获取嵌套字典中所有键的列表

时间:2014-10-02 18:12:18

标签: python dictionary nested

我想获得包含列表和词典的嵌套字典中所有键的列表。

我目前有这个代码,但似乎缺少在列表中添加一些键,并且重复添加一些键。

keys_list = []
def get_keys(d_or_l, keys_list):
    if isinstance(d_or_l, dict):
        for k, v in iter(sorted(d_or_l.iteritems())):
            if isinstance(v, list):
                get_keys(v, keys_list)
            elif isinstance(v, dict):
                get_keys(v, keys_list)
            else:
                keys_list.append(k)
    elif isinstance(d_or_l, list):
        for i in d_or_l:
            if isinstance(i, list):
                get_keys(i, keys_list)
            elif isinstance(i, dict):
                get_keys(i, keys_list)
    else:
        print "** Skipping item of type: {}".format(type(d_or_l))
    return keys_list

这只是一个空列表,并用键填充它。 d_or_l是一个变量,并将原始字典与其进行比较。

4 个答案:

答案 0 :(得分:4)

这应该做的工作:

def get_keys(dl, keys_list):
    if isinstance(dl, dict):
        keys_list += dl.keys()
        map(lambda x: get_keys(x, keys_list), dl.values())
    elif isinstance(dl, list):
        map(lambda x: get_keys(x, keys_list), dl)

为避免重复,您可以使用set,例如:

keys_list = list( set( keys_list ) )

示例测试用例:

keys_list = []
d = {1: 2, 3: 4, 5: [{7: {9: 1}}]}
get_keys(d, keys_list)
print keys_list
>>>> [1, 3, 5, 7, 9]

答案 1 :(得分:3)

目前,您的代码会忽略导致listdict值的键。删除第一个else循环中的for块,无论值是什么,都要添加密钥。

keys_list = []
def get_keys(d_or_l, keys_list):
    if isinstance(d_or_l, dict):
        for k, v in iter(sorted(d_or_l.iteritems())):
            if isinstance(v, list):
                get_keys(v, keys_list)
            elif isinstance(v, dict):
                get_keys(v, keys_list)
            keys_list.append(k)   #  Altered line
    elif isinstance(d_or_l, list):
        for i in d_or_l:
            if isinstance(i, list):
                get_keys(i, keys_list)
            elif isinstance(i, dict):
                get_keys(i, keys_list)
    else:
        print "** Skipping item of type: {}".format(type(d_or_l))
    return keys_list

get_keys({1: 2, 3: 4, 5: [{7: {9: 1}}]}, keys_list)返回[1, 3, 9, 7, 5]

为避免重复,您可以使用set datatype代替list

答案 2 :(得分:0)

不建议使用dict.iteritems更新@MackM对Python 3的响应(我更喜欢在.format{}样式中使用f字符串):

keys_list = []
def get_keys(d_or_l, keys_list):
    if isinstance(d_or_l, dict):
        for k, v in iter(sorted(d_or_l.items())):  #  Altered line to update deprecated method
            if isinstance(v, list):
                get_keys(v, keys_list)
            elif isinstance(v, dict):
                get_keys(v, keys_list)
            keys_list.append(k)   
    elif isinstance(d_or_l, list):
        for i in d_or_l:
            if isinstance(i, list):
                get_keys(i, keys_list)
            elif isinstance(i, dict):
                get_keys(i, keys_list)
    else:
        print(f'** Skipping item of type: {type(d_or_l)}')  #  Altered line to use f-strings
    return keys_list


unique_keys = list(set(get_keys(my_json_dict, keys_list)))  # Added line as example use case

答案 3 :(得分:0)

这是一个简单的解决方案:

def get_nested_keys(d, keys):
    for k, v in d.items():
        if isinstance(v, dict):
            get_nested_keys(v, keys)
        else:
            keys.append(k)

keys_list = []
get_nested_keys(test_listing, keys_list)
print(keys_list)

如果你也想知道键的层次结构,你可以像这样修改函数:

def get_nested_keys(d, keys, prefix):
    for k, v in d.items():
        if isinstance(v, dict):
            get_nested_keys(v, keys, f'{prefix}:{k}')
        else:
            keys.append(f'{prefix}:{k}')