Django指望多对多

I have the models:

我的模型:

class Article(models.Model):
    title = models.TextField(blank=True)
    keywords = models.ManyToManyField(Keyword, null=True, blank=True)

class Keyword(models.Model):
    keyword = models.CharField(max_length=355, blank=True)

I want to get a count of how many articles have each keyword. In essence I want to have a list of keywords where I can get each ones count to give it a relative weighting.

我想知道每个关键词有多少篇文章。本质上，我想要有一个关键字的列表，在这里我可以让每个关键字都有一个相对的权重。

I have tried:

我有尝试:

keyword_list=Article.objects.all().annotate(key_count=Count('keywords__keyword'))

but

但

keyword_list[0].key_count

just seems to give me the number of different keywords each article has? Is it somehow a reverse lookup?

似乎每一篇文章都有不同的关键词。是反向查找吗?

Any help would be much appreciated.

如有任何帮助，我们将不胜感激。

UPDATE

更新

So I got it working:

所以我让它起作用:

def keyword_list(request):
    MAX_WEIGHT = 5
    keywords = Keyword.objects.order_by('keyword')
    for keyword in keywords:
        keyword.count =  Article.objects.filter(keywords=keyword).count()
    min_count = max_count = keywords[0].count
    for keyword in keywords:
        if keyword.count < min_count:
            min_count = keyword.count
        if max_count > keyword.count:
            max_count = keyword.count 
    range = float(max_count - min_count)
    if range == 0.0:
        range = 1.0 
    for keyword in keywords:
        keyword.weight = (
            MAX_WEIGHT * (keyword.count - min_count) / range
        )
    return { 'keywords': keywords }

but the view results in a hideous number of queries. I have tried implementing the suggestions given here (thanks) but this is the only methid which seems to work at the moment. However, I must be doing something wrong as I now have 400+ queries!

但是视图会产生大量的查询。我已经尝试执行这里给出的建议(谢谢)，但这是目前唯一可行的方法。但是，我一定做错了什么，因为我现在有400多个查询!

UPDATE

更新

Wooh! Finally got it working:

Wooh !终于它工作:

def keyword_list(request):
    MAX_WEIGHT = 5
    keywords_with_article_counts = Keyword.objects.all().annotate(count=Count('keyword_set'))
    # get keywords and count limit to top 20 by count
    keywords = keywords_with_article_counts.values('keyword', 'count').order_by('-count')[:20]
    min_count = max_count = keywords[0]['count']
    for keyword in keywords:
        if keyword['count'] < min_count:
            min_count = keyword['count']
        if max_count < keyword['count']:
            max_count = keyword['count']             
    range = float(max_count - min_count)
    if range == 0.0:
        range = 1.0
    for keyword in keywords:
        keyword['weight'] = int(
            MAX_WEIGHT * (keyword['count'] - min_count) / range
        )
    return { 'keywords': keywords}

3 个解决方案

#1

Since you want the number of articles that have each keyword, you have to do it the other way:

既然你想要每个关键词都包含的文章的数量，你必须用另一种方式:

>>> Keyword.objects.all().annotate(article_count=models.Count('article'))[0].article_count
2

#2

This is the same as the answer from Vebjorn Ljosa, but with a little context, where article_set is the related_name of the reverse many-to-many relationship object.

这与Vebjorn Ljosa的答案相同，但是有一个小的上下文，其中article_set是反向多对多关系对象的related_name。

keywords_with_article_counts = Keyword.objects.all().annotate(article_count=Count('article_set'))

To illustrate your results, it would be easier to return the .values():

为了说明您的结果，可以更容易地返回.values():

keywords_with_article_counts.values('keyword', 'article_count')

Which would return a list of dictionaries that would look something like this:

它会返回一个字典列表看起来像这样:

[{'article_count': 36, 'keyword': u'bacon'}, 
 {'article_count': 4, 'keyword': u'unicorns'}, 
 {'article_count': 8, 'keyword': u'python'}]

#3

i don't know how you would do it efficiently but if you need to get it done.

我不知道你会怎么做，但如果你需要做的话。

keywords = Keyword.objects.all()
for keyword in keywords:
  print 'Total Articles: %d' % (Article.objects.filter(keywords=keyword).count())

#1