“method”和“method”的实例之间不支持“<”，Python，Djang

Question

我想做Winerama推荐人。我遇到了一个我解决不了的错误。当我试图转到“推荐列表”选项卡时，浏览器返回以下错误。在

错误

Environment:


Request Method: GET
Request URL: http://127.0.0.1:8000/recommendation/

Django Version: 2.0.7
Python Version: 3.7.0
Installed Applications:
['django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'bootstrap3',
 'reviews',
 'registration']
Installed Middleware:
['django.middleware.security.SecurityMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware']



Traceback:

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\core\handlers\exception.py" in inner
  35.             response = get_response(request)

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\core\handlers\base.py" in _get_response
  128.                 response = self.process_exception_by_middleware(e, request)

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\core\handlers\base.py" in _get_response
  126.                 response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\contrib\auth\decorators.py" in _wrapped_view
  21.                 return view_func(request, *args, **kwargs)

File "C:\Users\tymot\Desktop\Cd-12.50-20.08\env\my_app\winerama\reviews\views.py" in user_recommendation_list
  89.         reverse=True

Exception Type: TypeError at /recommendation/
Exception Value: '<' not supported between instances of 'method' and 'method'

我根据教程中的步骤创建了所有内容。在

文件模型.py

^{pr2}$

接下来我添加了管理员py并使用'/admin'中的用户创建3个klaster。在

from django.contrib import admin

from .models import Wine, Review, Cluster

class ReviewAdmin(admin.ModelAdmin):
    model = Review
    list_display = ('wine', 'rating', 'user_name', 'comment', 'pub_date')
    list_filter = ['pub_date', 'user_name']
    search_fields = ['comment']


class ClusterAdmin(admin.ModelAdmin):
    model = Cluster
    list_display = ['name', 'get_members']


admin.site.register(Wine)
admin.site.register(Review, ReviewAdmin)
admin.site.register(Cluster, ClusterAdmin)

我的文件视图.py

@login_required
def user_recommendation_list(request):
    # get request user reviewed wines
    user_reviews = Review.objects.filter(user_name=request.user.username).prefetch_related('wine')
    user_reviews_wine_ids = set(map(lambda x: x.wine.id, user_reviews))

    # get request user cluster name (just the first one righ now)
    user_cluster_name = \
        User.objects.get(username=request.user.username).cluster_set.first().name

    # get usernames for other memebers of the cluster
    user_cluster_other_members = \
        Cluster.objects.get(name=user_cluster_name).users \
            .exclude(username=request.user.username).all()
    other_members_usernames = set(map(lambda x: x.username, user_cluster_other_members))

    # get reviews by those users, excluding wines reviewed by the request user
    other_users_reviews = \
        Review.objects.filter(user_name__in=other_members_usernames) \
            .exclude(wine__id__in=user_reviews_wine_ids)
    other_users_reviews_wine_ids = set(map(lambda x: x.wine.id, other_users_reviews))

    # then get a wine list including the previous IDs, order by rating
    wine_list = sorted(
        list(Wine.objects.filter(id__in=other_users_reviews_wine_ids)),
        key=lambda x: x.average_rating,
        reverse=True
    )

    return render(
        request,
        'reviews/user_recommendation_list.html',
        {'username': request.user.username, 'wine_list': wine_list}
    )

我会指出，当我尝试使用简单版本时，一切正常。在

@login_required
def user_recommendation_list(request):
    # get this user reviews
    user_reviews = Review.objects.filter(user_name=request.user.username).prefetch_related('wine')
    # from the reviews, get a set of wine IDs
    user_reviews_wine_ids = set(map(lambda x: x.wine.id, user_reviews))
    # then get a wine list excluding the previous IDs
    wine_list = Wine.objects.exclude(id__in=user_reviews_wine_ids)

    return render(
        request,
        'reviews/user_recommendation_list.html',
        {'username': request.user.username,'wine_list': wine_list}
    )

我的错误在教程的这个stage (2.5)。第2.4阶段效果良好。 一切都表明视图.py. 在

任何帮助都将不胜感激。在

Answer 1

^{}函数接受返回值的函数key。似乎x.average_评级是方法，而不是值。所以你有两个选择

• 在x.average_评级后添加（）
• 将x.average评等转换为property

Answer 2

方法与方法返回的值之间存在差异。在您的Wine模型中，我们看到：

from django.db.models import Avg

class Wine(models.Model):
    name = models.CharField(max_length=200)

    def average_rating(self):
        return self.review_set.aggregate(
            mean=Avg('rating')
        )['mean']

    def __unicode__(self):
        return self.name

（我重写了它以进行效率查询，而不是Python/Django/Numpy-let来做这项工作）。在

如果现在有一个名为some_wine的Wine对象，那么这里的some_wine.average_rating并没有得到平均评级，因为这将返回一个方法，而是通过调用函数，所以some_wine.average_rating()。在

这里有一些选项：

1. 调用lambda表达式中的函数：

   wine_list = sorted(
       list(Wine.objects.filter(id__in=other_users_reviews_wine_ids)),
       key=lambda x: x.average_rating(),
       reverse=True
   )

2. 将方法定义为一个属性，在这种情况下，您不再调用函数，它是幕后调用的，因此：

   from django.db.models import Avg
   
   class Wine(models.Model):
     name = models.CharField(max_length=200)
   
     @property
     def average_rating(self):
         return self.review_set.aggregate(
             mean=Avg('rating')
         )['mean']
   
     def __unicode__(self):
         return self.name

3. 使用Wine.average_rating作为键，因为它是一个名为并带有实例的函数：

   wine_list = sorted(
       list(Wine.objects.filter(id__in=other_users_reviews_wine_ids)),
       key=Wine.average_rating,
       reverse=True
   )

4. 使用.annotate(..)对数据库中已有的Wine对象排序：

   wine_list = Wine.objects.filter(
       id__in=other_users_reviews_wine_ids
   ).annotate(
       mean=Avg('rating')
   ).order_by('-rating')

最新的方法可能是最有效的，因为数据库通常针对此类查询进行优化，而且这将通过一个单个查询来实现。在