Question

我正在尝试从列表中删除嵌套字典中的元素。因此，我有一个列表，其中有一个字典作为其元素之一，然后又有一个字典作为其键值之一。我知道我将在最上面的列表和最下面的字典中使用的键，而“中间”字典中的键是可变的。如果底部字典中的值不等于某个值，我想删除“中间”字典中的元素。

我认为我需要三个嵌套的for循环才能进入每个字典，但无法弄清楚如何删除“中间”字典中的元素。

到目前为止，我目前的尝试：

#remove unneccessary elements
def getCleanResults(the_dict1):
    for elem1 in the_dict1:
        the_dict2 = elem['dict1_key']
        for key, elem2 in the_dict2.keys():
            the_dict3 = elem2[key]
            for elem3 in the_dict3:
                if 'keyWanted' in elem3:
                    del elem2
    return the_dict

词典列表的模板有：

a = {
        {'key1':{'key1.1': {'delete': True, 'result': None},
        'key1.2': {'delete': False, 'result': None}},'key2': 3},
        {'key1':{'key1.1': {'delete': False, 'result': None},
        'key1.2': {'delete': True, 'result': None}},'key2': 5}
}

我要删除键keyx.x下的键'delete'，即True，根据上面的示例，该键将产生：

a = {
    {'key1':{'key1.2': {'delete': False, 'result': None}},'key2': 3},
    {'key1':{'key1.1': {'delete': False, 'result': None}},'key2': 5}
}

如上例所示，它们删除的键为key1.1和key1.2，因此是可变的。

Answer 1

我想，您的努力来自于python限制，以便在迭代过程中修改字典。要创建所需的内容，您需要复制一份字典-复制，迭代和删除原始密钥；或者或遍历原始密钥，然后将“过滤的”密钥放入新密钥。以后可以用

b = {k1: {k2: v2 for k2, v2 in v1.items() if not v2['delete']}
     for k1, v1 in a.items()}

编辑：由于某种原因，您提供的代码不一致，以上解决方案假设结构实际上是

a = {'key1.1': {'key1.1.1': {'delete': True, 'loremipsum': 'perhaps'},
                'key1.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
     'key1.2': {'key1.2.1': {'delete': False, 'loremipsum': 'perhaps'},
                'key1.2.2': {'delete': False, 'loremipsum': 'perhaps'}},
     'key2.1': {'key2.1.1': {'delete': False, 'loremipsum': 'perhaps'},
                'key2.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
     'key2.2': {'key2.2.1': {'delete': False, 'loremipsum': 'perhaps'},
                'key2.2.2': {'delete': True, 'loremipsum': 'perhaps'}}
}

b = {'key1.1': {'key1.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
     'key1.2': {'key1.2.1': {'delete': False, 'loremipsum': 'perhaps'},
                'key1.2.2': {'delete': False, 'loremipsum': 'perhaps'}},
     'key2.1': {'key2.1.1': {'delete': False, 'loremipsum': 'perhaps'},
                'key2.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
     'key2.2': {'key2.2.1': {'delete': False, 'loremipsum': 'perhaps'}}
}

Answer 2

如果对字典a使用正确的语法，则可以迭代第一层密钥，然后像您所说的那样，进行嵌套循环以迭代第二层密钥。 the_dict1 [key1] [key2]中的VALUE是最终的字典。我使用pprint帮助可视化字典。

def getCleanResults(the_dict1):
    # iterate through the first keys, cast as a tuple to make a copy of the keys iterator (allows for changing dictionary size during iteration)
    for key1 in tuple(the_dict1.keys()):
        the_dict2 = the_dict1[key1]
        # iterate through the second tier keys
        for key2 in tuple(the_dict2.keys()):
            if the_dict1[key1][key2]['delete']:
                del the_dict1[key1][key2]
    # no need to return a new dict, it was passed by reference, we modified the original dict. If you wanted a copy of the dict, make copy and return it instead.

正确格式的词典：

a = {
    'key1.1': {'key1.1.1': {'delete': True, 'loremipsum': 'perhaps'},
                'key1.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
    'key1.2': {'key1.2.1': {'delete': False, 'loremipsum': 'perhaps'},
            'key1.2.2': {'delete': False, 'loremipsum': 'perhaps'}},
    'key2.1': {'key2.1.1': {'delete': False, 'loremipsum': 'perhaps'},
            'key2.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
    'key2.2': {'key2.2.1': {'delete': False, 'loremipsum': 'perhaps'},
            'key2.2.2': {'delete': True, 'loremipsum': 'perhaps'}}
}

测试：

from pprint import pprint

getCleanResults(a)
print()
pprint(a)

输出：

{'key1.1': {'key1.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
 'key1.2': {'key1.2.1': {'delete': False, 'loremipsum': 'perhaps'},
            'key1.2.2': {'delete': False, 'loremipsum': 'perhaps'}},
 'key2.1': {'key2.1.1': {'delete': False, 'loremipsum': 'perhaps'},
            'key2.1.2': {'delete': False, 'loremipsum': 'perhaps'}},
 'key2.2': {'key2.2.1': {'delete': False, 'loremipsum': 'perhaps'}}}

Answer 3

您可以尝试这样。

代码段

import json

def delete_items(d):
    keys1 = d.keys()
    new_d = {}

    for key1 in keys1:
        keys2 = d[key1].keys();
        for key2 in keys2:
            if type(d[key1][key2]) is dict:
                if d[key1][key2]["delete"] == False:
                    if key1 in new_d:
                        new_d[key1][key2] = d[key1][key2]
                    else:
                        new_d[key1] = {key2: d[key1][key2]}
            else:
                if key1 in d:
                    new_d[key1] = {key2: d[key1][key2]}
                else:
                    new_d[key1][key2] = d[key1][key2]

    return new_d


a = {
        "key1.1": {
            "key1.1.1": {
                "delete": True,
                "loremipsum": "perhaps"
            },
            "key1.1.2": {
                "delete": False,
                "loremipsum": "perhaps"
            },
            "key1.1.3": 4
        },
        "key1.2": {
            "key1.2.1": {
                "delete": False,
                "loremipsum": "perhaps"
            },
            "key1.2.2": {
                "delete": False,
                "loremipsum": "perhaps"
            }
        },
        "key2.1": {
            "key2.1.1": {
                "delete": False,
                "loremipsum": "perhaps"
            },
            "key2.1.2": {
                "delete": False,
                "loremipsum": "perhaps"
            },
            "key2.1.3": 9
        },
        "key2.2": {
            "key2.2.1": {
                "delete": False,
                "loremipsum": "perhaps"
            },
            "key2.2.2": {
                "delete": True,
                "loremipsum": "perhaps"
            },
            "key2.2.3": 2
        }
    }

d = delete_items(a)
print(json.dumps(d, indent=4))

输出

{
    "key1.2": {
        "key1.2.1": {
            "delete": false,
            "loremipsum": "perhaps"
        },
        "key1.2.2": {
            "delete": false,
            "loremipsum": "perhaps"
        }
    },
    "key1.1": {
        "key1.1.3": 4,
        "key1.1.2": {
            "delete": false,
            "loremipsum": "perhaps"
        }
    },
    "key2.1": {
        "key2.1.3": 9
    },
    "key2.2": {
        "key2.2.3": 2
    }
}

根据列表中嵌套字典中的值删除字典元素

3 个答案: