用重复的键从字符串中创建dict/json [英] make a dict/json from string with duplicate keys Python

问题描述

我有一个可以解析为json或dict对象的字符串.我的字符串变量看起来像这样:

I have a string that could be parsed as a json or dict object. My string variable looks like this :

    my_string_variable = "{
                            "a":1,
                            "b":{
                                 "b1":1,
                                 "b2":2
                             },  
                            "b": { 
                                "b1":3, 
                                "b2":2,
                                "b4":8
                             } 
                           }"

当我执行json.loads(my_string_variable)时，我有一个字典，但是仅保留键"b"的第二个值，这是正常的，因为字典不能包含重复的键.

When I do json.loads(my_string_variable), I have a dict but only the second value of the key "b" is kept, which is normal because a dict can't contain duplicate keys.

拥有这样的defaultdict的最佳方法是什么:

What would be the best way to have some sort of defaultdict like this :

    result = {
               'a':1,
               'b': [{'b1':1,'b2':2}, { 'b1':3, 'b2':2,'b4':8 } ]
             }

我已经在寻找类似的问题，但是它们都将字典或列表作为输入，然后创建defaultdict以处理重复的键.

I have already looked for similar questions but they all deal with dicts or lists as an input and then create defaultdicts to handle the duplicate keys.

在我的情况下，我有一个字符串变量，我想知道是否有一种简单的方法可以实现这个目的:)

In my case I have a string variable and I would want to know if there is a simple way to achieve this :)

谢谢^^

推荐答案

可以完成以下操作.

import json

def join_duplicate_keys(ordered_pairs):
    d = {}
    for k, v in ordered_pairs:
        if k in d:
           if type(d[k]) == list:
               d[k].append(v)
           else:
               newlist = []
               newlist.append(d[k])
               newlist.append(v)
               d[k] = newlist
        else:
           d[k] = v
    return d

raw_post_data = '{"a":1, "b":{"b1":1,"b2":2}, "b": { "b1":3, "b2":2,"b4":8} }'
newdict = json.loads(raw_post_data, object_pairs_hook=join_duplicate_keys)
print (newdict)

请注意，上面的代码取决于值类型if type(d[k]) == list.因此，如果原始字符串本身提供了一个列表，则可能需要一些错误处理才能使代码健壮.

Please note that above code depends on value type, if type(d[k]) == list. So if original string itself gives a list then there could be some error handling required to make the code robust.

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