使用生成器的Python 3.3解决方案:

def flattenit(pyobj, keystring=''):
   if type(pyobj) is dict:
     if (type(pyobj) is dict):
         keystring = keystring + "_" if keystring else keystring
         for k in pyobj:
             yield from flattenit(pyobj[k], keystring + k)
     elif (type(pyobj) is list):
         for lelm in pyobj:
             yield from flatten(lelm, keystring)
   else:
      yield keystring, pyobj

my_obj = {'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y': 10}}, 'd': [1, 2, 3]}

#your flattened dictionary object
flattened={k:v for k,v in flattenit(my_obj)}
print(flattened)

# result: {'c_b_y': 10, 'd': [1, 2, 3], 'c_a': 2, 'a': 1, 'c_b_x': 5}

使用发电机:

def flat_dic_helper(prepand,d):
    if len(prepand) > 0:
        prepand = prepand + "_"
    for k in d:
        i = d[k]
        if isinstance(i, dict):
            r = flat_dic_helper(prepand + k,i)
            for j in r:
                yield j
        else:
            yield (prepand + k,i)

def flat_dic(d):
    return dict(flat_dic_helper("",d))

d = {'a': 1, 'c': {'a': 2, 'b': {'x': 5, 'y' : 10}}, 'd': [1, 2, 3]}
print(flat_dic(d))


>> {'a': 1, 'c_a': 2, 'c_b_x': 5, 'd': [1, 2, 3], 'c_b_y': 10}

上面的答案真的很管用。我只是想加上我写的unflatten函数:

def unflatten(d):
    ud = {}
    for k, v in d.items():
        context = ud
        for sub_key in k.split('_')[:-1]:
            if sub_key not in context:
                context[sub_key] = {}
            context = context[sub_key]
        context[k.split('_')[-1]] = v
    return ud

注意:这并没有解释键中已经存在的'_'，就像扁平化的对应物一样。

这里有一个优雅的、就地替换的算法。使用Python 2.7和Python 3.5进行测试。使用点字符作为分隔符。

def flatten_json(json):
    if type(json) == dict:
        for k, v in list(json.items()):
            if type(v) == dict:
                flatten_json(v)
                json.pop(k)
                for k2, v2 in v.items():
                    json[k+"."+k2] = v2

例子:

d = {'a': {'b': 'c'}}                   
flatten_json(d)
print(d)
unflatten_json(d)
print(d)

输出:

{'a.b': 'c'}
{'a': {'b': 'c'}}

我在这里发布了这段代码以及匹配的unflat_json函数。

如果你使用pandas，有一个函数隐藏在pandas.io.json中。_normalize1调用nested_to_record来完成这个操作。

from pandas.io.json._normalize import nested_to_record    

flat = nested_to_record(my_dict, sep='_')

1在熊猫0.24版本。X及以上版本使用panda .io.json.normalize(不带_)

def flatten(dictionary, prefix = '', separator = '_'):
    out_dict = {}
    if type(dictionary) != dict:
        out_dict[prefix] = dictionary
        return out_dict
    elif dictionary is None:
        return None
    for k in dictionary.keys():
        if prefix:
            prefix_n = prefix + f'{separator}{k}'
        else:
            prefix_n = k
        out_dict.update(flatten_new(dictionary[k], prefix_n))
    return out_dict

输出:

{'a': 1, 'c_a': 2, 'c_b_x': 5, 'c_b_y': 10, 'd': [1, 2, 3]}