當(dāng)“節(jié)點”值相等時，我試圖將id具有相同的鍵添加uuid.uuid4()到內(nèi)部字典中，并uuid.uuid4()在找到不同的 uuid 時添加一個新鍵。假設(shè) 2 個鍵（在本例中為“節(jié)點”）具有相同的值，例如 -> 節(jié)點：“班加羅爾”，因此我想為其生成相同的 ID，并為每個其他不同的節(jié)點生成一個新 ID。這是我現(xiàn)在正在處理的代碼：import uuidimport jsonnode_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))這是我想要的回應(yīng)："[ { "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" } ] }]"但目前它的生成是這樣的："[ { "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" } ] }]"如何記住鍵的值并在字典中為其應(yīng)用相同的 ID？
1 回答
一只斗牛犬
TA貢獻(xiàn)1784條經(jīng)驗 獲得超2個贊
看起來您希望uuid對于相同的“節(jié)點”值是相同的。因此，與其生成它，不如將其存儲到 dict
node_uuids = defaultdict(lambda: uuid.uuid4())
然后，在你的內(nèi)循環(huán)中，而不是
inner_dict['id'] = str(uuid.uuid4())
你寫
inner_dict['id'] = node_uuids[inner_dict['node']]
一個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))
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