python - 将项目添加到字典列表中具有相同值的字典

Question

我正在尝试在“节点”值相等时将具有相同 uuid.uuid4() 的键 id 添加到内部字典中，并且添加一个新的 uuid.uuid4() 当找到不同的 uuid 时。

假设 2 个键(在本例中为“节点”)具有相同的值，如 -> 节点:“类加罗尔”，因此我想为它生成相同的 ID，并为每个其他不同的节点生成一个新的 ID。

这是我现在正在处理的代码:

import uuid
import json

node_list = [
  {
    "nodes": [
      {
         "node": "Kunal",
         "label": "PERSON"
      },
      {
         "node": "Bangalore",
         "label": "LOC"
      }
    ]
  },
  {
    "nodes": [
      {
         "node": "John",
         "label": "PERSON"
      },
      {
         "node": "Bangalore",
         "label": "LOC"
      }
    ]
  }
]

for outer_node_dict in node_list:
      for inner_dict in outer_node_dict["nodes"]:
            inner_dict['id'] = str(uuid.uuid4()) # Remember the key's value here and apply this statement somehow?

print(json.dumps(node_list, indent = True))

---

这是我想要的响应:

"[
  {
    "nodes": [
      {
         "node": "Kunal",
         "label": "PERSON",
         "id": "fbf094eb-8670-4c31-a641-4cf16c3596d1"
      },
      {
         "node": "Bangalore",
         "label": "LOC",
         "id": "24867c2a-f66a-4370-8c5d-8af5b9a25675"
      }
    ]
  },
  {
    "nodes": [
      {
         "node": "John",
         "label": "PERSON",
         "id": "5eddc375-ed3e-4f6a-81dc-3966590e8f35"
      },
      {
         "node": "Bangalore",
         "label": "LOC",
         "id": "24867c2a-f66a-4370-8c5d-8af5b9a25675"
      }
    ]
  }
]"

---

但目前它的生成是这样的:

"[
 {
  "nodes": [
   {
    "node": "Kunal",
    "label": "PERSON",
    "id": "3cce6e36-9d1c-4058-a11b-2bcd0da96c83"
   },
   {
    "node": "Bangalore",
    "label": "LOC",
    "id": "4d860d3b-1835-4816-a372-050c1cc88fbb"
   }
  ]
 },
 {
  "nodes": [
   {
    "node": "John",
    "label": "PERSON",
    "id": "67fc9ba9-b591-44d4-a0ae-70503cda9dfe"
   },
   {
    "node": "Bangalore",
    "label": "LOC",
    "id": "f83025a0-7d8e-4ec8-b4a0-0bced982825f"
   }
  ]
 }
]"

如何记住key的值并在字典中应用相同的ID？

Answer 1

对于相同的“节点”值，您似乎希望 uuid 相同。所以，不是生成它，而是将它存储到字典中

node_uuids = defaultdict(lambda: uuid.uuid4())

然后，在你的内部循环中，而不是

inner_dict['id'] = str(uuid.uuid4())

你写

inner_dict['id'] = node_uuids[inner_dict['node']]

一个完整的工作示例如下:

from collections import defaultdict

import uuid
import json

node_list = [
  {
    "nodes": [
      {
         "node": "Kunal",
         "label": "PERSON"
      },
      {
         "node": "Bangalore",
         "label": "LOC"
      }
    ]
  },
  {
    "nodes": [
      {
         "node": "John",
         "label": "PERSON"
      },
      {
         "node": "Bangalore",
         "label": "LOC"
      }
    ]
  }
]

node_uuids = defaultdict(lambda: uuid.uuid4())

for outer_node_dict in node_list:
      for inner_dict in outer_node_dict["nodes"]:
            inner_dict['id'] = str(node_uuids[inner_dict['node']])

print(json.dumps(node_list, indent = True))