通过Django ORM获取两个领域的所有可用组合？

Question

我有这个Django模型:

class Lecture(models.Model):
    lecture_number = models.CharField(_('lecture number'), max_length=20)
    title = models.CharField(_('title'), max_length=100)
    term = models.IntegerField(_('term'), choices=TERM, db_index=True)
    year = models.IntegerField(_('year'), db_index=True)

term可以是1或2(春天或秋天).现在我想列出有讲座的所有术语,例如((2007, 2), (2008, 1), (2008, 2), (2009, 1), (2009, 2), (2010, 1)).该列表不必排序.是否可以有效地通过ORM生成此列表？我找到的最佳解决方案是:

term_list = set(Lecture.objects.values_list('year', 'term'))

但是,ORM调用仍会返回每个结果Lecture并在Python中减少它,因此它可能会因为大量的Lectures而变慢.

Answer 1

如果您拥有数据库中的所有数据,那么它将起作用.但如果没有数据,你应该自己生成.

In [1]: import itertools
In [2]: list(itertools.product((2006,2007,2008), (1,2)))
Out[2]: [(2006, 1), (2006, 2), (2007, 1), (2007, 2), (2008, 1), (2008, 2)]

直接来自数据库(但如果不在数据库中则不会返回所有组合):

In [1]: from django.db import connection

In [2]: Lecture.objects.values_list('year','term').distinct()
Out[2]: [(2001, 1), (2001, 2), (2002, 1), (2002, 2), (2003, 1), (2003, 2), (2004, 1), (2004, 2)]

In [3]: connection.queries
Out[3]: 
[{'sql': u'SELECT DISTINCT "backend_lecture"."year", "backend_lecture"."term" FROM    "backend_lecture" LIMIT 21', 'time': '0.001'}]