如何在深度嵌套的字典中出现所选键的任何位置用其值替换键:值对? [英] How can I replace a key:value pair by its value wherever the chosen key occurs in a deeply nested dictionary?

问题描述

这是如何用来自同一键值对的值替换嵌套字典中的键值对? 答案仅适用于一次性嵌套的字典.它是遍历所有嵌套字典值? 我无法解决这个问题.

This is a spin-off from How can I replace a key-value pair in a nested dictionary with the value from the same key-value pair? where the answer is working only in a one-time-nested dictionary. And it is a spin-off from Loop through all nested dictionary values? which I could not get to work on this problem.

我有一本嵌套多次的字典.

I have a dictionary that is nested many times.

dict_nested = {
    "key_":{
        "key0a":{
            "key1a":{
                "sub_key2a":"sub_value2a",
                "sub_key2b":"sub_value2b"},
            "key1b":"value1b"},
        "key0b":{
            "key_XYZ":{
                "key1a":{
                    "sub_key2a":"sub_value2a",
                    "sub_key2b":"sub_value2b"},
                "key1b":"value1b"}
            }
        }
    }

之后:

结果应该是这样的:

After:

The result should look like this:

dict_nested_new = {
    "key_":{
        "key0a":{
            "sub_key2a":"sub_value2a",
            "sub_key2b":"sub_value2b",
            "key1b":"value1b"},
        "key0b":{
            "key_XYZ":{
                "sub_key2a":"sub_value2a",
                "sub_key2b":"sub_value2b",
                "key1b":"value1b"}
            }
        }
    }

在迭代时修改 Python dict

当我遍历字典的项目以删除/替换时，出现错误

Modifying a Python dict while iterating over it

When I looped through the items of the dictionary to delete / replace, I got the error

RuntimeError: 字典在迭代过程中改变了大小

RuntimeError: dictionary changed size during iteration

需要以某种方式避免.

如何在 "key1a":SOME_VALUE 键值对每次出现在字典中的某个位置时用它的值替换它?

How can I replace the "key1a":SOME_VALUE key-value pair with its value each time it occurs somewhere in the dictionary?

推荐答案

据我所知，您希望在嵌套的 dict 中递归搜索键并提升其值.

As I understand it, you want to recursively search for a key in a nested dict and promote its value.

这可能不是非常有效，但它应该有效.它也没有真正探索以列表作为值的字典，但您的示例数据没有它们，所以我没有实现.

This might not be super efficient, but it should work. It also does not really explore dictionaries with lists as values but your example data does not have them so I did not implement that.

import copy
import json

def find_replace(this_dict, target_key):
    ## optional depending on if you care that you mutate this_dict
    this_dict = copy.deepcopy(this_dict)

    for key in this_dict:
        # if the current value is a dict, dive into it
        if isinstance(this_dict[key], dict):
            this_dict[key] = find_replace(this_dict[key], target_key)

        # if the current key is our target merge the values and remove the key
        if key == target_key:
            this_dict = {**this_dict, **this_dict[key]}
            del this_dict[key]

    return this_dict

dict_nested = {
    "key_":{
        "key0a":{
            "key1a":{
                "sub_key2a":"sub_value2a", 
                "sub_key2b":"sub_value2b"
            }, 
            "key1b":"value1b"
        },
        "key0b":{
            "key_XYZ":{
                "key1a":{
                    "sub_key2a":"sub_value2a", 
                    "sub_key2b":"sub_value2b",
                    "key1a": {
                        "sub_key3a":"sub_value3a", 
                        "sub_key3b":"sub_value3b"
                    }, 
                }, 
                "key1b":"value1b"
            }
        }
    }
}

dict_nested_new = find_replace(dict_nested, "key1a")
print(json.dumps(dict_nested_new, indent=4))

应该给你:

{
    "key_": {
        "key0a": {
            "key1b": "value1b",
            "sub_key2a": "sub_value2a",
            "sub_key2b": "sub_value2b"
        },
        "key0b": {
            "key_XYZ": {
                "key1b": "value1b",
                "sub_key2a": "sub_value2a",
                "sub_key2b": "sub_value2b",
                "sub_key3a": "sub_value3a",
                "sub_key3b": "sub_value3b"
            }
        }
    }
}

请注意，我使用子嵌套键匹配添加了额外的嵌套级别，只是为了显示该场景.额外的优化，如支持列表和避免更新以合理费用提供的未更改密钥:-P

Note that I added an addition level of nesting with a sub-nested key match just to show that scenario. Additional optimizations like support for lists and avoiding updates to unchanged keys available for a reasonable fee :-P

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