Python - List:如何从带有嵌套属性的字典列表中删除空属性？

Question

因此，我有下面的字典列表，我已经用blank、None或empty values初始化了这些字典。检查列表的每个字典并检查一个属性是否符合这些条件并删除它们，正确的方法是什么。

members = [{
        "member_id": 0,
        'title':  "Manager",
        'full_name': "John Doe",
        'first_name': None,
        'last_name': None,
        'age': 0,
},{
        "member_id": 1,
        'title':  " ",
        'full_name': "Jane Doe",
        'first_name': "Jane",
        'last_name': None,
        'age': 10,
},{...}]

[v for k, v in enumerate(members) if members[k] is not None]  

我期望得到以下结果：

members = [{
            'title':  "Manager",
            'full_name': "John Doe",
    },{
            "member_id": 1,
            'full_name': "Jane Doe",
            'first_name': "Jane",
            'age': 10,
    },{...}]

但似乎我的代码正在返回相同的列表。我在这里做错什么了。谢谢

编辑：，如果我有嵌套的属性，比如地址，我该如何做同样的事情呢？

    members = [{
        "member_id": 0,
        'title':  "Manager",
        'full_name': "John Doe",
        'first_name': None,
        'last_name': None,
        'age': 0,
        'address': {
            'street_number': 123, 
            'street_name': None, 
            'city': "SF", 
            'state': "CA", 
            'zip': None, 
            'full_address': '123 Main Street, San Francisco, CA 94103'
        }
},{
        "member_id": 1,
        'title':  " ",
        'full_name': "Jane Doe",
        'first_name': "Jane",
        'last_name': None,
        'age': 10,
        'address': {
            'street_number': None, 
            'street_name': None, 
            'city': None, 
            'state': None, 
            'zip': None, 
            'full_address': '123 Main Street, San Francisco, CA 94103'
        }
}]

Answer 1

一本字典可以用另一本字典来制作，而用字典的理解来过滤掉“假”值。

{k: v for k, v in d.items() if v}

看起来，您希望其他值像0和' '一样被过滤掉，您可以将它们添加到条件中。

{k: v for k, v in d.items() if v not in (None, ' ', 0)}

然后，您可以使用列表理解与此相同的字典理解，以获得您所需要的。

[{k: v for k, v in d.items() if v not in (None, ' ', 0)} for d in members]

编辑：

用于筛选多个字典级别的递归函数。

def filter_dictionary(value):
    if isinstance(value, dict):
        return {k: filter_dictionary(v) for k, v in value.items() if v not in (None, ' ', 0)}
    elif value not in (None, ' ', 0):
        return value

[filter_dictionary(d) for d in members]

Answer 2

一种使用递归的解决方案：

members = [{
        "member_id": 0,
        'title':  "Manager",
        'full_name': "John Doe",
        'first_name': None,
        'last_name': None,
        'age': 0,
        'address': {
            'street_number': 123,
            'street_name': None,
            'city': "SF",
            'state': "CA",
            'zip': None,
            'full_address': '123 Main Street, San Francisco, CA 94103'
        }
},{
        "member_id": 1,
        'title':  " ",
        'full_name': "Jane Doe",
        'first_name': "Jane",
        'last_name': None,
        'age': 10,
        'address': {
            'street_number': None,
            'street_name': None,
            'city': None,
            'state': None,
            'zip': None,
            'full_address': '123 Main Street, San Francisco, CA 94103'
        }
}]

def traverse(m):
    if isinstance(m, list):
        return list(v for v in (traverse(i) for i in m) if v)
    elif isinstance(m, dict):
        return dict((k, v) for k, v in ((k, traverse(v)) for k, v in m.items()) if v)
    else:
        return None if isinstance(m, str) and not m.strip() else m

filtered_members = traverse(members)

from pprint import pprint
pprint(filtered_members)

指纹：

[{'address': {'city': 'SF',
              'full_address': '123 Main Street, San Francisco, CA 94103',
              'state': 'CA',
              'street_number': 123},
  'full_name': 'John Doe',
  'title': 'Manager'},
 {'address': {'full_address': '123 Main Street, San Francisco, CA 94103'},
  'age': 10,
  'first_name': 'Jane',
  'full_name': 'Jane Doe',
  'member_id': 1}]

Answer 3

像这样吗？

members = [{
        "member_id": 0,
        'title':  "Manager",
        'full_name': "John Doe",
        'first_name': None,
        'last_name': None,
        'age': 0,
},{
        "member_id": 1,
        'title':  " ",
        'full_name': "Jane Doe",
        'first_name': "Jane",
        'last_name': None,
        'age': 10,
}]

for member in members:
    toRemoveKeys = []

    for key, value in member.items():
        if value in (0, None, " "):
            toRemoveKeys.append(key)

    for key in toRemoveKeys:
        del member[key]

print(members)