我創建了一個命名元組的字典串列，以事件型別為鍵。
[{'EVENT_DELETE': DeleteRequestDetails(rid=53421, user='user1', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)},{'EVENT_DELETE': DeleteRequestDetails(rid=13423, user='user2', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)},{'EVENT_DELETE': DeleteRequestDetails(rid=98343, user='user2', type='EVENT_DELETE', reviewed=1, approved=0, completed=0)}]
回傳/列印僅包含“已批準= 1”或“已審核= 1”和“已批準= 0”的結果的最pythonic方法是什么？
uj5u.com熱心網友回復：
不確定您到底想要什么輸出格式，但這里有一個approved == 1.
>>> from collections import namedtuple
>>> DeleteRequestDetails = namedtuple('DeleteRequestDetails', 'rid user type reviewed approved completed')
>>> lst = [{'EVENT_DELETE': DeleteRequestDetails(rid=53421, user='user1', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)},{'EVENT_DELETE': DeleteRequestDetails(rid=13423, user='user2', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)},{'EVENT_DELETE': DeleteRequestDetails(rid=98343, user='user2', type='EVENT_DELETE', reviewed=1, approved=0, completed=0)}]
>>> [v for d in lst for v in d.values() if v.approved == 1]
[DeleteRequestDetails(rid=53421, user='user1', type='EVENT_DELETE', reviewed=1, approved=1, completed=0),
DeleteRequestDetails(rid=13423, user='user2', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)]
uj5u.com熱心網友回復：
這是一個可能的解決方案：
# data = [{'EVENT_DELETE': DeleteRequestDetails(rid=53421, user='user1', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)},{'EVENT_DELETE': DeleteRequestDetails(rid=13423, user='user2', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)},{'EVENT_DELETE': DeleteRequestDetails(rid=98343, user='user2', type='EVENT_DELETE', reviewed=1, approved=0, completed=0)}]
res = filter(
lambda _: _.approved == 1 or (_.reviewed == 1 and _.approved == 0),
(b for a in data for b in a.values())
)
print(next(res))
# DeleteRequestDetails(rid=53421, user='user1', type='EVENT_DELETE', reviewed=1, approved=1, completed=0)
基本上，我們只是內置filter函式并使用自定義 (lambda-) 函式。
請注意，過濾器回傳一個類似生成器的物件。list如果您需要多次訪問資料，您可以將其包裝起來。
聚苯乙烯
正如@PatrickArtner 指出的那樣，擁有一個包含 1 個值的字典串列是沒有意義的。這樣的事情更有意義：
data = {
"EVENT_DELETE": [
DeleteRequestDetails(rid=53421, user='user1', type='EVENT_DELETE', reviewed=1, approved=1, completed=0),
DeleteRequestDetails(rid=13423, user='user2', type='EVENT_DELETE', reviewed=1, approved=1, completed=0),
DeleteRequestDetails(rid=98343, user='user2', type='EVENT_DELETE', reviewed=1, approved=0, completed=0)
]
}
這使得解決方案也更加清晰和簡單，因為不需要遍歷值，例如
res = filter(
lambda _: _.approved == 1 or (_.reviewed == 1 and _.approved == 0),
data["EVENT_DELETE"]
)
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