从路径列表转换为 Flare json 格式？

Question

我在 python 中有如下所示的数据：

[['a', 'b', 'c', 50],

 ['a', 'b', 'd', 100],

 ['a', 'b', 'e', 67],

 ['a', 'g', 'c', 12],

 ['q', 'k', 'c', 11],

 ['q', 'b', 'p', 11]]

其中列表的每个元素都是一个完整的分层路径，最后一个元素是路径的大小。要在 D3 中进行可视化，我需要将数据采用耀斑数据格式 - 见此处：

https://github.com/d3/d3-hierarchy/blob/master/test/data/flare.json

所以一小段看起来像这样

{

 "name": "root",

 "children": [

  {

   "name": "a",

   "children": [

    {

     "name": "b",

     "children": [

      {"name": "c", "value": 50},

      {"name": "d", "value": 100},

      {"name": "e", "value": 67},

     ]

    },

    {

     "name": "g",

     "children": [

      {"name": "c", "value": 12},

     ]

    },

等等……

从我一直在查找的内容来看，我认为该解决方案是递归的，并且会json在 Python 字典上使用该库，但我似乎无法让它发挥作用。任何帮助是极大的赞赏。

Answer 1

这是使用递归的解决方案： def add_to_flare(n, flare):     children = flare["children"]     if len(n) == 2:         children.append({"name": n[0], "value": n[1]})     else:         for c in children:             if c["name"] == n[0]:                 add_to_flare(n[1:], c)                 return         children.append({"name": n[0], "children": []})         add_to_flare(n[1:], children[-1]) flare = {"name": "root", "children": []}  for i in data:     add_to_flare(i, flare)为了很好地显示它，我们可以使用这个json库：import jsonprint(json.dumps(flare, indent=1)){ "name": "root",  "children": [  {   "name": "a",    "children": [    {     "name": "b",      "children": [      {       "name": "c",        "value": 50      },       {       "name": "d",        "value": 100      },       {       "name": "e",        "value": 67      }     ]    },     {     "name": "g",      "children": [      {       "name": "c",        "value": 12      }     ]    }   ]  },   {   "name": "q",    "children": [    {     "name": "k",      "children": [      {       "name": "c",        "value": 11      }     ]    },     {     "name": "b",      "children": [      {       "name": "p",        "value": 11      }     ]    }   ]  } ]}

Answer 2

试试这个：master = []for each in your_list:    head = master    for i in range(len(each)):        names = [e['name'] for e in head]        if i == len(each) - 2:            head.append({'name': each[i], 'value': each[i+1]})            break        if each[i] in names:            head = head[names.index(each[i])]['children']        else:            head.append({'name': each[i], 'children': []})            head = head[-1]['children']结果：[{'children': [{'children': [{'name': 'c', 'value': 50},                             {'name': 'd', 'value': 100},                             {'name': 'e', 'value': 67}],                'name': 'b'},               {'children': [{'name': 'c', 'value': 12}], 'name': 'g'}],  'name': 'a'}, {'children': [{'children': [{'name': 'c', 'value': 11}], 'name': 'k'},               {'children': [{'name': 'p', 'value': 11}], 'name': 'b'}],  'name': 'q'}]请注意name和children在本词典中被翻转，因为它是无序的。但最终的结构是一样的。把它放在根目录下以获得你的目标：my_dict = {'name':'root', 'children': master}

Answer 3

假设您的列表列表存储在 variable 中l，您可以执行以下操作：o = []for s in l:&nbsp; &nbsp; c = o&nbsp; &nbsp; for i, n in enumerate(['root'] + s[:-1]):&nbsp; &nbsp; &nbsp; &nbsp; for d in c:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if n == d['name']:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; break&nbsp; &nbsp; &nbsp; &nbsp; else:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; c.append({'name': n})&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; d = c[-1]&nbsp; &nbsp; &nbsp; &nbsp; if i < len(s) - 1:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if 'children' not in d:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; d['children'] = []&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; c = d['children']&nbsp; &nbsp; &nbsp; &nbsp; else:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; d['value'] = s[-1]这样o[0]就变成了：{'children': [{'children': [{'children': [{'name': 'c', 'value': 50},&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {'name': 'd', 'value': 100},&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {'name': 'e', 'value': 67}],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;'name': 'b'},&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {'children': [{'name': 'c', 'value': 12}],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;'name': 'g'}],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;'name': 'a'},&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {'children': [{'children': [{'name': 'c', 'value': 11}],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;'name': 'k'},&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {'children': [{'name': 'p', 'value': 11}],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;'name': 'b'}],&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;'name': 'q'}],&nbsp;'name': 'root'}