如何获取字典列表而不是在python中使用collection.defaultdict [英] How to get list of dict instead of using collection.defaultdict in python

问题描述

我正尝试将列表递归转换为嵌套字典，如下所示:-

I am trying to convert a list into nested dictionaries recursively as following :-

输入:-

parse_list = ['A','B','C','D'] 

必需的输出:-

data = [
    {'name': 'A',
     'childs': [
         {'name': 'B',
          'childs': [
              {'name': 'C',
               'childs': [
                   {'name': 'D',
                    'childs': none }]}]}]}]

我的代码:-

from collections import defaultdict
class_dict =defaultdict(list)
data =defaultdict(list)
parse_list.reverse()

def create_dict(parse_list,class_dict):
    for index ,listitem in enumerate(parse_list):
        new_class_dict = defaultdict(list)
        new_class_dict.__setitem__('name', listitem)
        new_class_dict['childs'].append(class_dict)
        class_dict = new_class_dict
    return class_dict

data = create_dict(parse_list,class_dict) 
import yaml
with open('data.yml', 'w') as outfile:
    outfile.write( yaml.dump(data, default_flow_style=False))

但是由于defaultdict(list)，我总是在yaml中获得许多额外的缩进，这是不期望的.还有其他方法可以获取[{....}]而不是使用collection.defaultdict.

However because of defaultdict(list), I always get many extra indent in yaml, which are not expected. Is there any other way to get the [{....}] instead of using collection.defaultdict.

推荐答案

以递归方式

parse_list = ['A','B','C','D']



def createdict(l, d):
    if len(l) == 0:
        return None
    d = dict()
    d['name'] = l[0]
    d['childs'] = [createdict(l[1:], d)]
    return d

resultDict = createdict(parse_list, dict())

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