如何使用 python 将唯一 ID 添加到 JSON children?
How can I use python to add unique IDs to JSON children?
我有一个 json 文件,里面有很多 children,像这样:
{
"tree": {
"name": "Top Level",
"children": [
{
"name": "[('server', 'Cheese')]",
"children": [
{
"name": "[('waiter', 'mcdonalds')]",
"percentage": "100.00%",
"duration": 100,
"children": [
{
"name": "[('server', 'kfc')]",
"percentage": "15.73%",
"duration": 100,
"children": [
{
"name": "[('server', 'wendys')]",
"percentage": "12.64%",
"duration": 100
},
{
"name": "[('boss', 'dennys')]",
"percentage": "10.96%",
"duration": 100
}
]
},
{
"name": "[('cashier', 'chickfila')]",
"percentage": "10.40%",
"duration": 100,
"children": [
{
"name": "[('cashier', 'burger king')]",
"percentage": "11.20%",
"duration": 100
}
]
}
]
}
]
}
]
}
}
我想为每个 child 添加一个唯一的 ID,对应于他们所在的级别,所以它最终看起来像这样,其中每个 ID 可以告诉有多少 parent 数据你有多深入 json(例如,21.2.3.102 将是第 102 个 child 第 3 个 child 第 2 个 child 第 21 个 parent):
{
"tree": {
"name": "Top Level",
"id": 1
"children": [
{
"name": "[('server', 'Cheese')]",
"id": 1.1
"children": [
{
"name": "[('waiter', 'mcdonalds')]",
"percentage": "100.00%",
"duration": 100,
"id": 1.1.1
"children": [
{
"name": "[('server', 'kfc')]",
"percentage": "15.73%",
"duration": 100,
"id": 1.1.1.1
"children": [
{
"name": "[('server', 'wendys')]",
"percentage": "12.64%",
"duration": 100,
"id":1.1.1.1.1
},
{
"name": "[('boss', 'dennys')]",
"percentage": "10.96%",
"duration": 100,
"id":1.1.1.1.2
}
]
},
{
"name": "[('cashier', 'chickfila')]",
"percentage": "10.40%",
"duration": 100,
"id":1.1.1.2
"children": [
{
"name": "[('cashier', 'burger king')]",
"percentage": "11.20%",
"duration": 100,
"id":1.1.1.2.1
}
]
}
]
}
]
}
]
}
}
对于包含很多 children 的非常长的 json 文件,是否有简化的方法?
请
谢谢!
您可以使用递归遍历,其中 d
- 您来自 json 的字典:
def walk(d, level="1"):
d["id"] = level
for i, child in enumerate(d.get("children", []), 1):
walk(child, level + "." + str(i))
walk(d["tree"])
我有一个 json 文件,里面有很多 children,像这样:
{
"tree": {
"name": "Top Level",
"children": [
{
"name": "[('server', 'Cheese')]",
"children": [
{
"name": "[('waiter', 'mcdonalds')]",
"percentage": "100.00%",
"duration": 100,
"children": [
{
"name": "[('server', 'kfc')]",
"percentage": "15.73%",
"duration": 100,
"children": [
{
"name": "[('server', 'wendys')]",
"percentage": "12.64%",
"duration": 100
},
{
"name": "[('boss', 'dennys')]",
"percentage": "10.96%",
"duration": 100
}
]
},
{
"name": "[('cashier', 'chickfila')]",
"percentage": "10.40%",
"duration": 100,
"children": [
{
"name": "[('cashier', 'burger king')]",
"percentage": "11.20%",
"duration": 100
}
]
}
]
}
]
}
]
}
}
我想为每个 child 添加一个唯一的 ID,对应于他们所在的级别,所以它最终看起来像这样,其中每个 ID 可以告诉有多少 parent 数据你有多深入 json(例如,21.2.3.102 将是第 102 个 child 第 3 个 child 第 2 个 child 第 21 个 parent):
{
"tree": {
"name": "Top Level",
"id": 1
"children": [
{
"name": "[('server', 'Cheese')]",
"id": 1.1
"children": [
{
"name": "[('waiter', 'mcdonalds')]",
"percentage": "100.00%",
"duration": 100,
"id": 1.1.1
"children": [
{
"name": "[('server', 'kfc')]",
"percentage": "15.73%",
"duration": 100,
"id": 1.1.1.1
"children": [
{
"name": "[('server', 'wendys')]",
"percentage": "12.64%",
"duration": 100,
"id":1.1.1.1.1
},
{
"name": "[('boss', 'dennys')]",
"percentage": "10.96%",
"duration": 100,
"id":1.1.1.1.2
}
]
},
{
"name": "[('cashier', 'chickfila')]",
"percentage": "10.40%",
"duration": 100,
"id":1.1.1.2
"children": [
{
"name": "[('cashier', 'burger king')]",
"percentage": "11.20%",
"duration": 100,
"id":1.1.1.2.1
}
]
}
]
}
]
}
]
}
}
对于包含很多 children 的非常长的 json 文件,是否有简化的方法?
请
谢谢!
您可以使用递归遍历,其中 d
- 您来自 json 的字典:
def walk(d, level="1"):
d["id"] = level
for i, child in enumerate(d.get("children", []), 1):
walk(child, level + "." + str(i))
walk(d["tree"])