Question

我有一个redis排序集，其中包含对象ID及其各自的排序分数。我需要检索这些对象的Django查询集，并根据排序集的分数进行排序。最有效的方法是什么？我的DB是postgresql。

目前，我正在分别检索redis排序集和未排序查询集。接下来，我根据python中排序集的分数对查询集进行排序。可以在这里有意义地切割步骤，例如以某种方式在检索点等排序查询集？

    obj_ids_with_scr = get_ids_w_scr() #retrieves redis sorted set with scores
    obj_ids = map(itemgetter(0),obj_ids_with_scr) # filters sorted set for just the obj ids
    queryset = Widget.objects.filter(id__in=obj_ids) #unsorted queryset
    a = dict(obj_ids_with_scr) #turning the sorted set into a dictionary
    for obj_pk, sort_score in a:
        obj = queryset.get(id=obj_pk) #get object with id equalling obj_pk 
        sort_score = obj  #assign the object to the 'score' value of this key-value 
    result = a.values() #making a list of all values, that are now in sorted order
    return result

Answer 1

如果您使用的是postgres，则可以通过定义自己的here来模拟field function in postgres所讨论的mysql的# config/initializers/delayed_job_config.rb Delayed::Worker.destroy_failed_jobs = false Delayed::Worker.sleep_delay = 60 Delayed::Worker.max_attempts = 3 Delayed::Worker.max_run_time = 5.minutes Delayed::Worker.read_ahead = 10 Delayed::Worker.default_queue_name = 'default' Delayed::Worker.delay_jobs = !Rails.env.test? Delayed::Worker.raise_signal_exceptions = :term Delayed::Worker.logger = Logger.new(File.join(Rails.root, 'log', 'delayed_job.log'))语法：

order by field

然后用类似的东西打电话：

CREATE FUNCTION field(anyelement, VARIADIC anyarray) RETURNS numeric AS $$
  SELECT
    COALESCE(
     ( SELECT i FROM generate_subscripts($2, 1) gs(i)
       WHERE $2[i] = $1 ),
     0);
$$ LANGUAGE SQL STABLE

n.b。如果您这样选择，也可以优化您的python代码：

queryset = queryset.extra(
    select={'manual': "FIELD(id, %s)" % (','.join(map(str, obj_ids)),) },
    order_by=['manual']
)

批量返回widgets = Widgets.objects.in_bulk(obj_ids) sorted_widgets = [ widgets[x] for x in obj_ids if x in widgets ]映射dict到id个实例，这对您正在进行的排序类型很有用。您可以阅读更多相关信息here。

检索根据redis排序集排序的Django查询集的最佳方法

1 个答案: