访问嵌套字典中的键，值

问题描述

我知道有关访问嵌套字典中的键，值的问题曾经被问过，但是我的以下代码却遇到了一些麻烦:

I know questions regarding accessing key, value in a nested dictionary have been asked before but I'm having some trouble with the following piece of my code:

用于访问嵌套字典的键和值，如下所示:

For accessing the keys and values of a nested dictionary as follows:

example_dict = {'key_outer_01': {'key_inner_01': 'value_inner_01'},
                'key_outer_02': {'key_inner_02': 'value_inner_02'}}

我有以下代码:

def get_key_value_of_nested_dict(nested_dict):
    for key, value in nested_dict.items():
        outer_key = None
        inner_key = None
        inner_value = None
        if isinstance(value, dict):
            outer_key = key
            get_key_value_of_nested_dict(value)
        else:
            inner_key = key
            inner_value = value
        return outer_key, inner_key, inner_value

我得到的输出是:

    key_outer_01 None None

我在这里做错了什么?

What am I doing wrong here?

推荐答案

在递归调用中，您将设置 outer_key ， inner_key 和 inner_value >到None.但是在if isintance(value, dict)中，您仅将 outer_key 重命名为key.您可能要分配inner_key和inner_value的新值.

In your recursive call, you are setting outer_key, inner_key and inner_value to None. But in the if isintance(value, dict), you are only redifining outer_key to key. You might want to assign the new values of inner_key and inner_value.

为inner_key和inner_value分配新值！如:

example_dict = {'key_outer_01': {'key_inner_01': 'value_inner_01'},
                'key_outer_02': {'key_inner_02': 'value_inner_02'}}

def get_key_value_of_nested_dict(nested_dict):
    for key, value in nested_dict.items():
        outer_key = None
        inner_key = None
        inner_value = None
        if isinstance(value, dict):
            outer_key = key
            _, inner_key, inner_value = get_key_value_of_nested_dict(value)
        else:
            inner_key = key
            inner_value = value
        return outer_key, inner_key, inner_value

print get_key_value_of_nested_dict(example_dict)
# outputs : ('key_outer_01', 'key_inner_01', 'value_inner_01')

---

我相信我们需要更多信息，而所有极端情况都需要知道此代码是否正确.

---

I believe we need more information whereas all the edge-cases to know if this code is good or not.

我尝试了一些改进，告诉我它是否比原始代码更适合您的情况.

I tried for fun some kind of improvement, tell me if it suits your case better than your original code.

example_dict = {'key_outer_01': {'key_inner_01': 'value_inner_01', 'key_inner_02' : 'value_inner_02'},
                'key_outer_02': {'key_inner_02': 'value_inner_02'},
                'key_outer_03' : None}

def get_k_v_of_inner(nested_dict):
    for k, v in nested_dict.items():
        if isinstance(v, dict):
            for k_nested, v_nested in v.items():
                yield {k : {k_nested : v_nested}}
        else:
            # do_something, or simply ignore if not nested dict.
            pass

gen = get_k_v_of_inner(example_dict)

for c, value in enumerate(gen):
    print '#{} : {}'.format(c, value)

# outs : #0 : {'key_outer_01': {'key_inner_02': 'value_inner_02'}}
         #1 : {'key_outer_01': {'key_inner_01': 'value_inner_01'}}
         #2 : {'key_outer_02': {'key_inner_02': 'value_inner_02'}}

这篇关于访问嵌套字典中的键，值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！