可以在一行中完成三个迭代吗？ Python 3 for循环

Question

以下代码可以正常工作。但是，它可以在单个语句中完成三遍而不是调用player_apps吗？

for club in all_clubs:
    total_app = sum(player['season_apps'] for player in players_apps if player['player_club'] == club)
    total_sub_app = sum(player['season_sub_apps'] for player in players_apps if player['player_club'] == club)
    total_goals = sum(player['season_goals'] for player in players_apps if player['player_club'] == club)
    club_app_goal.append({'club' : club, 'total_app' : total_app, 'total_sub_app' : total_sub_app, 'total_goals' : total_goals})

Answer 1

使用.：

numpy

Answer 2

嗯，从技术上讲，使用functools.reduce是，但是它不会使代码更漂亮，而且我不确定它是否真的可以改善性能：

total_app, total_sub_app, total_goals = reduce(
    lambda a, b: tuple(x + y for x, y in zip(a, b)), 
    ((p['season_apps'], p['season_sub_apps'], p['season_goals']) for p in players_apps if p["club"] == club), 
    (0, 0, 0)
    )

更有趣的“优化”恕我直言，每个俱乐部仅对球员进行一次过滤：

for club in all_clubs:
    players = [player for player in players_apps if player['player_club'] == club]
    total_app = sum(player['season_apps'] for player in players)
    total_sub_app = sum(player['season_sub_apps'] for player in players)
    total_goals = sum(player['season_goals'] for player in players)

更具可读性，甚至可能更快（取决于您的数据）。

话虽如此，使用适当的关系数据库（关系数据库是专门为这种计算而设计的）或至少使用Numpy或Panda，您可能会受益。