我有一個字典列表。my_list = [ {"id": "UU7t", "updated_at": "2020-01-06_16-40-00", "summary": "Renewed"}, {"id": "yT8h", "updated_at": "2020-01-07_18-24-22", "summary": "Renewed"}, {"id": "i8Po", "updated_at": "2020-01-08_13-16-36", "summary": "Renewed"}, {"id": "yT8h", "updated_at": "2020-01-13_18-24-05", "summary": "Deleted"}, {"id": "7uYg", "updated_at": "2020-01-18_23-37-19", "summary": "Transferred"},]我想獲取刪除重復字典的列表，其中 id 相同但“updated_at”是最新的。所以，我的最終名單將是：my_list = [ {"id": "UU7t", "updated_at": "2020-01-06_16-40-00", "summary": "Renewed"}, {"id": "i8Po", "updated_at": "2020-01-08_13-16-36", "summary": "Renewed"}, {"id": "yT8h", "updated_at": "2020-01-13_18-24-05", "summary": "Deleted"}, {"id": "7uYg", "updated_at": "2020-01-18_23-37-19", "summary": "Transferred"},]什么是有效的方法？
4 回答
一只名叫tom的貓
TA貢獻1906條經驗 獲得超3個贊
您可以使用 adict來累積項目。
字典可以存儲idas 鍵和列表項作為值。如果不存在具有相同鍵的項，則僅在字典中插入一項；如果它確實比較updated_at值并在需要時更新字典。
def generate_new_list(my_list):
counts = {}
for d in my_list:
item_id = d['id']
if item_id in counts:
if d['updated_at'] > counts[item_id]['updated_at']:
counts[item_id] = d
else:
counts[item_id] = d
return list(counts.values())
還有一些注意事項：
如果您想保留原始順序，請確保您使用的是 Python 3.7（它保證 dicts 按插入順序排序）或使用 OrderedDict。使用標準字典，您必須首先彈出條目，因為替換不會更改字典順序（因此每個項目都將按照其 id 第一次出現的順序輸出），而ordereddict 對該用例有特殊支持（move_to_end） .
您還可以通過使用dict.get和“空對象模式”刪除特殊情況：
MISSING = {'updated_at': '0'} # pseudo-entry smaller than all possible
def generate_new_list(my_list):
counts = {}
for d in my_list:
if d['updated_at'] > counts.get(d['id'], MISSING):
counts[d['id']] = d
return list(counts.values())
一種非字典替代方案（盡管非常不保存順序）是按（id，updated_by）排序，按id分組，然后只保留最后一個條目。我不認為 stdlib 提供了開箱即用的最后一個操作（islice 不接受負索引），因此您要么必須手動執行此操作，要么首先將子條目具體化為列表。
慕哥9229398
TA貢獻1877條經驗 獲得超6個贊
兩種解決方案，一種使用字典，另一種通過排序和分組：
from itertools import groupby
my_list = [
{"id": "UU7t", "updated_at": "2020-01-06_16-40-00", "summary": "Renewed"},
{"id": "yT8h", "updated_at": "2020-01-07_18-24-22", "summary": "Renewed"},
{"id": "i8Po", "updated_at": "2020-01-08_13-16-36", "summary": "Renewed"},
{"id": "yT8h", "updated_at": "2020-01-13_18-24-05", "summary": "Deleted"},
{"id": "7uYg", "updated_at": "2020-01-18_23-37-19", "summary": "Transferred"},
]
def newest_id(seq):
"""Keep id with most recent updated_at
Return a list of kept items.
"""
td = {}
for e in seq:
key = e['id']
if key not in td or td[key]['updated_at'] < e['updated_at']:
td[key] = e
return list(td.values())
def newest_id2(seq):
"""Keep id with most recent updated_at
Return a sorted list of kept items.
"""
tl = sorted(seq, key=lambda e: (e['id'], e['updated_at']), reverse=True)
return [next(g) for _, g in groupby(tl, key=lambda e: e['id'])]
res1 = newest_id(my_list)
res2 = newest_id2(my_list)
# Check result
res1.sort(key=lambda e: e['id'], reverse=True)
print(res1 == res2)
慕田峪7331174
TA貢獻1828條經驗 獲得超13個贊
一種方法是改變 dict 的結構。
my_list = [
{"id": "UU7t", "updated_at": "2020-01-06_16-40-00", "summary": "Renewed"},
{"id": "yT8h", "updated_at": "2020-01-07_18-24-22", "summary": "Renewed"},
{"id": "i8Po", "updated_at": "2020-01-08_13-16-36", "summary": "Renewed"},
{"id": "yT8h", "updated_at": "2020-01-13_18-24-05", "summary": "Deleted"},
{"id": "7uYg", "updated_at": "2020-01-18_23-37-19", "summary": "Transferred"},
]
def getNewUpdated(myList):
newList = {}
for element in myList:
if (element["id"] not in newList):
newList[element["id"]] = element
elif (element["updated_at"] >= newList[element["id"]]["updated_at"]):
newList[element["id"]] = element
return newList
print(getNewUpdated(my_list))
在這里，我們正在重構dict，使“id”是key，所有元素都是“values”，然后迭代您提供的列表以檢查“id”是否已經存在于newList中，如果存在，則只需更新相同的記錄（前提是更新時間是新的），否則添加新記錄。
輸出是這樣的：
{
'i8Po': {'summary': 'Renewed', 'id': 'i8Po', 'updated_at': '2020-01-08_13-16-36'},
'yT8h': {'summary': 'Deleted', 'id': 'yT8h', 'updated_at': '2020-01-13_18-24-05'},
'7uYg': {'summary': 'Transferred', 'id': '7uYg', 'updated_at': '2020-01-18_23-37-19'},
'UU7t': {'summary': 'Renewed', 'id': 'UU7t', 'updated_at': '2020-01-06_16-40-00'}
}
眼眸繁星
TA貢獻1873條經驗 獲得超9個贊
使用pandas
import pandas as pd
df = pd.DataFrame(my_list)
df = df.sort_values(by="updated_at").drop_duplicates(subset=["id"], keep="last")
my_list = df.to_dict(orient="records")
輸出：
[{'id': 'UU7t', 'summary': 'Renewed', 'updated_at': '2020-01-06_16-40-00'},
{'id': 'i8Po', 'summary': 'Renewed', 'updated_at': '2020-01-08_13-16-36'},
{'id': 'yT8h', 'summary': 'Deleted', 'updated_at': '2020-01-13_18-24-05'},
{'id': '7uYg', 'summary': 'Transferred', 'updated_at': '2020-01-18_23-37-19'}]
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