根据列表中嵌套字典中的值删除字典元素答案

【问题标题】：Delete dictionary element based on a value in a nested dictionary in a list根据列表中嵌套字典中的值删除字典元素
【发布时间】：2019-06-04 04:35:50
【问题描述】：

我正在尝试使用列表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，然后又将字典作为其键值之一。我知道我将在最顶层列表和最底层字典中使用的键，而“中间”字典中的键是可变的。如果底部字典中的某个值不等于某个值，我想删除“中间”字典中的元素。

我想我需要三个嵌套的 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。

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

词典列表的模板我有：

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}
}

我想删除下面有'delete' 即True 的键keyx.x，根据上面的示例，这将产生：

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

如上例所示，他们删除的键是 key1.1 和 key1.2，因此是可变的。

【问题讨论】：

我认为您提供的嵌套字典有很多语法错误。你能修复你的字典，让它准确地代表你的数据结构，看看它在你的机器上是否有任何语法错误吗？另外，我认为您的意思是删除上一段中“删除”为True 的键？
a 不是有效的字典。请更正。
修复了a 字典。
它已被@Slam 修复，现在再次修复后它不再是有效的字典...您的密钥不能是字典，因为它不可散列。在你的情况下，你的键必须是一个字符串，而值是一个字典，它有自己的字符串键，另一个字典作为它自己的值。
@KaanTheGuru，a 仍然不是有效的字典，因为它没有键，只有值。

标签： python python-3.x dictionary nested

【解决方案1】：

我猜，您的挣扎来自 Python 对在迭代期间修改字典的限制。要制作您想要的内容，您需要制作您的 dict 的副本 — 制作副本、迭代并从原始副本中删除键；或迭代原始密钥并将“过滤”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，您提供的代码没有一致性，上面的解决方案是假设结构实际上是

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'}}
}

【讨论】：

【解决方案2】：

如果您对字典 a 使用正确的语法，您可以遍历第一层键，然后像您说的那样，做一个嵌套循环来遍历第二层键。 the_dict1[key1][key2] 中的 VALUE 是最终的字典。我使用 pprint 来帮助可视化字典。

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.

正确格式的字典：

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'}}}

【讨论】：

【解决方案3】：

你可以这样试试。

代码sn-p

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
    }
}

【讨论】：