英文:
Transform data with level column in python
问题
这是您所提供的数据，表示分层树结构：
[{"level": 0,"name": "python"},{"level": 1,"name": "food"},{"level": 2,"name": "banana"},{"level": 3,"name": "protein"},{"level": 2,"name": "apple"},{"level": 1,"name": "fuel"}]
我想要将其转换为以下结构：
[{"level": 0,"name": "python","children": [{"level": 1,"name": "food","children": [{"level": 2,"name": "banana","children": [{"level": 3,"name": "protein","children": []}]},{"level": 2,"name": "apple","children": []}]},{"level": 1,"name": "fuel","children": []}]}]
我使用Python，希望在Python中提供解决方案，无论是否使用外部库（包括pandas）。期待您的解决方案，提前感谢！
英文:
I have data like this which represents hierarchical tree structure:
[{"level":0,"name":"python"},{"level":1,"name":"food"},{"level":2,"name":"banana"},{"level":3,"name":"protein"},{"level":2,"name":"apple"},{"level":1,"name":"fuel"}]
I want to transform it into:
[{"level":0,"name":"python","children":[{"level":1,"name":"food","children":[{"level":2,"name":"banana","children":[{"level":3,"name":"protein","children":[]}]},{"level":2,"name":"apple","children":[]}]},{"level":1,"name":"fuel","children":[]}]}]
I am using python and would prefer the solution in python with or without using external libraries, (even using pandas). I would love to see the solutions, thank you in advance 
答案1
得分: 1
以下是代码部分的中文翻译：
def create_tree(nodes):# 为每个节点添加父节点信息for i, node in enumerate(nodes):if node['level'] == 0:node['parent'] = Noneelse:j = i - 1while j >= 0:if nodes[j]['level'] < node['level']:node['parent'] = nodes[j]['name']breakj -= 1node_dict = {}for node in nodes:node_dict[node['name']] = nodefor node in nodes:parent_name = node.get('parent')if parent_name is not None:parent = node_dict[parent_name]if 'children' not in parent:parent['children'] = []parent['children'].append(node)for node in nodes:if node.get('parent') is None:return nodereturn Nonenodes = [{"level": 0, "name": "python"},{"level": 1, "name": "food"},{"level": 2, "name": "banana"},{"level": 3, "name": "protein"},{"level": 2, "name": "apple"},{"level": 1, "name": "fuel"}]tree = create_tree(nodes)print(tree)
这是你所需的代码。
英文:
Here is one way:
def create_tree(nodes):# add parent information to each nodefor i, node in enumerate(nodes):if node['level'] == 0:node['parent'] = Noneelse:j = i - 1while j >= 0:if nodes[j]['level'] < node['level']:node['parent'] = nodes[j]['name']breakj -= 1node_dict = {}for node in nodes:node_dict[node['name']] = nodefor node in nodes:parent_name = node.get('parent')if parent_name is not None:parent = node_dict[parent_name]if 'children' not in parent:parent['children'] = []parent['children'].append(node)for node in nodes:if node.get('parent') is None:return nodereturn Nonenodes = [{"level":0,"name":"python"},{"level":1,"name":"food"},{"level":2,"name":"banana"},{"level":3,"name":"protein"},{"level":2,"name":"apple"},{"level":1,"name":"fuel"}]tree = create_tree(nodes)print(tree)
which gives what you wanted
{'level': 0, 'name': 'python', 'parent': None, 'children': [{'level': 1, 'name': 'food', 'parent': 'python', 'children': [{'level': 2, 'name': 'banana', 'parent': 'food', 'children': [{'level': 3, 'name': 'protein', 'parent': 'banana'}]}, {'level': 2, 'name': 'apple', 'parent': 'food'}]}, {'level': 1, 'name': 'fuel', 'parent': 'python'}]}
答案2
得分: 1
以下是代码部分的翻译：
我想你的列表中字典的顺序很重要，否则我将不知道子元素应该在哪个父元素下列出。我建议编写一个递归函数，以获取在第 n 级的最后一个元素，并将子元素附加在那里：input_list = [{"level": 0,"name": "python"},{"level": 1,"name": "food"},{"level": 2,"name": "banana"},{"level": 3,"name": "protein"},{"level": 2,"name": "apple"},{"level": 1,"name": "fuel"}]def get_last_elt_at_lvl(rec, lvl):if lvl == 0:return rec[-1]else:for i in range(len(rec)-1,-1,-1):if rec[i]['children']:r = get_last_elt_at_lvl(rec[-1]['children'], lvl-1)if r:return rreturn Noneoutput_list = []for d in input_list:if d["level"] == 0:output_list.append(d)else:last_elt = get_last_elt_at_lvl(output_list, d["level"]-1)children = last_elt.setdefault('children', [])children.append(d)print(json.dumps(output_list, indent=4))
输出结果：
[{"level": 0,"name": "python","children": [{"level": 1,"name": "food","children": [{"level": 2,"name": "banana","children": [{"level": 3,"name": "protein"}]},{"level": 2,"name": "apple"}]},{"level": 1,"name": "fuel"}]}]
希望这对你有所帮助！
英文:
I suppose the order of the dictionaries in your list matters, since otherwise I wouldn't know under which parent the children are supposed to be listed. I would suggest to write a recursive function to get the last element at level n and append the children there:
import jsoninput_list = [{"level":0,"name":"python"},{"level":1,"name":"food"},{"level":2,"name":"banana"},{"level":3,"name":"protein"},{"level":2,"name":"apple"},{"level":1,"name":"fuel"}]def get_last_elt_at_lvl(rec, lvl):if lvl == 0:return rec[-1]else:for i in range(len(rec)-1,-1,-1):if rec[i]['children']:r = get_last_elt_at_lvl(rec[-1]['children'], lvl-1)if r:return rreturn Noneoutput_list = []for d in input_list:if d["level"] == 0:output_list.append(d)else:last_elt = get_last_elt_at_lvl(output_list, d["level"]-1)children = last_elt.setdefault('children', [])children.append(d)print(json.dumps(output_list, indent=4))
Output:
[{"level": 0,"name": "python","children": [{"level": 1,"name": "food","children": [{"level": 2,"name": "banana","children": [{"level": 3,"name": "protein"}]},{"level": 2,"name": "apple"}]},{"level": 1,"name": "fuel"}]}]
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