根据其他具有ID和分数的列表订购Django查询集 [英] Ordering a Django queryset based on other list with ids and scores

问题描述

我在某些事情上有些固执，乍一看似乎很简单.

I'm a bit mentally stuck at something, that seems really simple at first glance.

我要获取要选择的 ids 列表和 scores 对其进行排序的列表.

I'm grabbing a list of ids to be selected and scores to sort them based on.

我当前的解决方法如下:

My current solution is the following:

ids = [1, 2, 3, 4, 5]

items = Item.objects.filter(pk__in=ids)

现在，我需要以某种方式添加基于分数的排序，因此，我将建立以下列表:

Now I need to add a score based ordering somehow so I'll build the following list:

scores = [
        {'id': 1, 'score': 15},
        {'id': 2, 'score': 7},
        {'id': 3, 'score': 17},
        {'id': 4, 'score': 11},
        {'id': 5, 'score': 9},
    ]

ids = [score['id'] for score in scores]

items = Item.objects.filter(pk__in=ids)

到目前为止还不错-但是我如何实际将分数添加为某种汇总，并根据分数对查询集进行排序?

So far so good - but how do I actually add the scores as some sort of aggregate and sort the queryset based on them?

推荐答案

对得分列表进行排序，并使用 in_bulk()获取查询集.

Sort the scores list, and fetch the queryset using in_bulk().

scores = [
    {'id': 1, 'score': 15},
    {'id': 2, 'score': 7},
    {'id': 3, 'score': 17},
    {'id': 4, 'score': 11},
    {'id': 5, 'score': 9},
]
sorted_scores = sorted(scores)  # use reverse=True for descending order
ids = [score['id'] for score in scores]
items = Item.objects.in_bulk(ids)

然后按照所需顺序生成项目列表:

Then generate a list of the items in the order you want:

items_in_order = [items[x] for x in ids]

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