blocks|key|1908675|text|如果没有大量的工作，在vanilla+Django中很难做到最近(最大，平均，等等都可以通过注释和聚合来完成)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1908676|我用一个定制的管理器来做这件事，比如：|1908677|class+AuthorManager(models.Manager):
++def+with_recent_rating(self):
++++return+super(AuthorManager,+self).get_query_set().extra(
++++++++select={
++++++++++++'recent_rating':+'''
++++++++++++++++SELECT+e.rating
++++++++++++++++FROM+myapp_entry+e
++++++++++++++++WHERE+e.authors_id+=+myapp_author.id
++++++++++++++++ORDER+BY+e.pub_date+DESC
++++++++++++++++LIMIT+1
++++++++++++++++''',
++++++++})|code-block|syntax|javascript|1908678|然后将以下内容添加到Author模型中：|1908679|class+Author():
++++...
++++objects+=+AuthorManager()|1908680|然后，当您想要具有评分的作者时，您只需查询：|1908681|authors+=+Author.objects.with_recent_rating().filter(...)|1908682|除了作者现在有了一个recent_rating字段之外，它实际上和任何其他的fetch一样快。|1908683|for+author+in+authors:
++++print+author.recent_rating|1908684|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Recent is hard to do in vanilla Django without a lot of work (max, average, etc. can all be done with annotations and aggregations).

I do this with a custom manager something like:

<pre><code>class AuthorManager(models.Manager):
 def with_recent_rating(self):
 return super(AuthorManager, self).get_query_set().extra(
 select={
 'recent_rating': '''
 SELECT e.rating
 FROM myapp_entry e
 WHERE e.authors_id = myapp_author.id
 ORDER BY e.pub_date DESC
 LIMIT 1
 ''',
 })
</code></pre>

Then add the following to the Author model:

<pre><code>class Author():
 ...
 objects = AuthorManager()
</code></pre>

Then when you want authors with the ratings you just query:

<pre><code>authors = Author.objects.with_recent_rating().filter(...)
</code></pre>

It's practically the same speed as any other fetch except now the authors have a recent_rating field.:

<pre><code>for author in authors:
 print author.recent_rating
</code></pre>

blocks|key|186479|text|另一种方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|186480|from+collections+import+defaultdict
from+datetime+import+datetime,+timedelta

week_ago+=+datetime.now()+-+timedelta(days=7)

author_recent_ratings+=+dict(Entry.objects.filter(pub_date__gt=week_ago)
++++++++++++++++++++++++++++++++++++++++++.order_by('pub_date')
++++++++++++++++++++++++++++++++++++++++++.select_related()
++++++++++++++++++++++++++++++++++++++++++.values_list('author',+'rating'))

recent_by_rating+=+defaultdict(list)
for+author,+rating+in+author_recent_ratings.iteritems():
++++recent_by_rating[rating].append(author)|code-block|syntax|javascript|186481|这是你可以做到的一种方法。基本上，您先按最近的条目(在本例中是上周的条目)排序，然后按最旧的条目排序，然后将值列表返回的列表转换为字典。发生的事情是，当它被转换成字典时，新的条目会重创旧的条目，所以你最终会得到一个以作者为关键字，以作者的评分为值的字典。|186482|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

<h2>Another approach</h2>

<pre><code>from collections import defaultdict
from datetime import datetime, timedelta

week_ago = datetime.now() - timedelta(days=7)

author_recent_ratings = dict(Entry.objects.filter(pub_date__gt=week_ago)
 .order_by('pub_date')
 .select_related()
 .values_list('author', 'rating'))

recent_by_rating = defaultdict(list)
for author, rating in author_recent_ratings.iteritems():
 recent_by_rating[rating].append(author)
</code></pre>

This is one way you could do it. Basically you order by most recent entries (in this case entries from the last week) then order by oldest first, and then you convert the list returned by values list into a dict. What happens is, as it's converted into a dict and the newer entries clobber the older ones so you end up with a dict that has authors as keys, with their ratings as the values.

blocks|key|1908719|text|实际上，你可以在你的模板中完全做到这一点。像这样的东西应该是有效的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1908720|**Views.py**
authors+=+Author.objects.all()

**Template**
{%25+regroup+authors+by+rating_set.all%7Clast+as+rating_list+%25}

{%25+for+rating+in+rating_list+%25}
++++{{+rating.grouper+}} 
++++{%25+for+author+in+rating.list+%25}
++++++++{{+author.name+}} 
++++{%25+endfor+%25}
{%25+endfor+%25}|code-block|syntax|javascript|1908721|基本上，此方法使用regroup模板标记按评级对所有作者进行分组。last过滤器应该会在每个作者的评分列表中为您提供最新的评分。在那之后，这只是一个基本的重组练习，通过评级将其分解，并显示每个评级的所有作者。|offset|length|style|CODE|1908722|https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#regroup|1908723|https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#last|1908724|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|0|9|7|X|4|0|0|2A|0|0|0|27|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@$I|16|J|17|1|18]]|A|$]]|$1|O|3|P|5|6|7|19|8|@]|9|@$I|1A|J|1B|1|1C]]|A|$]]|$1|Q|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|N]]|X|$5|T|U|V|A|$W|P]]]]

You could actually do this entirely in your template. Something like this should work:

<pre><code>**Views.py**
authors = Author.objects.all()

**Template**
{% regroup authors by rating_set.all|last as rating_list %}

{% for rating in rating_list %}
 &lt;b&gt;{{ rating.grouper }}&lt;/b&gt;&lt;br&gt;
 {% for author in rating.list %}
 {{ author.name }}&lt;br&gt;
 {% endfor %}
{% endfor %}
</code></pre>

Basically this method groups all of your authors by rating using the <code>regroup</code> template tag. The <code>last</code> filter should give you the most recent rating in the list of each author's ratings. After that it's just a basic regroup exercise to break it down by rating and display all the authors for each rating. 

<a href="https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#regroup" rel="nofollow">https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#regroup</a>

<a href="https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#last" rel="nofollow">https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#last</a>

Given the following simplified models from the Django docs, I would like to return a list of all of the authors grouped by the rating on their most recent entry, or the most recent prior to some date in the past.

<pre><code> class Author(models.Model):
 name = models.CharField(max_length=50)
 email = models.EmailField()

 class Entry(models.Model):
 headline = models.CharField(max_length=255)
 pub_date = models.DateTimeField()
 mod_date = models.DateTimeField()
 authors = models.ForeignKey(Author)
 rating = models.IntegerField()
</code></pre>

Eventually I would like to turn this into a python dictionary like: {1star:(author1,author2),2star:(author3,author4,author5)...}.

One thought is to return all of the entries, then use itertools.groupby to manipulate the large dataset. Can anyone suggest a cleaner alternative?

Django list all authors by most recent rating

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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考虑到Django文档中的以下简化模型，我想返回所有作者的列表，这些作者根据他们最近的条目或过去某个日期之前的评级分组。 class Author(models.Model):     name = models.CharField(max_length=50)     email = models.EmailFiel...

问Django按最新评分列出所有作者
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