我需要将一个JSON数组整理成一个层次结构，这里是我的JSON文件，它从未被订购但是遵循结构：
{
"name":"Folder 2",
"id":"zRDg",
"parent":"OY00",
"type":"folder"
},
{
"name":"Folder 1",
"id":"OY00",
"type":"folder"
},
{
"name":"Folder 3",
"id":"ZDE1",
"type":"folder"
},
{
"name":"DX00025.jpg",
"id":"9Xdd",
"parent":"OY00",
"type":"jpeg"
}
在这个JSON文件中，结构如下：
{
"name":"Folder 1",
"id":"OY00",
"type":"folder",
"children": [{
"name":"Folder 2",
"id":"zRDg",
"type":"folder"
},
{
"name":"DX00025.jpg",
"id":"9Xdd",
"type":"jpeg"
}]
},
{
"name":"Folder 3",
"id":"ZDE1",
"type":"folder"
}
我无法弄明白，因为我是python的新手，我的开始（错误）：
for index,item in result:
if item['parent']:
for item2 in result:
if item2['id'] == item['parent']:
item['children'] = item2
brake
这没关系，但问题是它不正确 python ，folder1 / folder2 / folder3 /不适用于此，我需要一个递归函数。我还应该包含这个结构更改，它可以是文件夹和文件夹/文件夹的任意组合。
答案 0 :(得分：3)
myJson = [
{
"name":"Folder 2",
"id":"zRDg",
"parent":"OY00",
"type":"folder"
},
{
"name":"Folder 1",
"id":"OY00",
"type":"folder"
},
{
"name":"Folder 3",
"id":"ZDE1",
"type":"folder"
},
{
"name":"DX00025.jpg",
"id":"9Xdd",
"parent":"OY00",
"type":"jpeg"
}
]
#this creates a dictionary that maps id names to JSON items.
#ex. itemsKeyedById["9Xdd"] gives the jpg item with id "9Xdd"
itemsKeyedById = {i["id"]: i for i in myJson}
#iterate through each item in the `myJson` list.
for item in myJson:
#does the item have a parent?
if "parent" in item:
#get the parent item
parent = itemsKeyedById[item["parent"]]
#if the parent item doesn't have a "children" member,
#we must create one.
if "children" not in parent:
parent["children"] = []
#add the item to its parent's "children" list.
parent["children"].append(item)
#filter out any item that has a parent.
#They don't need to appear at the top level,
#since they will appear underneath another item elsewhere.
topLevelItems = [item for item in myJson if "parent" not in item]
print topLevelItems
输出（由我添加缩进）：
[
{
'name': 'Folder 1',
'id': 'OY00',
'type': 'folder',
'children': [
{
'name': 'Folder 2',
'id': 'zRDg',
'parent': 'OY00',
'type': 'folder'
},
{
'name': 'DX00025.jpg',
'id': '9Xdd',
'parent': 'OY00',
'type': 'jpeg'
}
]
},
{
'name': 'Folder 3',
'id': 'ZDE1',
'type': 'folder'
}
]
它也适用于嵌套多个深度的项目。示例输入：
myJson = [
{
"name":"TopLevel folder",
"id":"0",
"type":"folder",
},
{
"name":"MidLevel folder",
"id":"1",
"type":"folder",
"parent":"0"
},
{
"name":"Bottom Level folder",
"id":"2",
"type":"folder",
"parent":"1"
},
{
"name":"Vacation Picture",
"id":"3",
"type":"jpg",
"parent":"2"
},
]
输出：
[
{
'type': 'folder',
'name': 'TopLevel folder',
'id': '0',
'children': [
{
'type': 'folder',
'name': 'MidLevel folder',
'parent': '0',
'id': '1',
'children': [
{
'type': 'folder',
'name': 'Bottom Level folder',
'parent': '1',
'id': '2',
'children': [
{
'type': 'jpg',
'name': 'Vacation Picture',
'parent': '2',
'id': '3'
}
]
}
]
}
]
}
]
答案 1 :(得分：3)
如何使用Python networkx库这样的东西？
import json
#networkx is a library for working with networks and trees
import networkx as nx
#The json_graph routines print out JSONic representations of graphs and trees
#http://networkx.github.com/documentation/latest/reference/readwrite.json_graph.html
from networkx.readwrite import json_graph
dd='[{ "name":"Folder 2", "id":"zRDg", "parent":"OY00", "type":"folder"},{ "name":"Folder 1", "id":"OY00", "type":"folder"},{"name":"Folder 3", "id":"ZDE1", "type":"folder"},{ "name":"DX00025.jpg", "id":"9Xdd", "parent":"OY00", "type":"jpeg"}]'
d=json.loads(dd)
#A tree is a directed graph - create one with a dummy root
DG=nx.DiGraph()
DG.add_node('root')
#Construct the tree as a directed graph and annotate the nodes with attributes
#Edges go from parent to child
for e in d:
DG.add_node(e['id'],name=e['name'],type=e['type'])
#If there's a parent, use it...
