我的表如下所示：User| id | name | | 1 | user1 || 2 | user2 || 3 | user3 || 4 | user4 || 5 | user5 || 6 | user6 || 7 | user7 || 8 | user8 |Event| id | name | created_by | published_by || 1 | event1 | 1 | 2 || 2 | event2 | 2 | 2 || 3 | event3 | 3 | 1 |Team| id | name | player1 | player2 | player3 | player4 || 1 | teamA | 1 | 2 | | || 2 | teamB | 5 | 6 | | |Schedule| id | event_id | player1 | player2 | team1 | team2 | when_do_they_play | round || 1 | 1 | 1 | 3 | | | 2019-09-20 11:22:33 | 1 || 1 | 1 | 2 | 4 | | | 2019-09-21 12:32:23 | 1 || 1 | 1 | 5 | 6 | | | 2019-09-22 22:42:03 | 4 || 2 | 2 | | | 1 | 2 | 2019-09-25 21:12:43 | 2 |我正在使用以下查詢從中選擇數(shù)據(jù)scheduled_games = ( db.session.query( e.name.label('event_name'), u1.name.label('player1'), c1.code.label('player1_country'), u2.name.label('player2'), c2.code.label('player2_country'), s.when_do_they_play, s.round.label('round'), t1.name.label('team1'), t2.name.label('team2') ) .outerjoin(u1, u1.id == s.player1) .outerjoin(u2, u2.id == s.player2) .outerjoin(e, e.id == s.event_id) .outerjoin(c1, c1.id == u1.country) .outerjoin(c2, c2.id == u2.country) .outerjoin(t1, t1.id == s.team1) .outerjoin(t2, t2.id == s.team2) .filter(s.when_do_they_play>=(datetime.utcnow() - timedelta(days=10)))\ .order_by(s.when_do_they_play) .order_by(s.round.asc()))
1 回答
慕俠2389804
TA貢獻(xiàn)1719條經(jīng)驗(yàn) 獲得超6個(gè)贊
您希望每個(gè)事件的回合由列表表示。在您的嘗試中，這些行：
if ldata[1] or ldata[3]:
scheduled_json[ldata[0]]['round'+str(ldata[6])]['player1']=ldata[1]
scheduled_json[ldata[0]]['round'+str(ldata[6])]['player2']=ldata[3]
else:
scheduled_json[ldata[0]]['round'+str(ldata[6])]['team1']=ldata[7]
scheduled_json[ldata[0]]['round'+str(ldata[6])]['team2']=ldata[8]
... 將覆蓋嵌套在密鑰中的任何現(xiàn)有密鑰'round' + str(ldata[6])。
而不是這個(gè)：
scheduled_json[ldata[0]]['round'+str(ldata[6])]={}
使用list：
scheduled_json[ldata[0]]['round'+str(ldata[6])] = []
然后為每個(gè)事件的每一輪附加一個(gè)新dict的：list
if ldata[1] or ldata[3]:
d = {'player1': ldata[1], 'player2': ldata[2]}
else:
d = {'team1': ldata[7], 'team2': ldata[8]}
d['when_do_they_play'] = ldata[5]
scheduled_json[ldata[0]]['round'+str(ldata[6])].append(d)
從風(fēng)格上講，您也可以對(duì)整體可讀性進(jìn)行一些改進(jìn)。例如，您的解決方案可以這樣重寫：
for ldata in scheduled_games:
event = ldata.event # the row proxy facilitates attribute access syntax
if event not in scheduled_json:
scheduled_json[event] = {}
round_key = f"round{ldata.round}" # have a look at f-string and str.format() for string concat
if round_key not in scheduled_json[event]:
print('init!')
scheduled_json[event][round_key] = []
if ldata.player1 or ldata.player2:
d = {'player1': ldata.player1, 'player2': ldata.player2}
else:
d = {'team1': ldata.team1, 'team2': ldata.team2}
d['when_do_they_play'] = ldata.when_do_they_play
scheduled_json[event][round_key].append(d)
您甚至可以更進(jìn)一步并使用 acollections.defaultdict這樣您就不需要實(shí)例化空集合：
from collections import defaultdict
scheduled_json = defaultdict(lambda: defaultdict(list))
for ldata in scheduled_games:
if ldata.player1 or ldata.player2:
d = {'player1': ldata.player1, 'player2': ldata.player2}
else:
d = {'team1': ldata.team1, 'team2': ldata.team2}
d['when_do_they_play'] = ldata.when_do_they_play
scheduled_json[ldata.event][f"round{ldata.round}"].append(d)
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