jquery - Django "Similar Objects"基于相似标签集

Question

假设您有 1k 个对象，每个对象都有 10-50 个标签。

对于任何给定的对象，必须有一个对象列表，这些对象具有与其相同的关键字，从多到少。

然而，在 Django 参数中，我不知道如何以这种方式进行查询。

django 中是否存在这样的东西，或者我应该在 django 的约束之外编写算法吗？

---

class Product(models.Model):

    product_id = models.IntegerField(
        unique=True,
    )

    slug = models.SlugField(
        unique=True,
        blank = True,
        null = True,   
    )

    meta_description = models.TextField(
        max_length = 160,
        blank = True,
        null = True,    
    )


    title = models.CharField(
        max_length = 160,
        blank = True,
        null = True,
    )


    description = models.TextField(
        blank = True,
        null = True,    
    )

    first_subject_heading = models.CharField(
        max_length = 160,
        blank = True,
        null = True,
    )

    description_main = models.TextField(
        blank = True,
        null = True,    
    )

    price = models.DecimalField(
        max_digits=6, 
        decimal_places=2,
        blank = True,
        null = True,        
    )

    published = models.DateTimeField(auto_now_add=True, blank=True)

    tags = TaggableManager(
        blank = True,
    )

    category = models.ManyToManyField(
        'ProductCategory',
        blank = True,
    )

    license_selection_model = models.IntegerField(
        default = 1,
        blank = True,
        null = True, 
    )

    minipic = models.ImageField(
        upload_to='minipics/',
        blank = True,
        null = True, 
    )

    def get_absolute_url(self):
        from django.core.urlresolvers import reverse
        return reverse('store.views.product', args=[self.slug])

    def save(self, *args, **kwargs):
        if not self.id and not self.slug:
            #Only set the slug when the object is created.
            self.slug = slugify(self.title) #Or whatever you want the slug to use
        super(Product, self).save(*args, **kwargs)

    def __str__(self):
        return "%i, %s"%(self.product_id, self.title)

这是使用 postgres 和 taggit 扩展(taggit 的“similars”选项有错误)

Answer 1

Django Taggit提供一个 TaggableManager帮助使用他们的 API。 它提供了一个similar_objects()方法，它返回一个对象列表，这些对象被标记为与特定对象相似，并按相似性降序排列。

来自docs:

similar_objects()
Returns a list (not a lazy QuerySet) of other objects tagged similarly to this one, ordered with most similar first. Each object in the list is decorated with a similar_tags attribute, the number of tags it shares with this object.

# returns list of objects tagged similarly
some_object.tags.similar_objects()