我有機型數:Django的許多一對多
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)
我想的很多文章如何讓每個關鍵字的計數。實質上,我想要一個關鍵字列表,我可以讓每個關鍵字都給予相對權重。
我曾嘗試:
keyword_list=Article.objects.all().annotate(key_count=Count('keywords__keyword'))
但
keyword_list[0].key_count
似乎只是給我的不同關鍵字每篇文章都有多少?它以某種方式反向查找?
任何幫助將不勝感激。
UPDATE
所以我得到它的工作:
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 }
但查看結果在查詢中的一個可怕的數字。我試圖執行這裏給出的建議(謝謝),但這是目前似乎工作的唯一方法。但是,由於我現在有400多個查詢,所以我一定在做錯事!
UPDATE
嗚!最後得到它的工作:
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}
爲什麼這下調了它似乎是正確的答案,或者我錯過了什麼? – solartic
這確實是正確的答案,但可能會使用一點解釋。 – jathanism
您可以將其簡化爲'Keyword.objects.all()。annotate(Count('article'))[0] .article__count' – guival