我正在嘗試對我在我正在處理的 django 應用程序中找到的這些數據進行排序和聚合。問題是我迷失了遍歷列表和存儲數據的最佳方式。這是我所擁有的一個例子：from score.models import LocDataq = [ {'ref': '002', 'loc': 'seattle', 'total': '200'}, {'ref': '002', 'loc': 'seattle', 'total': '100'}, {'ref': '123', 'loc': 'dallas', 'total': '100'}, {'ref': '452', 'loc': 'cleveland', 'total': '600'}, {'ref': '123', 'loc': 'dallas', 'total': '200'}, {'ref': '002', 'loc': 'seattle', 'total': '300'} ]我想最終得到的是下面的列表，它由 ref 字段聚合并使用 loc 維護此字段，但添加了 total 字段。這是所需的輸出。q = [ {'ref': '002', 'loc': 'seattle', 'total': '600'}, {'ref': '123', 'loc': 'dallas', 'total': '300'}, {'ref': '452', 'loc': 'cleveland', 'total': '600'}, ]有人能告訴我我有哪些工具可以做到這一點嗎？提前致謝！
2 回答
喵喔喔
TA貢獻1735條經驗 獲得超5個贊
可以先構造一個中間字典，然后根據所需的輸出對其進行轉換：
from collections import defaultdict
q = [
{'ref': '002', 'loc': 'seattle', 'total': '200'},
{'ref': '002', 'loc': 'seattle', 'total': '100'},
{'ref': '123', 'loc': 'dallas', 'total': '100'},
{'ref': '452', 'loc': 'cleveland', 'total': '600'},
{'ref': '123', 'loc': 'dallas', 'total': '200'},
{'ref': '002', 'loc': 'seattle', 'total': '300'}
]
temp_dict = defaultdict(int)
for entry in q:
temp_dict[(entry['ref'], entry['loc'])] += int(entry['total'])
result = [{'ref': k[0], 'loc': k[1], 'total': str(v)} for k, v in temp_dict.items()]
print(result)
# [{'ref': '002', 'loc': 'seattle', 'total': '600'},
# {'ref': '123', 'loc': 'dallas', 'total': '300'},
# {'ref': '452', 'loc': 'cleveland', 'total': '600'}]
慕萊塢森
TA貢獻1810條經驗 獲得超4個贊
您可以使用一個聚合collections.Counter：
from collections import Counter
from pprint import pprint
q = [
{"ref": "002", "loc": "seattle", "total": "200"},
{"ref": "002", "loc": "seattle", "total": "100"},
{"ref": "123", "loc": "dallas", "total": "100"},
{"ref": "452", "loc": "cleveland", "total": "600"},
{"ref": "123", "loc": "dallas", "total": "200"},
{"ref": "002", "loc": "seattle", "total": "300"},
]
counts = Counter()
for x in q:
ref, loc, total = x["ref"], x["loc"], x["total"]
counts[ref, loc] += int(total)
pprint(
[
{"ref": ref, "loc": loc, "total": str(total)}
for (ref, loc), total in counts.items()
]
)
#[{'loc': 'seattle', 'ref': '002', 'total': '600'},
# {'loc': 'dallas', 'ref': '123', 'total': '300'},
# {'loc': 'cleveland', 'ref': '452', 'total': '600'}]
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