将项目添加到字典列表中具有相同值的字典
当“节点”值相等时，我试图将id具有相同的键添加uuid.uuid4()到内部字典中，并uuid.uuid4()在找到不同的 uuid 时添加一个新键。
假设 2 个键（在本例中为“节点”）具有相同的值，例如 -> 节点：“班加罗尔”，因此我想为其生成相同的 ID，并为每个其他不同的节点生成一个新 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))
这是我想要的回应：
"[
{
"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"
}
]
}
]"
如何记住键的值并在字典中为其应用相同的 ID？
SMILET1回答
-
一只斗牛犬
看起来您希望uuid对于相同的“节点”值是相同的。因此，与其生成它，不如将其存储到 dictnode_uuids = defaultdict(lambda: uuid.uuid4())然后，在你的内循环中，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 defaultdictimport uuidimport jsonnode_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))
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Python