2013-11-25 159 views
1

比方说,我的模型是这样的:增量计数和计算

class Publisher(models.Model):  
    name = models.CharField(max_length=30) 
    code = models.SmallIntegerField(unique=True) 

class Book(models.Model): 
    date = models.DateField(auto_now_add=True) 
    publisher = models.ForeignKey(Publisher) 
    hardback = models.BooleanField() 
    large_print = models.BooleanField() 

对于给定的日期范围,我希望能够输出CSV其具有的每出版商书总数,每个布尔字段的百分比。

如:

Publisher-code Total %hardback %large_print: 

123 120 32 10 

(一)到目前为止,我的工作与每个出版商的书籍总数生成一个查询集

totalset = Publisher.objects.all() 
d1 = dict(totalset.annotate(total_books=Count('publisher')).values_list('code','total_books')) 

然后,视图获取每个布尔字段的字典转换查询集,例如

d2 = dict(totalset.filter(book__hardback=True).annotate(hardc=Count('book__hardback')).values_list('code','hardc')) 

然后得到了基于两套

d3 = {k: round(float(d2[k])*100/d1[k]) for k in d1.viewkeys() & d2.viewkeys()} 

我是新来这一切的交集计算百分比一个新的字典,所以我觉得这是非常令人费解。有没有更直接的方法??! (b)如果可以在数据库中做到这一点(例如使用某种模型属性),这是否比在Python中做数据库变得更加高效?

非常感谢

回答

0

对于纯SQL的解决方案,我可能会做一个查询这样的:

publishers = Publisher.objects.all().extra(
    select = { 
     'book_count': 'SELECT COUNT(*) FROM app_book \ 
         WHERE app_book.publisher_id = app_publisher.id', 
     'hardback_ratio': 'SELECT COUNT(*) * 100.0/SUM(COUNT(*)) OVER() \ 
          FROM app_book WHERE hardback = TRUE \ 
          AND app_book.publisher_id = app_publisher.id', 
     'largeprint_ratio': 'SELECT COUNT(*) * 100.0/SUM(COUNT(*)) OVER() \ 
          FROM app_book WHERE largeprint = TRUE \ 
          AND app_book.publisher_id = app_publisher.id', 
    } 
) 

请Django的查询集的extra()方法和count()阅读起来,over() SQL。这太低效了,因为数据库被扫描了3次,但这是我想的开始。

0

我实际上最终使用了发布模型的模型方法;如果有更好的方法,请告诉我!

def get_percentage(self, d1, d2, choose): 
    kwargs = {'book__date__range':[d1,d2], 'book__publisher':self} 
    kwargs2 = {'book__date__range':[d1,d2], 'book__publisher':self, choose:True} 
    total_count = Publisher.objects.filter(**kwargs).count() 
    if total_count == 0: 
     #otherwise perc returns a ZeroDivisionError 
     return total_count 
    count = Publisher.objects.filter(**kwargs2).count() 
    perc = round(float(count) * 100/float(total_count)) 
    return perc