},
        {"section_id": 2, "parent_sec_id": 0, "sec_name": "banana", "key2": "val2"},
        {"section_id": 3, "parent_sec_id": 1, "sec_name": "orange", "key3": "val3"},
        {"section_id": 4, "parent_sec_id": 2, "sec_name": "guava", "key4": "val4"},
        {"section_id": 5, "parent_sec_id": 3, "sec_name": "grape", "key5": "val5"},
    ]

    data = group_sections(data=list_dicts, parent_id="parent_sec_id", section_id="section_id")
    write_config_file(filename='config', data=data, name_key='sec_name')

配置文件：

apple
{
    'key1': 'val1'
    orange
    {
        'key3': 'val3'
        grape
        {
            'key5': 'val5'
        }
    }
}
banana
{
    'key2': 'val2'
    guava
    {
        'key4': 'val4'
    }
}

注意：这是一种迭代解决方案，而不是递归解决方案。