熊猫：树状结构的扁平化

Question

我有一个类别树，由以下内容表示。

import pandas as pd

asset_tree = [
    {'id': 1, 'name': 'Linear Asset', 'parent_id': -1},
    {'id': 2, 'name': 'Lateral', 'parent_id': 1},
    {'id': 3, 'name': 'Main', 'parent_id': 1},
    {'id': 4, 'name': 'Point Asset', 'parent_id': -1},
    {'id': 5, 'name': 'Fountain', 'parent_id': 4},
    {'id': 6, 'name': 'Hydrant', 'parent_id': 4}
]
tree = pd.DataFrame(asset_tree)
print(tree)

这给出了一个dataframe,如下所示：

   id          name  parent_id
0   1  Linear Asset         -1
1   2       Lateral          1
2   3          Main          1
3   4   Point Asset         -1
4   5      Fountain          4
5   6       Hydrant          4

树中最高的节点的parent_id等于-1，因此树可以用图形表示如下：

Linear Asset
   | - Lateral
   | - Main
Point Asset
   | - Fountain
   | - Hydrant

我需要生成以下DataFrame。

   id          name  parent_id  flat_name
0   1  Linear Asset         -1  Linear Asset
1   2       Lateral          1  Linear Asset : Lateral
2   3          Main          1  Linear Asset : Main
3   4   Point Asset         -1  Point Asset
4   5      Fountain          4  Point Asset : Fountain
5   6       Hydrant          4  Point Asset : Hydrant

该树是动态生成的，可以具有任意数量的级别，因此下面的树

asset_tree = [
    {'id': 1, 'name': 'Linear Asset', 'parent_id': -1},
    {'id': 2, 'name': 'Lateral', 'parent_id': 1},
    {'id': 3, 'name': 'Main', 'parent_id': 1},
    {'id': 4, 'name': 'Point Asset', 'parent_id': -1},
    {'id': 5, 'name': 'Fountain', 'parent_id': 4},
    {'id': 6, 'name': 'Hydrant', 'parent_id': 4},
    {'id': 7, 'name': 'Steel', 'parent_id': 2},
    {'id': 8, 'name': 'Plastic', 'parent_id': 2},
    {'id': 9, 'name': 'Steel', 'parent_id': 3},
    {'id': 10, 'name': 'Plastic', 'parent_id': 3}
]

应产生以下结果：

   id          name  parent_id  flat_name
0   1  Linear Asset         -1  Linear Asset
1   2       Lateral          1  Linear Asset : Lateral
2   3          Main          1  Linear Asset : Main
3   4   Point Asset         -1  Point Asset
4   5      Fountain          4  Point Asset : Fountain
5   6       Hydrant          4  Point Asset : Hydrant
6   7         Steel          2  Linear Asset : Lateral : Steel
7   8       Plastic          2  Linear Asset : Lateral : Plastic
8   9         Steel          3  Linear Asset : Main : Steel
9  10       Plastic          3  Linear Asset : Main : Plastic

Answer 1

这是网络问题，请尝试networkx:

将networkx导入为nx

# build the 
G = nx.from_pandas_edgelist(tree, source='parent_id', target='id',
                            create_using=nx.DiGraph)

node_names = tree.set_index('id')['name'].to_dict()

def get_path(node):
    for path in nx.simple_paths.all_simple_paths(G, -1, node):
        return ' : '.join(node_names.get(i) for i in path[1:])

tree['flat_name'] = tree['id'].apply(get_path)

Answer 2

可以使用递归查找父ID的路径：

asset_tree = [{'id': 1, 'name': 'Linear Asset', 'parent_id': -1}, {'id': 2, 'name': 'Lateral', 'parent_id': 1}, {'id': 3, 'name': 'Main', 'parent_id': 1}, {'id': 4, 'name': 'Point Asset', 'parent_id': -1}, {'id': 5, 'name': 'Fountain', 'parent_id': 4}, {'id': 6, 'name': 'Hydrant', 'parent_id': 4}]
n_tree = {i['id']:i for i in asset_tree}
def get_parent(d, c = []):
   if (k:=n_tree.get(d['parent_id'])) is None:
      return c + [d['name']]
   return get_parent(k, c+[d['name']])

r = [{**i, 'flat_name':' : '.join(get_parent(i)[::-1])} for i in asset_tree]
df = pd.DataFrame(r)