if 'parent' in e: DG.add_edge(e['parent'],e['id'])
#else create a dummy parent from the dummy root
else: DG.add_edge('root',e['id'])
#Get the tree as JSON
data = json_graph.tree_data(DG,root='root')
#and dump the data from the dummy root's children down...
json.dumps(data['children'])
'''
'[{"children": [{"type": "folder", "name": "Folder 2", "id": "zRDg"}, {"type": "jpeg", "name": "DX00025.jpg", "id": "9Xdd"}], "type": "folder", "name": "Folder 1", "id": "OY00"}, {"type": "folder", "name": "Folder 3", "id": "ZDE1"}]'
'''
答案 2 :(得分：2)
我对此案例的解决方案是这样的：
data = INPUT_LIST
class Item:
def __init__(self, _id, name, type, parent):
self._id = _id
self.name = name
self.type = type
self.parent = parent
self.children = []
def get_dict(self):
return {
'id': self._id,
'name': self.name,
'type': self.type,
'children': [child.get_dict() for child in self.children]
}
lookup = dict((item['id'], Item(item['id'], item['name'], item['type'], item['parent'] if 'parent' in item else None)) for item in data)
root = []
for _id, item in lookup.items():
if not item.parent:
root.append(item)
else:
lookup[item.parent].children.append(item)
dict_result = [item.get_dict() for item in root]
答案 3 :(得分：2)
这里有一个图表（可能只是一棵树），其中节点之间的连接由parent键表示。这称为adjacency list。
您希望通过children键生成一个包含指向其他节点的链接的树结构，该键是其他节点的列表。
要将邻接列表转换为树，首先需要通过其id来检索节点，因此第一步是创建每个节点的dict，以id为键。
然后你必须再次浏览列表并将孩子添加到他们的父母身上。
最后，制作一个没有父节点的节点列表（根节点）。
请注意，您根本不需要递归算法。
我们可以结合其中一些步骤来避免多次浏览列表。代码如下：
def nest_list(L):
"""Modify list of associative dicts into a graph and return roots of graph
Association is via a 'parent' key indicating a corresponding 'id'.
Items in the list *will be modified*.
Children will be placed in a list for a 'children' key
Items without children will have no 'children' key
'parent' keys will be removed.
Returned list is the full list of items which are not children of
any other items.
"""
idx = {}
children = []
root = []
# first pass: index items by id
# if the item has no parents, put it in the 'root' list
# otherwise add it to the 'children' list which we will
# process in a second pass
for item in L:
idx[item['id']] = item
if 'parent' in item:
children.append(item)
else:
root.append(item)
# second pass: find the parents of our children
for child in children:
parent = idx[child['parent']]
del child['parent']
try:
# retrieve the list of children for this
# parent
childlist = parent['children']
except KeyError:
# this item has received no children so far
# so we need to make a 'children' key and list
# for it.
parent['children'] = childlist = []
# append this child to the parent's list of children
childlist.append(child)
#all done: return new root list
return root
行动准则：
oldlist = [{
"name":"Folder 2",
"id":"zRDg",
"parent":"OY00",
"type":"folder"
},
{
"name":"Folder 1",
"id":"OY00",
"type":"folder"
},
{
"name":"Folder 3",
"id":"ZDE1",
"type":"folder"
},
{
"name":"DX00025.jpg",
"id":"9Xdd",
"parent":"OY00",
"type":"jpeg"
}]
expected = [{
"name":"Folder 1",
"id":"OY00",
"type":"folder",
"children": [{
"name":"Folder 2",
"id":"zRDg",
"type":"folder"
},
{
"name":"DX00025.jpg",
"id":"9Xdd",
"type":"jpeg"
}]
},
{
"name":"Folder 3",
"id":"ZDE1",
"type":"folder"
}]
from pprint import pprint
newlist = nest_list(oldlist)
pprint(newlist)
assert newlist == expected