'<';在';方法';和';方法';-Python,Django
我在试着做Winerama推荐人。我遇到了一个我无法解决的错误。当我尝试转到“推荐列表”选项卡时,浏览器返回以下错误 错误'<';在';方法';和';方法';-Python,Django,python,django,Python,Django,我在试着做Winerama推荐人。我遇到了一个我无法解决的错误。当我尝试转到“推荐列表”选项卡时,浏览器返回以下错误 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
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'
接下来,我添加了admin.py,并在“/admin”中创建了3个klaster用户
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
from django.db import models
from django.contrib.auth.models import User
import numpy as np
class Wine(models.Model):
name = models.CharField(max_length=200)
def average_rating(self):
all_ratings = [list(map(lambda x: x.rating, self.review_set.all()))]
return np.mean(all_ratings)
def __unicode__(self):
return self.name
class Review(models.Model):
RATING_CHOICES = (
(1, '1'),
(2, '2'),
(3, '3'),
(4, '4'),
(5, '5'),
)
wine = models.ForeignKey(Wine, on_delete=models.CASCADE)
pub_date = models.DateTimeField('date published')
user_name = models.CharField(max_length=100)
comment = models.CharField(max_length=200)
rating = models.IntegerField(choices=RATING_CHOICES)
class Cluster(models.Model):
name = models.CharField(max_length=100)
users = models.ManyToManyField(User)
def get_members(self):
return "\n".join([u.username for u in self.users.all()])
@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,一切都很好
@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一部分。阶段2.4运行良好。
一切都表明views.py中存在错误
非常感谢您的帮助。该函数接受返回值的函数键。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_评级转换为
该函数接受返回值的函数键。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_评级转换为
方法和方法返回的值之间存在差异。在您的葡萄酒
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
将方法定义为属性,在这种情况下,您不再调用函数,而是在幕后调用,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
使用.annotate(..)
,对数据库中已存在的Wine
对象进行排序:
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,而且这将通过单个查询完成。方法和方法返回的值之间存在差异。在您的葡萄酒
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
将方法定义为属性,在这种情况下,您不再调用函数,而是在幕后调用,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
使用.annotate(..)
,对数据库中已存在的Wine
对象进行排序:
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,而且这将通过单个查询来完成。看起来平均评级
是一个函数,而不是一个属性/属性。你能分享Wine
模型吗?当然,我在上面添加了整个models.py文件,这是一个糟糕的教程;作者似乎不知道如何进行跨模型查询。没有理由重复这些set/map/lambda的东西。例如,“简单版本”中的一个可以替换为Wine.objects.exclude(review\uu user\u name=request.user.username)
。但是另外,审查应该有一个用户外键,而不是将用户名存储为CharField。实际上,你应该找到一个更好的教程。还要注意,User.objects.get(username=request.User.username)
只是浪费处理器周期和开发人员的大脑时间-request。User
就是同一个对象。谢谢你的评论,我必须做一个更好的教程。也许你听说过一些类似的好教程(这意味着,展示机器学习是如何工作的)?在搜索互联网后,我还会添加一些有趣的链接。看起来平均评分
是一个函数,而不是一个属性/属性。你能分享Wine
模型吗?当然,我在上面添加了整个models.py文件,这是一个糟糕的教程;作者似乎不知道如何进行跨模型查询。没有理由重复这些set/map/lambda的东西。例如,“简单版本”中的一个可以替换为Wine.objects.exclude(review\uu user\u name=request.user.username)
。但是另外,审查应该有一个用户外键,而不是将用户名存储为CharField。实际上,你应该找到一个更好的教程。还要注意,User.objects.get(username=request.User.username)
只是浪费处理器周期和开发人员的大脑时间-request。User
就是同一个对象。谢谢你的评论,我必须做一个更好的教程。也许你听说过一些类似的好教程(这意味着,展示机器学习是如何工作的)?在搜索互联网后,我还会添加一些有趣的链接。或者,也许更优雅一些,当你已经有了一个函数时,根本不用lambda:key=Wine。average_rating
非常感谢,当然,添加“add()po x.average_rating”解决了整个问题。或者,也许更优雅一些,如果你已经有了一个函数,就不要使用lambda:key=Wine.average\u rating
非常感谢,当然添加“add()po x.average\u rating”解决了整个问题。
wine_list = Wine.objects.filter(
id__in=other_users_reviews_wine_ids
).annotate(
mean=Avg('rating')
).order_by('-rating')