如何基於另一棵樹構建一棵 Python anytree 樹？

Question

如果我有這棵樹：

# mytree.py
import anytree as at
import anytree.importer


data = {
    "a": "root",
    "children": [
        {
            "a": "sub0",
            "b": 3,
            "children": [{"a": "sub0A", "b": 9}, {"a": "sub0B", "b": 1}],
        },
        {"a": "sub1", "b": 5},
    ],
}
root = at.importer.DictImporter().import_(data)

python3 -i mytree.py

print(at.RenderTree(root, style=at.render.ContStyle()))
AnyNode(a='root')
├── AnyNode(a='sub0', b=3)
│   ├── AnyNode(a='sub0A', b=9)
│   └── AnyNode(a='sub0B', b=1)
└── AnyNode(a='sub1', b=5)

我怎樣才能建立這棵樹（不改變原來的）？

AnyNode(a='root')
├── AnyNode(a='sub0', c="small")
│   ├── AnyNode(a='sub0A', c="large")
│   └── AnyNode(a='sub0B', c="small")
└── AnyNode(a='sub1', c="small")

它具有相同的結構（“形狀”），但每個節點沒有b屬性，並且具有c屬性，如果原始節點的b小於 6，則為“小”，如果為 >=6，則為“大” .

我試圖用類似的東西遍歷原始樹

for on_this_level in at.LevelOrderGroupIter(root)

但無法讓它發揮作用。

Answer 1

創建另一棵樹作為副本，然后 - 直接更改其根descendants ，如下所示：

import copy

new_root = copy.deepcopy(root)
for dsc in new_root.descendants:
    dsc.c = 'large' if dsc.b >=6 else 'small'
    del dsc.b
    
print(at.RenderTree(new_root, style=at.render.ContStyle()))

---

AnyNode(a='root')
├── AnyNode(a='sub0', c='small')
│   ├── AnyNode(a='sub0A', c='large')
│   └── AnyNode(a='sub0B', c='small')
└── AnyNode(a='sub1', c='small')