我有一个用户表，有3种类型的用户学生，教师和俱乐部，我有一张大学表。 我想要的是特定大学里有多少用户。 我得到了我想要的输出，但输出非常慢。我有90k用户，它产生的输出需要几分钟才能产生结果。
我的用户模型： -
from __future__ import unicode_literals
from django.db import models
from django.contrib.auth.models import User
from cms.models.masterUserTypes import MasterUserTypes
from cms.models.universities import Universities
from cms.models.departments import MasterDepartments
# WE ARE AT MODELS/APPUSERS
requestChoice = (
('male', 'male'),
('female', 'female'),
)
class Users(models.Model):
id = models.IntegerField(db_column="id", max_length=11, help_text="")
userTypeId = models.ForeignKey(MasterUserTypes, db_column="userTypeId")
universityId = models.ForeignKey(Universities, db_column="universityId")
departmentId = models.ForeignKey(MasterDepartments , db_column="departmentId",help_text="")
name = models.CharField(db_column="name",max_length=255,help_text="")
username = models.CharField(db_column="username",unique=True, max_length=255,help_text="")
email = models.CharField(db_column="email",unique=True, max_length=255,help_text="")
password = models.CharField(db_column="password",max_length=255,help_text="")
bio = models.TextField(db_column="bio",max_length=500,help_text="")
gender = models.CharField(db_column="gender",max_length=6, choices=requestChoice,help_text="")
mobileNo = models.CharField(db_column='mobileNo', max_length=16,help_text="")
dob = models.DateField(db_column="dob",help_text="")
major = models.CharField(db_column="major",max_length=255,help_text="")
graduationYear = models.IntegerField(db_column='graduationYear',max_length=11,help_text="")
canAddNews = models.BooleanField(db_column='canAddNews',default=False,help_text="")
receivePrivateMsgNotification = models.BooleanField(db_column='receivePrivateMsgNotification',default=True ,help_text="")
receivePrivateMsg = models.BooleanField(db_column='receivePrivateMsg',default=True ,help_text="")
receiveCommentNotification = models.BooleanField(db_column='receiveCommentNotification',default=True ,help_text="")
receiveLikeNotification = models.BooleanField(db_column='receiveLikeNotification',default=True ,help_text="")
receiveFavoriteFollowNotification = models.BooleanField(db_column='receiveFavoriteFollowNotification',default=True ,help_text="")
receiveNewPostNotification = models.BooleanField(db_column='receiveNewPostNotification',default=True ,help_text="")
allowInPopularList = models.BooleanField(db_column='allowInPopularList',default=True ,help_text="")
xmppResponse = models.TextField(db_column='xmppResponse',help_text="")
xmppDatetime = models.DateTimeField(db_column='xmppDatetime', help_text="")
status = models.BooleanField(db_column="status", default=False, help_text="")
deactivatedByAdmin = models.BooleanField(db_column="deactivatedByAdmin", default=False, help_text="")
createdAt = models.DateTimeField(db_column='createdAt', auto_now=True, help_text="")
modifiedAt = models.DateTimeField(db_column='modifiedAt', auto_now=True, help_text="")
updatedBy = models.ForeignKey(User,db_column="updatedBy",help_text="Logged in user updated by ......")
lastPasswordReset = models.DateTimeField(db_column='lastPasswordReset',help_text="")
authorities = models.CharField(db_column="departmentId",max_length=255,help_text="")
class Meta:
managed = False
db_table = 'users'
我正在使用的查询产生了所需的输出但是sloq是： -
universities = Universities.objects.using('cms').all()
for item in universities:
studentcount = Users.objects.using('cms').filter(universityId=item.id,userTypeId=2).count()
facultyCount = Users.objects.using('cms').filter(universityId=item.id,userTypeId=1).count()
clubCount = Users.objects.using('cms').filter(universityId=item.id,userTypeId=3).count()
totalcount = Users.objects.using('cms').filter(universityId=item.id).count()
print studentcount,facultyCount,clubCount,totalcount
print item.name
答案 0 :(得分：2)
您应该使用注释来获取每所大学的计数和条件表达式，以根据条件获得计数(docs)
Universities.objects.using('cms').annotate(
studentcount=Sum(Case(When(users_set__userTypeId=2, then=1), output_field=IntegerField())),
facultyCount =Sum(Case(When(users_set__userTypeId=1, then=1), output_field=IntegerField())),
clubCount=Sum(Case(When(users_set__userTypeId=3, then=1), output_field=IntegerField())),
totalcount=Count('users_set'),
)
答案 1 :(得分：1)
首先，一个明显的优化。在循环中，您基本上执行了四次相同的查询：对不同的userTypeId进行三次过滤，一次没有一次。您可以在一个COUNT(*) ... GROUP BY userTypeId查询中执行此操作。
...
# Here, we're building a dict {userTypeId: count}
# by counting PKs over each userTypeId
qs = Users.objects.using('cms').filter(universityId=item.id)
counts = {
x["userTypeId"]: x["cnt"]
for x in qs.values('userTypeId').annotate(cnt=Count('pk'))
}
student_count = counts.get(2, 0)
faculty_count = counts.get(1, 0)
club_count = count.get(3, 0)
total_count = sum(count.values()) # Assuming there may be other userTypeIds
...
但是，您仍在进行1 + n次查询，其中n是您在数据库中拥有的大学数量。如果数量很少，这很好，但如果数量很高，则需要进一步聚合，加入Universities和Users。我带来的初稿是这样的：
# Assuming University.name is unique, otherwise you'll need to use IDs
# to distinguish between different projects, instead of names.
qs = Users.objects.using('cms').values('userTypeId', 'university__name')\
.annotate(cnt=Count('pk').order_by('university__name')
for name, group in itertools.groupby(qs, lambda x: x["university__name"]):
print("University: %s" % name)
cnts = {g["userTypeId"]: g["cnt"] for g in group}
faculty, student, club = cnts.get(1, 0), cnts.get(2, 0), cnts.get(3, 0)
# NOTE: I'm assuming there are only few (if any) userTypeId values
# other than {1,2,3}.
total = sum(cnts.values())
print(" Student: %d, faculty: %d, club: %d, total: %d" % (
student, faculty, club, total))
我可能在那里打错了，但希望这是正确的。就SQL而言，它应该发出类似
的查询SELECT uni.name, usr.userTypeId, COUNT(usr.id)
FROM some_app_universities AS uni
LEFT JOUN some_app_users AS usr ON us.universityId = uni.id
GROUP BY uni.name, usr.userTypeId
ORDER BY uni.name
考虑阅读aggregations and annotations上的文档。并且一定要查看Django ORM发出的原始SQL（例如使用Django Debug Toolbar）并分析它在数据库上的工作情况。例如，如果您正在使用PostgreSQL，请使用EXPLAIN SELECT。根据您的数据集，您可能会从那里的某些索引中受益（例如，在userTypeId列上）。
哦，并且在旁注...这是偏离主题的，但在Python中，使用lowercase_with_underscores使变量和属性成为一种习惯。在Django中，模型类名称通常是单数，例如User和University。