我将以我绝对不是开发人员的身份开头，但是我被丢了这份工作，我只是迷路了。这是我第一次使用python，也是7年来的第一次编码，而且进展不太顺利。
我拥有的JSON是一个组织树，其中每个级别都可能在其下面有子级。
我需要在Jupyter Notebook中用Python编写脚本以将其展平为这种格式，或者类似的方式，其中每个新子代都是新行。
change.after这是JSON：
public override static var activityCategory: UIActivityCategory {
return .share;
}
我已经尝试了各种for循环，并且已经搜寻了几天互联网，但是我认为我缺少一些非常基础的知识来完成这项工作。我会非常感激某人可以提供的任何帮助。
这是我的首发：
level1 | level2 | level3
org1
org1 org2
org1 org2 org3
或
[{
"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": []
}]
}]
}]
}]
答案 0 :(得分：0)
您可以使用堆栈处理嵌套结构：
s'last_value循环中，从堆栈中取出顶部元素。在该组织中进行您需要做的事情，例如记录名称。在组织路径中产生一行，并添加当前组织名称。'0' -> '1'
'L' -> '1'
'0' -> 'H'
'L' -> 'H'
键中的所有元素以及上级组织的组织路径添加到堆栈中。需要逆转，因为从堆栈中获取元素会以相反的顺序进行。您仍然想为该作业使用堆栈（而不是队列），因为我们要深度优先地输出信息。
这看起来像这样：
while stack:然后您可以遍历上面的函数以获取所有路径：
Children答案 1 :(得分：0)
您可以递归地迭代数据。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']]
相同，使用yield：
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, [])))
如果需要所有部分列表，则解决方案较短：
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']]
答案 2 :(得分：0)
json递归树可以具有多个根，并且叶子不必强制指定void子代。例如，这是一棵树，具有两个根“ a”和“ b”，并且节点仅具有“层”数据，该数据是节点深度（“子级”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'}
]
}
]
代码必须返回叶子并完成分支路径，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：