python - 从嵌套字典中递归删除 None 值或 None 键

Question

问题:

如何遍历字典并从中删除 None 键或值？

这是我尝试过的:

代码:

import copy


def _ignore(data):
    copied_data = copy.deepcopy(data)

    print('-------------------------')
    print(f'copied_data: {copied_data}')
    print('-------------------------')

    if isinstance(copied_data, list):
        print(f'item is instance of list: {copied_data}')

        for idx, item in enumerate(data):
            if isinstance(item, list):
                return _ignore(item)

            elif isinstance(item, dict):
                return _ignore(item)

            elif item is None:
                del copied_data[idx]

    elif isinstance(copied_data, dict):
        print(f'item is instance of dict: {copied_data}')

        for key, item in data.items():
            if isinstance(item, list):
                return _ignore(item)

            elif isinstance(item, dict):
                return _ignore(item)

            elif item is None:
                del copied_data[key]

    return copied_data


if __name__ == '__main__':
    data = {'key1': None, 'key2': {'k1': None, 'k2': 2}, 'key3': [None, 1, 2, 3, 4]}

    print(f'output: {_ignore(data=data)}')

输入:

{'key1': None, 'key2': {'k1': None, 'k2': 2}, 'key3': [None, 1, 2, 3, 4]}

输出:

-------------------------
copied_data: {'key1': None, 'key2': {'k1': None, 'k2': 2}, 'key3': [None, 1, 2, 3, 4]}
-------------------------
item is instance of dict: {'key1': None, 'key2': {'k1': None, 'k2': 2}, 'key3': [None, 1, 2, 3, 4]}
-------------------------
copied_data: {'k1': None, 'k2': 2}
-------------------------
item is instance of dict: {'k1': None, 'k2': 2}
output: {'k2': 2}

Keep in mind that the code also should remove None values in a nested list or nested dict and we can't remove an element from dict while we are iterating over it.

更新:代码应支持嵌套的 dict 或 list

这是另一个输入示例:

{'key1': None, None:1 ,'key2': {'k1': {'k3': [None, 1, 23], 'k4': None}, 'k2': 2},
        'key3': [{'key1': None, 'key2': [None, 1, 2, 3], 'key3': {'k1': 1}}, 1, 2, 3, 4]} 

谢谢。

Answer 1

你没有做你声称要做的事情，因为据我所知，你只是删除了 None 值。这个字典:

{'key1': None, 'key2': {'k1': None, 'k2': 2}, 'key3': [None, 1, 2, 3, 4]}

只有 None 值，没有键。我要在这里坚持键和值的 Python 定义。 None 键看起来像:

{None: "value"}

但是，您可以像这样轻松地做到这一点:

def recursive_filter(item, *forbidden):
    if isinstance(item, list):
        return [recursive_filter(entry, *forbidden) for entry in item if entry not in forbidden]
    if isinstance(item, dict):
        result = {}
        for key, value in item.items():
            value = recursive_filter(value, *forbidden)
            if key not in forbidden and value not in forbidden:
                result[key] = value
        return result
    return item

你可以这样使用:

clean = recursive_filter(dirty, None)

或者如果你想过滤掉更多:

clean = recursive_filter(dirty, *iterable_of_forbidden_things)
clean = recursive_filter(dirty, None, other_forbidden_thing)

如果你真的只关心值，那么你可以去掉对键的检查。