我将以此为序，我不是开发人员，但是我被抛弃了这个任务，我只是迷失了。这是我第一次使用python和7年以上的第一次编码，而且进展不顺利。
我拥有的JSON是一个组织树，其中每个级别可能在其下面有子级。
我需要在Jupyter Notebook中用Python编写一个脚本，将其压缩成这种格式，或类似的地方，每个新子节点都是新行。
level1 | level2 | level3
org1
org1 org2
org1 org2 org3
这是JSON：
[{
"Id": "f035de7f",
"Name": "Org1",
"ParentId": null,
"Children": [{
"Id": "8c18a70d",
"Name": "Org2",
"ParentId": "f035de7f",
"Children": []
}, {
"Id": "b4514099",
"Name": "Org3",
"ParentId": "f035de7f",
"Children": [{
"Id": "8abe58d1",
"Name": "Org4",
"Children": []
}]
}, {
"Id": "8e35bdc3",
"Name": "Org5",
"ParentId": "f035de7f",
"Children": [{
"Id": "331fffbf",
"Name": "Org6",
"ParentId": "8e35bdc3",
"Children": [{
"Id": "3bc3e085",
"Name": "Org7",
"ParentId": "331fffbf",
"Children": []
}]
}]
}]
}]
我已经尝试了各种for循环，并且已经在互联网上搜索了好几天，但我认为我缺少一些非常基本的知识来使这项工作。我非常感谢有人能给予的任何帮助。
这是我的先发：
for item in orgs_json:
orgs_json_children = item["Children"]
orgs_list.append(orgs_json_children)
要么
wanted = ['Children', 'Name']
for item in orgs_json[0]:
details = [X["Name"] for X in orgs_json]
for key in wanted:
print(key, ':', json.dumps(details[key], indent=4))
# Put a blank line at the end of the details for each item
print()
您可以使用堆栈处理嵌套结构：
while stack:循环中，从堆栈中取出顶部元素。做你需要做的组织，比如记录名字。从组织路径生成一行，并添加当前组织名称。Children密钥中的所有元素与父组织的组织路径一起添加到堆栈中。需要反转，因为从堆栈中获取元素会以相反的顺序给出它们。您仍然希望为此作业使用堆栈（而不是队列），因为我们希望先深度输出信息。
这看起来像这样：
def flatten_orgs(orgs):
stack = [(o, ()) for o in reversed(orgs)] # organisation plus path
while stack:
org, path = stack.pop() # top element
path += (org['Name'],) # update path, adding the current name
yield path # give this path to the caller
# add all children to the stack, with the current path
stack += ((o, path) for o in reversed(org['Children']))
然后，您可以循环上面的函数以获取所有路径：
>>> for path in flatten_orgs(orgs_json):
... print(*path, sep='\t')
...
Org1
Org1 Org2
Org1 Org3
Org1 Org3 Org4
Org1 Org5
Org1 Org5 Org6
Org1 Org5 Org6 Org7
您可以递归迭代数据。前缀表示到目前为止看到的名称列表，数据表示您仍然必须学习的词典列表。
data = [{
"Id": "f035de7f",
"Name": "Org1",
"ParentId": None,
"Children": [{
"Id": "8c18a70d",
"Name": "Org2",
"ParentId": "f035de7f",
"Children": []
}, {
"Id": "b4514099",
"Name": "Org3",
"ParentId": "f035de7f",
"Children": [{
"Id": "8abe58d1",
"Name": "Org4",
"Children": []
}],
}, {
"Id": "8e35bdc3",
"Name": "Org5",
"ParentId": "f035de7f",
"Children": [{
"Id": "331fffbf",
"Name": "Org6",
"ParentId": "8e35bdc3",
"Children": [{
"Id": "3bc3e085",
"Name": "Org7",
"ParentId": "331fffbf",
"Children": []
}],
}],
}],
}]
def flatten(data, prefix):
if not data:
return [prefix]
result = []
for org in data:
name = org["Name"]
result.extend(flatten(org["Children"], prefix + [name]))
return result
print(flatten(data, []))
# [['Org1', 'Org2'], ['Org1', 'Org3', 'Org4'], ['Org1', 'Org5', 'Org6', 'Org7']]
同样，使用产量：
def flatten(data, prefix):
if not data:
yield prefix
for org in data:
name = org["Name"]
yield from flatten(org["Children"], prefix + [name])
print(list(flatten(data, [])))
如果您需要所有部分列表，则解决方案更短：
def flatten(data, prefix):
yield prefix
for org in data:
name = org["Name"]
yield from flatten(org["Children"], prefix + [name])
print(list(flatten(data, [])))
# [[], ['Org1'], ['Org1', 'Org2'], ['Org1', 'Org3'], ['Org1', 'Org3', 'Org4'], ['Org1', 'Org5'], ['Org1', 'Org5', 'Org6'], ['Org1', 'Org5', 'Org6', 'Org7']]
一个json递归树可以有几个根，叶子不应该强制指定void childrenrens。例如，这是一个具有两个根'a'和'b'的树，以及仅具有节点深度''level'数据的节点（'children'是可选的）：
json_struct = [
{
'level': 'a0',
'children': [{'level': 'a0.1', 'children':
[{'level': 'a0.1.1', 'children': []}]},
{'level': 'a0.2', 'children': [
{'level': 'a0.2.1', 'children': [
{'level': 'a0.2.1.1'},
{'level': 'a0.2.1.2'},
{'level': 'a0.2.1.3'},
{'level': 'a0.2.1.4', 'children': [{'level': 'a0.2.1.4.1'}, {'level': 'a0.2.1.4.2'}]}
]
}
]
},
{'level': 'a0.3', 'children': []},
{'level': 'a0.4', 'children': [{'level': 'a0.4.1'}, {'level': 'a0.4.2', 'children': []}]}
]
},
{
'level': 'b0',
'children': [{'level': 'b0.1', 'children': [{'level': 'b0.1.1'}]},
{'level': 'b0.2', 'children': [{'level': 'b0.2.1', 'children': [
{'level': 'b0.2.1.1'},
{'level': 'b0.2.1.2'},
{'level': 'b0.2.1.3', 'children': [{'level': 'b0.2.1.3.1'}, {'level': 'b0.2.1.3.2'}]},
{'level': 'b0.2.1.4'}
]
}]},
{'level': 'b0.3'}
]
}
]
代码必须返回叶子和完整的分支路径，直到每次离开：
def flatten_json_tree(nodes, lower_nodes_key='children', path=[]):
if not nodes: # void node
yield path # so it is a leaf
for node in nodes: # go on into each branch of the tree
level = node['level'] # get node datas
try:
lower_nodes = node[lower_nodes_key] # search for lower nodes
except KeyError:
lower_nodes = [] # no lower nodes
yield from flatten_json_tree(lower_nodes, lower_nodes_key, path + [level]) # continue to explore the branch until leaf
if __name__ == "__main__":
for path in list(flatten_json_tree(json_struct)):
leaf = path[-1:][0]
complete_path = ''
for node in path:
complete_path += node + (' -> ' if node is not leaf else '')
print("LEAF: {:20s} PATH: {}".format(leaf, complete_path))
它显示：