Question

我必须递归查找其值不包含其他字典的字典，然后基于该字典创建一个新字典，当前，我的递归函数能够处理字典词典

b= {'order': {'id': u'62211'}, 'no': 0, 'data': 340, 'instruction': {'id': u'99AA', 'version': u'1', 'validation': ('nota', None), 'time': 9001, 'instructions': [], 'restart': True}}

temp_1={}
import copy
def getKeys(d):


    for k,v in d.items():
        if isinstance(v, dict):

            getKeys(v)
        else:
            temp_1[k]=v

    return temp_1



a= getKeys(b)
Output
{'no': 0, 'version': u'1', 'time': 9001, 'validation': ('nota', None), 'data': 340, 'id': u'62211', 'restart': True, 'instructions': []}

但是在某些情况下，字典会出现在列表或元组中时，请不确定如何以递归方式处理它。

ab={'apple':9,'ball':{'cat':8,'www':('data',{'new':'hello','ppp':('dat',[{'jjj':'world'}])})}}

expected output:

   { 'apple':9,'cat':8,'new':'hello','jjj':'world'}

Answer 1

def get_keys(x, tmp=None):
    if tmp is None: tmp = {}
    if isinstance(x, dict):
        for key,value in x.items():
            if isinstance(value, dict):
                get_keys(value, tmp)
            elif isinstance(value, (tuple, list, set):
                for item in value:
                    get_keys(item, tmp)
            else:
                tmp[key] = value
    elif isinstance(x, (tuple, list set)):
        for item in x:
            get_keys(item, tmp)
    return tmp


b= {'order': {'id': u'62211'}, 'no': 0, 'data': 340, 'instruction': {'id': u'99AA', 'version': u'1', 'validation': ('nota', None), 'time': 9001, 'instructions': [], 'restart': True}}
ab={'apple':9,'ball':{'cat':8,'www':('data',{'new':'hello','ppp':('dat',[{'jjj':'world'}])})}}

a = get_keys(ab)
aa = get_keys(b)
print(a)
print(aa)

Answer 2

您只需要处理iterable个项目（而不仅仅是dict）。这里是一个例子：

import collections

example_data = {
    'apple': 9,
    'ball': {
        'cat': 8,
        'www': (
            'data',
            {
                'new': 'hello',
                'ppp': (
                    'dat',
                    [{'jjj': 'world'}]
                )
            }
        )
    }
}


def get_keys(data, result=None):
    result = result or dict()
    # dict processing
    if isinstance(data, dict):
        for key, value in data.items():
            # not str, but iterable - call get_keys one more time
            if type(value) != str and isinstance(value, collections.Iterable):
                result = get_keys(value, result)
            else:
                # value isn't iterable - just set to result
                result[key] = value
    # list, tuple, set processing
    elif isinstance(data, (list, tuple, set)):
        for item in data:
            # list item is not str, but iterable - call get_keys one more time
            if type(item) != str and isinstance(item, collections.Iterable):
                result = get_keys(item, result)

    return result


print(get_keys(example_data))

希望这会有所帮助。

Answer 3

您可以使用一个函数来处理dict，并使用另一个函数来处理tuple和list。根据元素的类型，您可以选择要应用的功能：

result = {}

def process_list(l):
    for x in l:
        try:
            process[type(x)](x)
        except KeyError:
            pass

def process_dict(d):
    for k, v in d.items():
        try:
            process[type(v)](v)
        except KeyError:
            result[k] = v

process = {
    list: process_list,
    tuple: process_list,
    dict: process_dict
}

然后仅致电process_dict(ab)。

递归函数进入字典的字典

3 个答案: